Abstract Algebra 2

Math 103B

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Review

Monoid

A monoid has two properties,

• $\mathbf{Associative}:x\cdot (y\cdot z)=(x\cdot y)\cdot z$
• $Identity/Neutral\mathbf{Element}:\exists e\in G:x\cdot e=e\cdot x=x$

Adding a third property

• $\mathbf{Inverse}:\forall x\in G\exists y\in G:x\cdot y=y\cdot x=e$
  • ($y=x^{-1}$)

Groups

A Group is a Monoid with Inverses

Abelian Group

A group with Communativity is an Abelian Group

• $\mathbf{Commutativity}:\forall x,y\in G:x\cdot y=y\cdot x$

Examples:

• $(\mathbb{Z},+)$ - Abelian Group
• $(\mathbb{Z},\cdot )$ - Monoid (Abelian Monoid because $n\cdot m=m\cdot n$)
  • 0 is not invertible
• $(\mathbb{Z}\ast n,+\ast mod(n))$ = {0, 1, …, n - 1} - Abelian Group
• $(\mathbb{Z}\ast n,\cdot \ast mod(n))$ - Monoid
• $(\mathbb{R},+)$ - Abelian Group
• $(\mathbb{R},\cdot )$ - Monoid
• $(M_n(\mathbb{R}),+)$ - Abelian Group
  • \begin{pmatrix} 2 & 0\\ 1 & -1 \\end{pmatrix} + \\begin{pmatrix} 3 & 1\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} 5 & 1\\ 1 & 0 \\end{pmatrix}
  • Identity Element is the 0 matrix
  • Inverse - make each element in matrix negative
• $(M_n(\mathbb{R}),\cdot )$ - Monoid
  • \begin{pmatrix} 2 & 0\\ 1 & -1 \\end{pmatrix} \cdot \\begin{pmatrix} 3 & 1\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} 6 & 2\\ 3 & 0 \\end{pmatrix}
  • Identity Element is the identity matrix
• $(f:\mathbb{R}\to \mathbb{R},+)$ - Abelian Group
  • $(f+g)(x)=f(x)+g(x)$
• $(f:\mathbb{R}\to \mathbb{R},\cdot )$ - Monoid
  • $(f\cdot g)(x)=f(x)g(x)$
  • Identity Element: $f(x)=1$
  • $f(x)=0$ is not invertable
  • Invertible elements have no roots (do not intersect x axis)
• $(f:\mathbb{R}\to \mathbb{R},\circ )$ - Monoid
  • Not invertible, $f(x)=x^2$ is not invertible (Must be bijective)

Rings

A ring is a set $(R,+,\cdot )$ with operations +, ⋅ such that

1. $(R,+)$ Abelian Group
2. $(R,\cdot )$ Monoid
3. $$\begin{gathered} \mathbf{Distributivity}:\forall x,y,z\in R:x\cdot (y+z)=x\cdot y+x\cdot z \\ (x+y)\cdot z=x\cdot z+y\cdot z \end{gathered}$$

A ring is called commutative if $\forall x,y\in R:x\cdot y=y\cdot x$, $(R,\cdot )$ is a commutative monoid

• Matrix Multiplication / Addition is a non Communative Ring
• Continuous Functions form a communative ring
• Function Composition is not a ring

Subring Test

A subset $S\subseteq R$ is a subring if it is closed under $+,\cdot$ and

• $(S,+)$ is a subgroup of $(R,+)$
  • S is closed under subtraction $(x,y\in S⟹x-y\in S)$
• $(S,\cdot )$ is a submonoid of $(R,\cdot )$
  • S is closed under multiplication $(x,y\in S⟹x\cdot y\in S)$
• $1\in S$

Properties inherited to subrings

1. Commutativity
2. Domain (no zero-divisors) Being a field / division ring is not inherited

Remark: We also know $0\in S$. Since $1\in S$, so $-1\in S$

Example:

1. $\mathbb{Z}\subset \mathbb{Q}\subset \mathbb{R}\subset \mathbb{C}$ are subrings!

2. $Z_n=0,1,\cdots ,n-1\subset \mathbb{Z}=\cdots ,-1,0,1,\cdots$

   $0-1=-1\notin \mathbb{Z}\_n$ not a subring

3. \mathbb{Z}\[\\sqrt{2}\] = {a+b\sqrt{2}\ |\ a, b \in \mathbb{Z} } \subset \mathbb{R}

   Subring test: 1=1+0\sqrt2 \in \mathbb{Z}\[\\sqrt2\]

   Multiplication: (a+b\sqrt2)(c+d\sqrt2) = ac+ad\sqrt2+b\sqrt2c+ b\sqrt2\cdot d\sqrt2 = (ac+2bd)+(ad+bc)\sqrt2 \in \mathbb{Z}\[\\sqrt2\]

   Subtraction: (a+b\sqrt2)-(c+d\sqrt2)=(a-c)+(b-d)\sqrt2 \in Z\[\\sqrt2\]

   \therefore \mathbb{Z}\[\\sqrt2\] \subset \mathbb{R} it is a subring!

4. \mathbb{Z}\[i\] = {\ a+bi\ |\ a,b\in\mathbb{Z}} \subset \mathbb{C} - Gaussian Integers (subring)

\\begin{matrix} & \Bbb{C} & \\ / & & \backslash \\ \| & & \Bbb{R} \\ \| & & | \\ \\Bbb{Z}\[i\] & &\Bbb{Z}\[\\sqrt2\] \\ \\backslash & & / \\ &\Bbb{Z} & \\end{matrix}

Remark: “lines are transitive” in this diagram, namely, if $S\subseteq R$ is a subring and $T\subseteq S$ is a subring, then $T\subseteq R$ is a subring

$U{T}_n(\mathbb{R})=\left(\begin{array}{cccc}\ast & \cdots & \cdots & \ast \\ 0& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & ⋮\\ 0& \cdots & 0& \ast \end{array}\right)\subset M_n(\mathbb{R})$ is a subring 2. B(\mathbb{R}) {f: \mathbb{R}\rightarrow \mathbb{R} bounded continuous functions}= {}\subset C(\mathbb{R}) = {f: \mathbb{R}\rightarrow \mathbb{R} continuous functions}

Polynomial Rings

“Polynomial Ring over R” \Longleftrightarrow \ R\[x\]

R\[x\] = {r_0+r_1x+r_2x^2+\cdots+r_nx^n\ |\ r_0,\cdots,r_n\in \mathbb{R}}

Polynomial Addition Formula

$$\begin{gathered} r_0+r_{1}x+\cdots +r_n{x}^n \\ {s}_0+s_{1}x+\cdots +s_m{x}^m \\ implies(r_0+r_{1}x+\cdots +r_n{x}^n)+(s_0+s_{1}x+\cdots +s_m{x}^m) \\ =(r_0+s_0)+(r_1+s_1)x+\cdots +(r_n+s_n)x^n+s\_n+1x^{n+1}+\cdots +s_m{x}^m \end{gathered}$$

Polynomial Multiplication Formula

$$\begin{gathered} p(x)=\sum \_{i=0}^n{r}_i{x}^i=r_0+r_{1}x+\cdots +r_n{x}^n \\ q(x)=\sum \_{j=0}^m{s}_j{x}^j=r{s}_0+s_{1}x+\cdots +s_m{x}^m \\ p(x)\cdot q(x)=(r_0+r_{1}x+\cdots +r_n{x}^n)\cdot (s_0+s_{1}x+\cdots +s_m{x}^m) \\ =r_0+s_0+(r_1{s}_0+r_0{s}_1)x+(r_0{s}_2+r_1{s}_1+r_2{s}_0)x^2+\cdots \\ =(\sum \_{i=0}^n{r}_i{x}^i)(\sum \_{j=0}^m{s}_j{x}^j) \\ =\sum \_{d=0}^{n+m}(\sum \_i+j=d{r}_i{s}_j)x^d) \end{gathered}$$

Remark: (R\[x\], +, \cdot) is a ring (assuming R is an arbitary ring)

Remark: A ring is communitive if $x\cdot y=y\cdot x\forall x,y$

• $\mathbb{Z},\mathbb{Q},\mathbb{R},\mathbb{C}$ - Commutative Rings
• $M_2(\mathbb{R})$ - Non Commutative

When is R\[x\] commutative? ⇒ if R\[x\] is commutative then R is, conversely, if R is commutative then R\[x\] is commutative

Using the formula for polynomial multiplication, we find that

$$\begin{gathered} p(x)\cdot q(x)=q(x)\cdot p(x) \\ sum\_{d=0}^{n+m}(\sum \_i+j=d{r}_i{s}_j)x^d)=\sum \_{d=0}^{n+m}(\sum \_j+i=d{s}_j{r}_i)x^d) \end{gathered}$$

Therefore, R\ is \ Commutative\ \Longleftrightarrow\ R\[x\]\\ is\ Commutative

Remark: Is \mathbb{R}\[x\] \subset C(\mathbb{R}) a subring? ⇒ No, $e^x,sin(x),\cdots \in C(\mathbb{R})$ but \notin \mathbb{R}\[x\]

Remark: R\[x\] is defined for an arbitary ring R. We can have a ring, \mathbb{Z} \rightarrow \mathbb{Z}\[x\] \rightarrow (\mathbb{Z}\[x\])\[y\] = \mathbb{Z}\[x,y\] . Which gives us a polynomial of y with coefficents from x or vice versa.

Direct Sums

Suppose that $R,S$ are rings. Define the direct sum $R\oplus S:$

$$R\oplus S=(r,s)|r\in R,s\in S$$

Addition:

$$(r_1,s_1)+(r_2,s_2)=(r_1+\_R{r}_2,s_1+\_S{s}_2)$$

Multiplication:

$$(r_1,s_1)\cdot (r_2,s_2)=(r_1{\cdot }_R{r}_2,s_1{\cdot }_S{s}_2)$$

$0\_R\oplus S=(0_R,0_S),1\_R\oplus S=(1_R,1_S)$

Examples:

1. M_2(\mathbb{R}) \oplus \mathbb{Z}\[x\]
2. $\mathbb{Z}\_2\oplus \mathbb{Z}\_3$
3. $\mathbb{Z}\oplus \mathbb{Z}$
4. $R_1\oplus R_2\oplus \cdots \oplus R_n$ ← Iterative direct sum

Remarks on Direct Sums:

1. If $R_1\subseteq S_1,R_2\subseteq S_2$ are subrings, then $R_1\oplus R_2\subseteq S_1\oplus S_2$ is a subring
2. $R\oplus Siscommutative⟺R,Sarecommutative$
3. $(r,r)|r\in R\subset R\oplus R$ is a subring

Ring Proofs

$r\cdot 0=r\cdot (0+0)=r\cdot 0+r\cdot 0⟹0=r\cdot 0$

Commutativity

Recall: R is commutative if $\forall r,s\in R,r\cdot s=s\cdot r$

Remark: If R is commutative and $S\subseteq R$ is a subring, then S is also cummutative. In particular, \mathbb{C}, \mathbb{R}, \mathbb{Q}, \mathbb{Z}, \mathbb{Z}\[\\sqrt2\], \mathbb{Q}\[i\] are all commutative

Example: $\mathbb{R}\oplus \mathbb{Z}\_7$ is commutative whereas $\mathbb{Q}\oplus M_2(\mathbb{C})$ is not

Definition: Let R be a ring. The center of R is:

$$Z(R)=r\in R|\forall x\in R,r\cdot x=x\cdot r$$

In other words, Z(R) is the set of elements in R that commute with any other element of R.

Remark:

1. We always have $0,1\in Z(R)$ this is because $$\begin{gathered} 1\cdot r=r\cdot 1=r \\ 0\cdot r=r\cdot 0=0 \end{gathered}$$
2. If $r,s\in Z(R)$ then $r-s\in Z(R)$ Pf. Let $x\in R$ be arbitary, we need to show that $(r-s)\cdot x=x\cdot (r-s)$ But, $(r-s)\cdot r=r\cdot x-s\cdot x=x\cdot r-x\cdot S=x\cdot (r-s)$ so, $r-s\in Z(R)$
3. If $r,s\in Z(R)$ then $r\cdot s\in Z(R)$ Pf. Let $x\in R$ be arbitary, we need to show that $(r\cdot s)\cdot r=x\cdot (r\cdot s)$ Indeed, $(r\cdot s)\cdot x$ $=r\cdot (s\cdot x)$ ← Associativity of $\cdot$ $=r\cdot (x\cdot s)$ ← $s\in Z(R),s\cdot x=x\cdot s$ $=(r\cdot x)\cdot s$ ← Associativity of $\cdot$ $=(x\cdot r)\cdot s$ ← $r\in Z(R),r\cdot x=x\cdot r$ $=x\cdot (r\cdot s)$← Associativity of $\cdot$

Corollary: $Z(R)\subseteq R$ (the center) is a subring

Examples:

1. If R is commutative then $Z(R)=R$
2. $R=M_n(\mathbb{C})$, then Z(R) = {\begin{pmatrix} \\alpha & \cdots & 0 \\ \\vdots & \ddots & \vdots \\ 0& \cdots & \alpha \\ \\end{pmatrix}|\ \alpha \in \mathbb{C}}
3. If $R=M_n(S),$ (where S is another ring), then Z(R)={\begin{pmatrix} \\alpha & \cdots & 0 \\ \\vdots & \ddots & \vdots \\ 0& \cdots & \alpha \\ \\end{pmatrix}|\ \alpha \in Z(S)}
4. Z(R\[x\])=(Z(R))\[x\] ← The center of a polynomial ring is a polynomial ring over the center of R
5. $Z(R\oplus S)=Z(R)\oplus Z(S)$ ← The center of the direct sum of two wrings is the direct sum of the centers

Units

Let R be a ring. The set of units in R is:

$$U(R)=R^x=a\in R|\exists b\in R:ab=ba=1$$

Where $a$ and $b$ are units or invertible elements. We denote $b=a^{-1}$ or $b$ is the inverse of $a$

Remark:

1. $U(R)\subseteq R$
2. $1\in U(R)$ since $1\cdot 1=1\cdot 1=1$
3. $0\notin U(R)$ since $0\cdot b=b\cdot 0=0\ne 1$
4. If $a\in U(R)$ then $a^{-1}$ is unique: if $a{b}_1=b_{1}a=1$ and $a{b}_2=b_{2}a=1$ then: $b_1=b_1(a{b}_2)=b_{2}a{b}_2=(b_{1}a)b_2=b_2$
5. One must be careful, it is possible that $ab=1$ but $a$ is not a unit

Proposition: $(U(R),\cdot )$ is a group

Proof: First, let us show that $U(R)$ is closed under ⋅ If $a_1,a_2\in U(R)$ Then: $(a_1{a}_2)(a_2^{-1}{a}_1^{-1})=a_1(a_2{a}_2^{-1})a_1^{-1}=a_1{a}_1^{-1}=1$

$(a_2^{-1}{a}_1^{-1})(a_1{a}_2)=a_2^{-1}(a_1^{-1}{a}_1)a_2=a_2^{-1}{a}_2=1$

$⟹a_1{a}_2$ is a unit $((a_1{a}_2{)}^{-1}=a_2^{-1}{a}_1^{-1})$

Neutral Element ( Identity element ) is $1\in U(R)$ by the above remark

If $a\in U(R)$ then, by the definition of a unit, there exits an $a^{-1}\in R$ such that $a{a}^{-1}=a^{-1}a=1$ but it follows that $a^{-1}\in U(R)((a^{-1}{)}^{-1})=a)$

Therefore $(U(R),\cdot )$ is a group. (Associativity of ⋅is ensured by the fact that R is a ring, hence $(R,\cdot )$ is associative.

Examples:

1. $R=\mathbb{R},\mathbb{C},\mathbb{Q}$ $R^x=R\backslash 0$

$$\begin{gathered} {\mathbb{R}}^x=\mathbb{R}\backslash 0 \\ mathbb{C}^x=\mathbb{C}\backslash 0 \\ mathbb{Q}^x=\mathbb{Q}\backslash 0 \end{gathered}$$

1. $R=\mathbb{Z}$ (What are the units in the ring of integers) $R^x=\pm 1$ $\mathbb{Z}\subset \mathbb{Q}$ ← (Every nonzero element of $\mathbb{Z}$ is invertible in $\mathbb{Q}$)

(For example, $2\in \mathbb{Z}$, not invertible within $\mathbb{Z}$ because $\frac{1}{2}\notin \mathbb{Z}$. However, if we think of $2\in \mathbb{Q}$, it is invertible within $\mathbb{Q}$ because $\frac{1}{2}\in \mathbb{Q}$)

1. $R=\mathbb{Z}\_n=0,1,\cdots ,n-1$ ← What are the units of the integers mod n $R^x=x\in \mathbb{Z}\ast n|\exists xy=1\ast modn$ Exactly the coprime to n residues: $R^x=\mathbb{Z}\_n^x=1\le a\le n-1|gcd(a,n)=1$ The cardinality of $\mathbb{Z}\_n^x$ is $\varphi (n)$ ← Eulers totient function

$n=6$ $\mathbb{Z}\_6=0,1,2,3,4,5⟹\mathbb{Z}\_6^x=1,5$

$n=7$ $\mathbb{Z}\_7=0,1,2,3,4,5,6⟹\mathbb{Z}\_7^x=1,2,3,4,5,6$

Remark: (From 103A): $\mathbb{Z}\_n^x=\mathbb{Z}\_n\backslash 0⟺$ $n$ is prime The units of the Integers mod n is equal to the integers up to n excluding zero if n is a prime number

Question: How to invert mod n?

$17^{-1}(mod48)=?$

$17,48$ are coprime

$(48,17)=(17,14)=(14,3)=(3,2)=(2,1)$

$48=2\cdot 17+14$ ← $14$ is the “Residue” $17=1\cdot 14+3$

$14=4\cdot 3+2$

$3=1\cdot 2+1$ ← Once we have 1, we now invert

$1=3-2=3-(14-4\cdot 3)=-14+5\cdot 3=-14+5(17-14)=-6\cdot 14+5\cdot 17=-6\cdot (48-2\cdot 17)+5\cdot 17=-6\cdot 48+17\cdot 17$

So, $17\cdot 17=1(mod48)$ so, $17^{-1}=17$ in $\mathbb{Z}\_48$

2. $R=M_n(\mathbb{R})$ $R^x=A\in M_n(\mathbb{R})|\exists B\in M_n(\mathbb{R}):AB=BA=I_n$ ← invertible matrices $=A\in M_n(\mathbb{R})|det(A)\ne 0=G{L}_n(\mathbb{R})$
3. R=\mathbb{Z}\[i\]={\ a+bi\ |\ a,b\in Z\ } \mathbb{Z}\[i\]^x=\ ?

$=a+bi|\exists c,d\in \mathbb{Z}:(a+bi)(c+di)=1\iff (a-bi)(c-di)=1$ Using: $a+bi\to a-bi$ or, $\bar{z}\bar{w}=\bar{z}\cdot \bar{w}$ $(a+bi)(a-bi)(c+di)(c-di)=1\cdot 1=1$ $=(a^2+b^2)\cdot (c^2+d^2)=1$ ← $(a^2+b^2)\in \mathbb{Z}$, $(c^2+d^2)\in \mathbb{Z}$ So, $a^2+b^2=1⟺a=0,b=\pm 1$ or $a=\pm 1,b=0$ $c^2+d^2=1⟺c=0,d=\pm 1$ or $c=\pm 1,d=0$ For conclusion, \mathbb{Z}\[i\]^x={\pm1, \pm i}

Definition: A ring is called a division ring if $R^x=R\backslash 0$

There are inverses for every element $\in R$ except 0

Definition: A commutative division ring is called a field

In other words, a field is a system $(R,0,1,\pm ,\cdot ,÷)$ with “all reasonable axioms” where we can divide except for $x÷0$

Examples:

$\mathbb{Q},\mathbb{R},\mathbb{C}-$ Fields

$\mathbb{Z}\_p-$ Field (p prime)

\mathbb{Z},\ \mathbb{Z}\[i\], \mathbb{Z}\_6, \cdots not fields

$\mathbb{H}-$ division ring that is not a field

Remark: $U(R\oplus S)=U(R)\times U(S)$

In other words, $(r,s)\in R\oplus S$ is a unit (in $R\oplus S$) iff $r\in U(R)$ and $s\in U(S)$

Proof:

$(\Rightarrow )$

If $(r,s)\in U(R\oplus S)$ then, $\exists (r^{\prime },s^{\prime })\in R\oplus S:$

$(r,s)(r^{\prime },s^{\prime })=(r^{\prime }{s}^{\prime })(r,s)=(1_R,1_S)\in R\oplus S$

$\iff r{r}^{\prime }=r^{\prime }r=1_R,s{s}^{\prime }=s^{\prime }s=1_S$

$(\Leftarrow )$

Exercise: (if $r\in U(R),s\in U(S)$ then $(r,s)\in U(R\oplus S)$)

Example: How many units are there in $\mathbb{Z}\_7\oplus \mathbb{Z}\_8$ ?

$U(\mathbb{Z}\_7\oplus Z_8)=U(\mathbb{Z}\_7)\times U(\mathbb{Z}\_8)=6\times 4=24$ units

Definition: An element $a\in R$ is a zero-divisor if there exists $b\in R$ such that $ab=0$ or $ba=0$

Examples:

1. In $\mathbb{Z}\_6,$ $$\begin{gathered} 2\cdot 3=0, \\ 4\cdot 3=0 \end{gathered}$$ 2, 3, 4 are zero divisors and 1, 5 are units

2. In $M_2(\mathbb{R}),$

   0&0\\1&0 \\end{pmatrix} \\begin{pmatrix} 0&1\\0&0 \\end{pmatrix}= \\begin{pmatrix} 0&0\\0&0 \\end{pmatrix}$$ # $$\begin{pmatrix} 0&1\\0&0 \\end{pmatrix} \\begin{pmatrix} 0&0\\1&0 \\end{pmatrix} \\begin{pmatrix} 1&0\\0&0 \\end{pmatrix}$$

Proposition: If $a\in U(R),$ then $a$ is not a zero divisor

Proof:

Assume on the contrary that $a\in U(R)$ yet a zero divisor. So: $\exists b\in R$ such that

Case 1: $ab=0⟹b=a^{-1}ab=a^{-1}\cdot 0=0$

Case 2: $ba=0⟹b=ba{a}^{-1}=0\cdot a^{-1}=0$

In any case, $b=0$, a contradiction

Note: Units are never zero divisors

Remark: In general, there might be elements that are neither units nor zero divisors

Note: Units are generators in $\mathbb{Z}\_n$

Domains

Definition: A ring R is a domain if it has no zero divisors. R is an integral domain if it is a communative domain

Examples:

1. $\mathbb{Z}$ is an integral domain
2. If R is a division ring then it is a domain

(Because in a division ring, non-zero elements are units, but we saw that units cannot be zero-divisors!)

1. If R is a field then it is an integral domain

\\begin{matrix} Field & \implies & Integral\ Domain \\ \Downarrow & & \Downarrow \\ Division Ring & \implies & Domain \\end{matrix}

Properties of Domains

1. Domains are “canellative” If $ab=ac,a\ne 0⟹b=c$

Proof:

Suppose that $ab=ac,a\ne 0$ in a domain. Then, $a(b-c)=ab-ac=0$. In a domain, if a product of two elements is 0 then one of them must be 0. Since $a\ne 0$, it follows that $b-c=0$. Namely, $b=c$

Remark: If your ring is not a domain, do not expect it to be cancellative

In $\mathbb{Z}\_6$, $3\cdot 2=3\cdot 4$ but $3\ne 4$

Proof: Suppose R is a domain $S\subseteq R$ subring

Let $a,b\in S$ be non-zero, let us prove that $ab\ne 0$. But $a,b\in R$ are non-zero, and R is a domain, so $ab\ne 0$

If in addition R is commutative, then so is S

Remark: A subring of a field need not be a field Ex. Not a field → $\mathbb{Z}\subset R$ ← Field

Corollary: \mathbb{Z}, \mathbb{Z}\[{\sqrt2}\], \mathbb{Z}\[{i}\], \cdots \subset \mathbb{C} therefore they are integral domains

Proposition: If R is a domain then R\[x\] is also a domain

(And if R is an integral domain then so is R\[x\]

Corollary: If F is a field then F\[x\] is an integral domain.

Moreover, F\[x,y\],\ F\[x,y,z\],\cdots are integral domains

Remark: Any subring of a field is an integral domain and conversely, if R is an integral domain then it is a subring of a field

Ex. \mathbb{Z} \subset \mathbb{Q},\ \mathbb{Z}\[i\] \subset \mathbb{Q}\[i\], \cdots

Remark: Any subring of a division ring is a domain. It is tempting to conjugate that any domain is a subring of a division ring. But that’s not true!

Homomorphisms

Homomorphism

Definition: A homomorphism is a “Structure-Preserving Map”

Group Homomorphism:

$$\begin{gathered} f:G\to H \\ f(g_1\cdot g_2)=f(g_1)\cdot f(g_2) \end{gathered}$$

(It follows that $f(e_G)=e_H$, identity mapped to identity)

Example:

$\ast \ast f:\mathbb{Z}\to \mathbb{Z}\_n\ast \ast$

Addition group of integers → $(0,1,\cdots ,n-1,+\_modn)$

$f(a)=a\_(modn)$

Definition: A function $f:R\to S$ (where $R$ and $S$ are rings) is a ring homomorphism if:

1. $f(r_1+\_R{r}_2)=f(r_1)+\_Sf(r_2)\forall r_1,r_2\in R$ (Namely, $f$ is a group homomorphism $f:(R,+)\to (s,+)$)
2. $f(r_1{\cdot }_R{r}_2)=f(r_1){\cdot }_{S}f(r_2)\forall r_1,r_2\in R$
3. $f(1_R)=1_S$

Remark: It is guarenteed that if $f:R\to S$ is a ring homomoprhism, then $f(0_R)=0_S$

Examples:

1. For any ring R, the identity map $$\begin{gathered} i{d}_R:R\to R \\ r\mapsto r \end{gathered}$$ is a ring homomorphism
2. If $S\subseteq R$ is a subring, then the inclusion map: $$\begin{gathered} i_S:S\to R \\ s\mapsto s \end{gathered}$$ is a ring homomorphism. E.g., $$\begin{gathered} f:Q\to R \\ f(\alpha )\mapsto \alpha \end{gathered}$$
3. $$\begin{gathered} f:\mathbb{Z}\to \mathbb{Z}\ast n=0,1,\cdots ,n-1 \\ f(a)=a\ast (modn) \end{gathered}$$

• $f(a+b)=f(a)+f(b)$
• $$\begin{gathered} f(a\cdot b)=f(a)\cdot f(b): \\ (a\cdot b)\ast modn=a\ast modn\cdot b\_modn \end{gathered}$$
• $f(1)=1$

$E.g.f:\mathbb{Z}\to \mathbb{Z}\ast 7,(12\cdot 13)\ast \mathrm{mod}7$

$$\begin{gathered} ⟹12\_\mathrm{mod}7\cdot 13\_\mathrm{mod}7) \\ ⟹156\mathrm{mod}7=30\mathrm{mod}7 \\ ⟹2=2 \end{gathered}$$

1. A non-example: $f:\mathbb{Z}\to \mathbb{Z}\oplus \mathbb{Z}$ by $f(a)=(a,0)$ $f$ is $+$- preserving, $\cdot$ - preserving, but not a ring homomorphism since $f(1)=(1,0)\ne (1,1)$, a violation of the 3rd axiom
2. f:\mathbb{C}\rightarrow M_2(\mathbb{R}) \\ f(a+bi)= \\begin{pmatrix} a & b \\ -b & a \\end{pmatrix}\\$$f(2-3i)=\begin{pmatrix}2&-3\\3&2\end{pmatrix}\\ $f((a_1+b_{1}i)+\ast complex(a_2+b_{2}i))=f(a_1+b_{1}i)+\ast matrixmultf(a_2+b_{2}i)$ f((a_1a_2-b_1b_2)+(a_1b_2+b_1a_2)i)$$\implies \begin{pmatrix} a_1a_2-b_1b_2 & a_1b_2+b_1a_2\\ -(a_1b_2+b_1a_2) & a_1a_2-b_1b_2 \\end{pmatrix} vs 2x2 matrix multiplication \begin{pmatrix} a_1 & b_1\\ -b_1 & a_1 \\end{pmatrix}\begin{pmatrix} a_2 & b_2\\ -b_2 & a_2 \\end{pmatrix}=\begin{pmatrix} a_1a_2-b_1b_2 & a_1b_2+b_1a_2\\ -b_1a_2-a_1b_2 & -b_1b_2+a_1a_2 \\end{pmatrix} As we can see they both yield the same result f(1)=1\_{M_2(\mathbb{R})}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \\end{pmatrix} Indeed, $1+0i$ → f(1) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\end{pmatrix}
3. Let R be a ring f:R\[x\]\\rightarrow R Polynomial Ring over R → R $f(a_0+a_{1}x+\cdots +a_n{x}^n)=a_0$ That’s always a ring homomorphism (Exercise: Check)

Definition: A ring homomorphism $f:R\to S$

1. Injective homo/monomorphism if $f(r_1)=f(r_2)\forall r_1,r_2\in R⟹r_1=r_2$ ← (1:1)
2. Surjective homo/monomorphism if $\forall s\in S:\exists r\in R:f(r)=s$ ← (Onto)
3. Isomorphism: Injective + Surjective ← (Bijective)

Examples:

1. $$\begin{gathered} f:\mathbb{Z}\to \mathbb{Z}\ast n \\ f(a)=a\ast \mathrm{mod}n \end{gathered}$$

Surjective

Not Injective because

$$\begin{gathered} f(1)=f(n+1) \\ f(0)=f(n) \end{gathered}$$

1. f(a+bi)=\begin{pmatrix}a&b\\-b &a\end{pmatrix}$ $f(a_1+b_1i)=f(a_2+b_2i)$$ $$\begin{pmatrix} a_1 & b_1\\ -b_1 & a_1 \\end{pmatrix} = \\begin{pmatrix} a_2 & b_2\\ -b_2 & a_2 \\end{pmatrix} \\ \implies a_1=a_2, b_1=b_2 \implies a_1+b_1i=a_2+b_2i$$ Injective Not Surjective because $$\begin{pmatrix} 1 & 1\\ 1 & 1 \\end{pmatrix}$$ does not have a source under $f$
2. f:\mathbb{Z} \rightarrow \mathbb{Z}*2\[x\] $f(a)=a\ast \mathrm{mod}2$

Not Injective 2 and 4 both mapped to 0

Not Surjective x doesn’t have a source

1. For any ring $R$, $i{d}_R:R\to R$ ← an isomorphism (Call an isomorphsim from a ring to itself “automorphism”)
2. $f:\mathbb{C}\to \mathbb{C}$ $f(a+bi)=a-bi$ aka “complex conjugation” is a ring/field automorphism

Ring Homomorphisms

$f:R\to S:$

$$\begin{gathered} f(r_1+r_2)=f(r_1)+f(r_2) \\ f(r_1{\cdot }_R{r}_2)=f(r_1){\cdot }_{S}f(r_2) \\ f(1_R)=1_S \end{gathered}$$

Remark - Basic Properties of Homomorphsims:

1. $f(0_R)=0_S,f(1_R)=1_S$
2. $f(-r)=-f(r)$
3. f(r\cdot r\cdot \cdots \cdot r)=f(r)\cdot \cdots \cdot f(r)$$ ← n times

Recall that a ring homomorphism which is injective and surjective (namely bijective) is called an isomorphsim.

Defintion: Two rings $R,S$ are isomorphic to each other if there is an isomorphsim $f:R\to S$ Denoted $R\cong S$

Remark:

1. If $f:R\to S$ is an isomorphism, then the inverse function: $f^{-1}:S\to R$ is also an isomorphism
2. If $f:R\to S,g:S\to T$ are isomorphic ($R,S,T$ rings) Then, $g\circ f:R\to T$ is $g\circ f(r)=g(f(r))$ Proof: >$(g\circ f)(r_1+r_2)=g(f(r_1+r_2))=g(f(r_1)+f(r_2))=g(f(r_n))+g(f(r_2))=(g\circ f)(r_1)+(g\circ f)(r_2)$

Multiplication is very similar $(g\circ f)(1_R)=g(f(1_R))=g(1_S)=1_T$

3. Every ring is isomorphic to itself (find an iso) Properties of rings, preserved under isomorphisms

   Examples:

• Cardinality
• Commutativity
• Being a domain, field, division ring
• Having isomorphic addition groups
• Having isomorphic groups of units (if $R\cong S$ then $R^x\cong S^x$)
• Existence of an element $x\ne 0$ for which $x^2=0$ (If $f:R\to S$ is an isomorphism and $0\ne x\ne R$ such that $x^2=0$, then $0\ne f(x)\in S$ and $f(x{)}^2=f(x^2)=f(0)=0$)

Remark: An example of a function that preserves $+,\cdot$ but not 1 (therefore not a homomorphism):

\\ \ \ \ \ f((a,b))=(0,b)$$ **Examples:** 1. $\mathbb{Z}\_6 \cong \mathbb{Z}\_8$? ←(No, different cardinalities) 1. $\mathbb{Q} \cong \mathbb{R}$? ←(No, different cardinalities - $\mathbb{Q}$ is countable, $\mathbb{R}$ is uncountable) 1. $\mathbb{Z} \cong \mathbb{Q}$? (No, $\mathbb{Q}$ is a field but $\mathbb{Z}$ is not) 1. $M_2(\mathbb{R})\cong \mathbb{C} \oplus \mathbb{C}$? (No, LHS is not commutative but RHS is commutative) $\mathbb{C} \oplus \mathbb{C}$ ← not a domain $(0,1)\cdot (1,0) = (0,0)$ there are zero-divisors **Remark:** The direct sum of two rings is never a domain 1. $\mathbb{Z}\cong \mathbb{Z} \oplus \mathbb{Z}$? ← No, $\mathbb{Z}$ is a domain but $\mathbb{Z} \oplus \mathbb{Z}$ is not 1. $\mathbb{Z} \oplus \mathbb{Z} \cong \mathbb{Z}\oplus \mathbb{Z}\oplus \mathbb{Z}$ ← No, calculating the units for both sides, we see that the units are not the same $U(\mathbb{Z} \oplus \mathbb{Z}) = U(\mathbb{Z}) \oplus U(\mathbb{Z}) = 4$ $U(\mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}) = U(\mathbb{Z}) \oplus U(\mathbb{Z})\oplus U(\mathbb{Z}) = 8$ 1. $\mathbb{R}\cong \mathbb{C}$? ← No, suppose we had $f:\mathbb{C}\rightarrow \mathbb{R}$ iso $f(i)^2=f(i^2)=f(-1)=-f(1)=-1$ So, there is a real number $f(i)$ whose square is $-1$. That is false. 1. $\mathbb{Z}\[i\] \cong \mathbb{Z}\[\\sqrt2\]$? (hint: use hw 2) 1. $\mathbb{Z}\_{15}\cong \mathbb{Z}\_5\oplus\mathbb{Z}\_3$ ← (yes, isomorphic because of Chinese remainder theorem) **Definition:** Chinese Remainder Theorem - If $n, m$ are coprime numbers $(n,m>1)$ then $\mathbb{Z}\_{nm}\cong \mathbb{Z}\_n \oplus \mathbb{Z}\_m$ ($\mathbb{Z}\_{24}\not\cong \mathbb{Z}*2 \oplus \mathbb{Z}*{12}$) because $2,12$ are not coprime Shift back to homomorphisms, not necessarily isomorphic Let $f:R\rightarrow S$ be a ring homomorphism $Im(f)={s\in S\ |\ \exists\ r \in R: f(r)=s} \subseteq S$ $Ker(f)={r\in R\ |\ f(r)=0_S } \subseteq R$ **Remark:** * $Im(f) \subseteq S$ is a subring of $S$ * $Ker(f) \subseteq R$ is closed under $+, \cdot$ but is not a subring ( If $1_R\in Ker(f),$ then $f(1_R)=0_S$, but $f(1_R)=1_S,$ we obtain $1_S=0_S$, impossible!) **Remark:** A ring homomorphism is surjective $**\\Longleftrightarrow$ $Im(F)=S$** A ring homomorphism is injective $\Longleftrightarrow Ker(f)={0_R}$ **Proof:** ($\Rightarrow$) Suppose $f:R\rightarrow S$ is an injective homomorphism. Obviously, $f(0_R)=0_S$ so $0_R\in Ker(f)$. Suppose that $0\neq r\in R$, if $r\in Ker(f)$ then: $f(r)=0_S=f(0_R),$ contradicting injectivity. Therefore $Ker(f)={0_R}$ ($\Leftarrow$) Suppose $Ker(f)={0_R},$ let us prove that f is injective Suppose $f(r_1)=f(r_2),$ let us show $r_1=r_2$$f(r_1)=f(r_2)\implies f(r_1)-f(r_2)=0_S \implies f(r_1-r_2)=0_S\implies r_1-r_2 \in Ker(f)={0_R}$ so, $r_1-r_2 = 0_R\implies r_1=r_2$ **Exercise:** $f: \mathbb{R} \oplus \mathbb{R} \rightarrow M_2(\mathbb{R}) \\ f((a,b))=\begin{pmatrix} 2a-b & -2a+tb \\ a-b & -a+2b \\end{pmatrix}$ This is a ring homomorphism Calculate $Ker(f)$ and deduce whether f is injective or not $f: \mathbb{Z}\_{15}\rightarrow \mathbb{Z}*5 \\ f(a)=a*{\mod 5}$ this is a ring, calculate $Ker(f)$ ### Injective / Surjective A ring homomorphism $f:R\rightarrow S$ is surjective if $Im(f) =S$ (Namely if every $s\in S$ has a source) A ring homomorphism $f:R\rightarrow S$ is injective if it is $1:1$, equivalently if $Ker(f) = {0_R}$ or ${r\in R\ |\ f(r)=0_S}$ **Example:** Let $f:\Bbb{C} \rightarrow M_2(\Bbb{R})$ be given by

f(a+bi)=\begin{pmatrix} a-3b & -2b \ 5b & a+ 3b \end{pmatrix}

This is indeed a ring homomorphism. Is it injective? **Solution:** Calculate $ker(f)$ $$\lbrace a+bi\ |\ \\begin{pmatrix} a-3b & -2b \\ 5b & a+3b \\end{pmatrix} =0\_{M_2(\Bbb{R})} \\rbrace$$ $\implies \begin{matrix} a-3b = 0 \implies \underline{a=0} \\ -2b = 0 \implies \underline{b = 0}\\ 5b=0 \\ a+3b = 0 \end{matrix}$ $\implies$Therefore the only complex number $a+bi$ which is mapped to 0 is 0 $\implies ker(f) = {0}$, thus, $f$ is injective F is not surjective: 1. $\begin{pmatrix}0&0\\1&0\end{pmatrix}\notin Im(f)$ Otherwise $\begin{pmatrix}0&0\\1&0\end{pmatrix} = \begin{pmatrix}a-3b & -2b \\ 5b & a+3b\end{pmatrix}\implies \begin{matrix}b=0\\ b= \frac{1}{5}\end{matrix}\implies contradiction$ 1. $f$ is injective, so if $f$ is surjective, then it is not an isomorphism. But, $\Bbb{C}\not\cong M_2(\Bbb{R})$, $\Bbb{C}$ is commutative, $M_2(\Bbb{R})$ is not. # Chinese Remainder Theorem (CRT) **Theorem:** If $n,m > 1$ are coprime numbers, then: $\Bbb{Z}\_{nm}\cong \Bbb{Z}\_n \oplus \Bbb{Z}\_m$ (a ring isomorphism) **Remark:** For example, $\Bbb{Z}*{15}\cong \Bbb{Z}*3 \oplus \Bbb{Z}*5$, but $\Bbb{Z}*{120}\cong \Bbb{Z}*{10} \oplus \Bbb{Z}*{12}$, as 10 and 12 are not coprime **Proof:** Suppose that $n,m>1$ are coprime. Define a function: $f:\Bbb{Z}*{nm} \rightarrow \Bbb{Z}*n \oplus \Bbb{Z}*m$ by $f(a) = (a*{(mod\ n)}, a*{(mod\ m)})\ \forall\ a\in \Bbb{Z}*{nm}$ This is a ring homomorphism: $f(1)=1,\ f(a+b)=f(a)+f(b),\ f(ab)=f(a)f(b)$ Exercise: Verify **Proof:** Let us show that f is surjective Pick $s \in \Bbb{Z}\_n,\ t \in \Bbb{Z}*m$. We aim to show that $(s,t)\in Im(f),$ namely, that there is $x\in \Bbb{Z}*{nm}$ For which $f(x)=(s,t)=(x\_{mod\ n}, x\_{mod\ m})$ Namely: $\begin{matrix} x=s(mod\ n)\\ x=t(mod\ m) \\end{matrix}$ Since $n, m$ are coprime, by the Euclidean Algorithm. There is a linear combination: $un+vm=1\ |\ u,v\in \Bbb{Z}$ (For example, 5,7 are coprime we have a linear combination $3\cdot 5+(-2)\cdot 7 = 1$) Consider $x=t\cdot un+s\cdot vm \in \Bbb{Z}$, (consider $x\_{(mod\ nm)}$) We claim that $x\equiv s\_{(mod\ n)},\ x\equiv t\_{(mod\ m)}$ $x\_{(mod n)}\equiv tun+svm\_{(mod\ n)}$ ← $tun=0$ $\equiv svm\_{(mod\ n)} \equiv s\cdot (1-un)*{(mod\ n)} \equiv s-sun*{(mod\ n)}$ ← $sun=0$ $\equiv s\_{(mod\ n)}$ $x\_{(mod m)}\equiv tun+svm\_{(mod\ m)}$ ← $svm=0$ $\equiv tun\_{(mod\ m)} \equiv t\cdot (1-vm)*{(mod\ m)} \equiv t-tvm*{(mod\ m)}$ ← $tvm=0$ $\equiv t\_{(mod\ m)}$ Therefore, $x\equiv s\_{(mod\ n)}$, $x\equiv t\_{(mod\ m)}$ as required. It follows that $f:\Bbb{Z}*{nm} \rightarrow \Bbb{Z}*{n} \oplus \Bbb{Z}\_{m}$ is a surjective ring homomorphism Since $|\Bbb{Z}*{nm}| = n\cdot m = |\Bbb{Z}*{n}\oplus \Bbb{Z}\_{m}|,$ and f is surjective, it is in fact bijective. Hence, $f$ is a ring homomorphism $\ \square$ **Example:** Find $x\in \Bbb{Z}*{35}$ such that $x\equiv 3*{(mod\ 5)}, x\equiv 6\_{(mod\ 7)}$ Step 1: Find a linear combination $35=7\cdot 5$ ← 7 and 5 are coprime $u\cdot 7+ v\cdot 5 = 1\ | u=3,\ v=-2$ ← Linear Combination $(7,5)=(5,2)=(2,1)$ $x=t\cdot un + s\cdot vm=3\cdot (-2)\cdot 7+6\cdot 3 \cdot 5=-42+90=48\_{(mod\ 35)}=13=x$ $13\_{(mod\ 5)} \equiv3$ $13\_{(mod\ 7)} \equiv 6$ **Application:** How many units are there in $\Bbb{Z}\_{35}$? $5,\ 7$ are coprime $\implies \Bbb{Z}*{35}\cong \Bbb{Z}*{5} \oplus \Bbb{Z}\_{7}$ But that means that $\Bbb{Z}*{35}^x \cong (\Bbb{Z}*{5}\oplus \Bbb{Z}*{7})^x=\Bbb{Z}*{5}^x\oplus \Bbb{Z}\_{7}^x=4\cdot 6 = 24$ **Recall:** Units in $\Bbb{Z}\_n$ are exactly residues that are coprime with n **Notation:** $\phi(n) =$ number of residues mod n which are coprime with n **Recall:** If $f: R\rightarrow S$ is a ring homomorphism, then: * $\text{Im} f = {s\in S:\exists r\in R: f(r)=s} \subseteq S$ * $\text{ker} f ={r\in R: f(r)=0_s}\subseteq R$ $f$ is injective (one-to-one) $\Longleftrightarrow$ $\ker f={0_R}$ $f$ is surjective (onto) $\Longleftrightarrow$ $\text{Im} f= S$ ## Properties of Im, kernel of $f$ 1. $\text{Im} f\subseteq S$ is a subring 1. $\ker f\subseteq R$ 1. is an additive subgroup 1. contains $0_R$ 1. $\forall x\in \ker f, r\in R, xr, rx\in\ker f$ 1. $1_R\not\in\ker f$ # Ideals **Definition:** A subset $I\vartriangleleft R$ of a ring is called an **ideal** if: 1. It is an additive subgroup of $R$ (it contains 0 and is closed under $\pm$) 1. $I$ is closed under multiplication by an arbitrary element of $R$.

\forall x\in I, r\in R,\ rx,xr\in I

1. It is a proper subset since $1_R\not\in I$ 1. If $1\in I$ then $\forall\ r \in R, r= r\cdot 1 \in I\implies R = I$ $R$ is not an ideal of itself, it is an “improper ideal” Examples: 1. $f:\mathbb{Z}\rightarrow\mathbb{Z}$ $a\mapsto a\pmod n$ $\ker f ={a\in\mathbb{Z}:a\equiv0\pmod n}={...,-2n, -n,0,n,2n,...}= n\Bbb{Z}$ $f$ is an ideal What do ideals of $\mathbb{Z}$ look like? Any $I\vartriangleleft \mathbb{Z}$ is an additive subgroup of $(\mathbb{Z},+)$ and is cyclic: $m\mathbb{Z}$ for some integer $m$ 1. Let $F$ be a field. $f:F\[x\]\\rightarrow F$ $f(p(x))=p(0)=a_0+a_1(0)+a_2(0)^2+\cdots+a_n(0)^n=a_0$ Then $f$ is a ring homomoprhism. $\ker f ={p(x)\in F\[x\]:p(x)=a_1x+a_2x^2+\cdots+a_nx^n}$, the set where the leading coefficient is equal to 0. Given $\alpha\in F,$ $f:F\[x\]\\rightarrow F$ $f(p(x))=p(\alpha)$ If $\alpha=1,$ $f(p(x))=\sum\_{k=0}^na_k$, the sum of all coefficients. $\ker f={p(x)\in F\[x\]:p(\alpha)=0}$ 1. $f:\mathbb{Z}\[i\]\\rightarrow\mathbb{Z}\_5$ $f(a+bi)=(a+2b)\pmod 5$ **Exercise**: Verify $f$ is a ring homomorphism. Is there a homomorphism $g:\mathbb{Z}\[i\]\\rightarrow\mathbb{Z}\_7$? (more difficu $\ker f={a+bi:a,b\in\mathbb{Z}:a+2b\equiv0\pmod 5}$ (equivalently $a\equiv3b\pmod 5)$ For example, $4+3i\in\ker f$ 1. Suppose $F$ is a field. What are the ideals of $F$? ${0_F}\vartriangleleft F$ Let $I\vartriangleleft F$ be an ideal of a field. If $0\not=x\in I.$ Then, $x^{-1}\in I$. But then $x^{-1}x=1\in I$ which is a contradiction. Thus, ${0_F}$ is the only ideal of a field. **Corollary:** If $I\vartriangleleft R$, then $I$ contains no units. **Exercise:** If $F$ is a field, then the only ideal of $M_n(F) \text{ is } {0}$ For the rest of the lecture, $R$ is a commutative ring. (some of this can be found in section 5.4 of the textbook) **Definition:** Let $S\subseteq R$ be a subset. The ideal **generated** by $S$, denoted $\langle\ S\ \rangle \vartriangleleft R,$ is

\boxed{\langle\ S\ \rangle = \left{ \sum_{i=1}^m r_is_i:r_i\in R,s_i \in S\right}}

**Exercise:** Prove $\langle\ S\ \rangle \vartriangleleft R$ is an ideal (but possibly improper) **Remark:** $<S>$ is the smallest ideal of $R$ that contains $S$. If $J\vartriangleleft R, S\subseteq J\Rightarrow I\subseteq J$ **Examples:** 1. Ideals of $\mathbb{Z}$ are of the form ${a\in\mathbb{Z}:a\equiv0\pmod m}={xn:x\in \mathbb{Z}}=<n>$ 1. $\<6,10>\vartriangleleft \mathbb{Z}$ $\<6,10>={r_1\cdot6+r_2\cdot10:r_1,r_2\in\mathbb{Z}}\subseteq\<2>$ since all the elements are even numbers We want to show that $\<2>\subseteq\<6,10>$ by showing $2\in\<6,10>$ $2=(-3)(6)+(2)(10)\in\<6,10>$ Thus, $\<6,10>=\<2>$ In general for $\mathbb{Z}$, $\<a,b>=\<\\gcd(a,b)>$ **Exercise**: Prove it 1. $F\[x\]$ ${p(x)\in F\[x\]:\text{ zero constant term}}=\langle\ x\ \rangle={p(x)\cdot x:p(x)\in F\[x\]}$ **Summary:** Ideal: $I \triangleleft R$ * Additive subgroup * Closed under multiplication with arbitrary elements of R * Proper Ideal generated by a set $S\subseteq R$ - subset

\langle\ S\ \rangle = \left{ \sum_{i=1}^m r_is_i:r_i\in R,s_i \in S\right}

Examples: 1. $\langle\ 2\ \rangle \triangleleft \Bbb{Z} = {\ r\cdot 2\ |\ r\in \Bbb{Z}\ }= 2\mathbb{Z} = {\text{ Even Integers }}$ 1. $\langle\ n\ \rangle \triangleleft \Bbb{Z} \text{ is } n\Bbb{Z} = {\text{ all integers divisible by n}}$ 1. $\langle\ 6,\ 8\ \rangle \triangleleft \Bbb{Z} = {r_1\cdot 6 +r_2\cdot 8\ |\ r_1,r_2 \in \Bbb{Z}\ } = \langle\ 2\ \rangle$ 1. $\langle\ n,\ m\ \rangle = \langle\ \text{gcd}(n,m)\ \rangle$ 1. If $n,m$ are coprime, $\langle\ n, m\ \rangle = \Bbb{Z}$ ← “Improper Ideal” 1. $\langle\ x\ \rangle \triangleleft F\[x\] = {\ f(x)\cdot x\ |\ f(x)\in F\[x\]\\ }={\text{ polynomials with free term 0}}={\ \text{ker }(F\[x\]\\to F\ |\ x=0)}$ ## Principal Ideal Definition: \*\*\*\*An ideal $I \triangleleft R$ is principal if it is generated by a single element (R is assumed to be a commutative ring)

I=\langle\ r\ \rangle={a\cdot r\ |\ a\in R\ }=R\cdot r

**Proposition:** Suppose that $\langle\ r\ \rangle=\langle\ s\ \rangle$ in $R$. Then, $r=s\cdot u$ where $u\in R^x$. Conversely, if $r,s\in R \text{ and } r=s\cdot u \text{ for } u\in R^x,\text{ then } \langle\ r\ \rangle =\langle\ s \rangle$ **Examples:** (Prop): 1. $\langle\ 2\ \rangle = \langle\ -2\ \rangle \text{ in } \Bbb{Z}$ ($-1$ is the only $\neq 1$ element in $\Bbb{Z}$ ) 1. In $\Bbb{Z_6}:$ $\langle \ 2 \ \rangle = {\ 0,\ 2,\ 4\ } = \langle\ 4\ \rangle = {\ 0,\ 4,\ 2\ }$ 1. In $\Bbb{R}\[x\],\ \langle\ 2x+1\ \rangle = \langle\ -x-\frac{1}{2}\ \rangle \implies (2x+1)=-2(-x-\frac{1}{2})$ is a unit in $\Bbb{R}\[x\]$ **Proof:** (of proposition): Suppose $r,s\in R$ and $r=u\cdot s$ for $u\in R ^x$. Let us prove $\langle\ r\ \rangle = \langle\ s\ \rangle$. Recall that $\langle\ S\ \rangle$ is the smallest ideal containing S. Thus,

\begin{cases} \text{ if } r\in \langle\ s \ \rangle\implies \langle\ r\ \rangle \subseteq \langle\ s\ \rangle \ \text{ if } s\in \langle\ r \ \rangle\implies \langle\ s\ \rangle \subseteq \langle\ r\ \rangle \end{cases} \ \implies \text{It suffices to show: } r\in \langle\ s\ \rangle,s\in \langle\ r\ \rangle \ \implies r=u\cdot s \in \langle\ s\ \rangle={a\cdot s:\ a\in R} \ s=u^{-1}r\in \langle\ r\ \rangle,\text{ hence }\langle\ s\ \rangle=\langle\ r\ \rangle

**Proof of Converse:** Suppose that $r,s\in R \ st. \langle\ r\ \rangle =\langle\ s\ \rangle$. $r\in \langle\ r\ \rangle=\langle\ s\ \rangle={\ a\cdot s \in r\ }$ Thus, $\exists \ a\ st.\ r=a\cdot s$ . Similarly,

\exists\ b\in R\ st.\ s=b\cdot r\ r=a\cdot s=a\cdot b \cdot r\ (1- a \cdot b)\ r=0

Since $R$ has no zero divisors and $r\neq 0$, then $1-ab=0\implies ab=1$ Thus, $a\in R^x$ Hence $r=a\cdot s$ for $a\in R^x$ **Definition:** A ring is a principal integral domain (PID) if it is an integral domain if every ideal is a principal **Examples:** 1. $\Bbb{Z}$. Every ideal of $\Bbb Z$ is an additive subgroup. Subgroups of $(\Bbb Z, +)$ are cyclic $\implies \langle\ n\ \rangle = n\Bbb Z$. So, ${\ \text{ideals of }\Bbb Z\ }={\ n\Bbb Z: n\in \Bbb Z \ }$ 1. Any field is a PID **Note:** The only ideal of a field is ${\ 0\ }$ which is clearly principal **Non-Example:** $\Bbb Z \[x\]$ is an integral domain WTS: $\exists$ ideal that is not principal Take $I \triangleleft \Bbb Z\[x\]$ to be $I = \langle\ 2,\ x\ \rangle \triangleleft \mathbb{Z}\[x\]$ $\langle\ 2,\ x\ \rangle={\ f(x)\cdot2+g(x)\cdot x\ |\ f,g\in \mathbb{Z}\[x\]\\ }$ Show $I$ isn’t principal Assume to the contrary that $\langle\ 2,\ x\ \rangle=\langle\ g(x)\ \rangle\ |\ g(x)\in \mathbb{Z}\[x\]$ $2=h_1(x)g(x)$ for some $h_1(x)\in \mathbb{Z}\[x\]$ So degree of $g=0,\ g(x)=n,$ a constant Also, $x=h_2(x)g(x) = h_2(x)\ n \implies deg(h_2)\leq 1$ $h_2(x)=a+bx \implies x=an+bnx \implies an=0, \ bn=1 \implies n=\pm 1$ $\implies \langle \ g(x)\ \rangle = \mathbb{Z}\[x\]$ $1=p(x)\cdot 2 + q(x)\cdot x \implies 1$ is even Free term $p(x)$ is even and Free term $q(x)$ is zero This is a Contradiction so we conclude that $\langle\ 2,\ x\ \rangle$ is not principal $F\[x\]$, F is a field, is always a PID * $\mathbb{Z}\[i\]-\text{PID}$ * $\mathbb{Z}\[\\sqrt2\]-\text{PID}$ * $\mathbb{Z}\[\\sqrt{-2}\]-\text{PID}$ * $\mathbb{Z}\[\\sqrt{-5}\]-\text{Not a PID}$ Ideal is principal:

I=\langle\ r\ \rangle={a\cdot r\ |\ a\in R\ }=R\cdot r\ (\ \langle\ 3\ \rangle\triangleleft \mathbb{Z}, \langle\ 3\ \rangle={\ a\cdot 3\ |\ a\in \mathbb{Z}\ }=\mathbb{Z}\cdot 3 = 3\ )

**Definition:** PID - Integral domain in which every ideal is principal Examples of PID: $\mathbb{Z},\ \mathbb{Z}\[i\],\ F-\text{Field}$ Non-Examples of PID: $\mathbb{Z}\[x\], \mathbb{Z}\[\\sqrt{-5}\], ...$ Let R be a commutative ring (integral domain) **Definition:** For $r,s\in R$, then $r\ |\ s$ $(\text{r divides s})$ if $\exists\ a\in R\ st.\ s=a\cdot r$ **Examples:** 1. In $\mathbb{Z}$, $3|6$ 1. In $R\[x\],\ x-1\ |\ x^3-1$ Exercise: If $f,g\in F\[x\]\\text{ and } f\ |\ g$ then $deg(f)\leq deg(g) \text{ (converse false)}$ In an integral domain R, 1. $r\ |\ s \iff \langle\ s\ \rangle \subseteq \langle\ r\ \rangle$ 1. $\langle\ r\ \rangle=\langle\ s\ \rangle\iff r=u\cdot s \text{ for } u\in R^x$ **Remark:** In an integral domain R: 1. $r\ |\ s \iff \langle\ s\ \rangle \subseteq\ \langle\ r\ \rangle$ 1. $\braket{\ r\ } = \langle \ s\ \rangle \iff r=u\cdot s \text{ for some }u\in R^x$ ### Division Algorithm for $F\[x\]$ Let $f(x),g(x)\in F\[x\]$, then $\exists\ q(x), r(x) \in F\[x\]\\ st: f(x)=q(x)g(x)+r(x)$ and either $r(x)=0$ or $deg(r)\<deg(g)$ **Proof:** Let $f(x)=\sum\_{i=0}^na_ix^i,\ g(x)=\sum\_{j=0}^mb_jx^j$ So $deg(f)=n\text{ and } deg(g)=m\ st.\ a_n\neq 0, b_m\neq 0$ Case 1: $n\<m: (\text{base case})$ Take $q(x)=0,\ r(x)=f$ Then $f=q\cdot g+r=0+r$ $deg(r)=deg(f)=n\<m=deg(g)$ Case 2: $n\geq m:$ Perform induction on n

\text{Take }q(x)=\frac{a_n}{b_m}x^{n-m},\text{ Then:}\ q(x)g(x)=\frac{a_n}{b_m}x^{n-m}\cdot \sum_{i=0}^mb_ix^i=\sum_{i=0}^{m}\frac{a_nb_i}{b_m}x^{i+n-m} \= \frac{a_m\cdot \cancel{b_m}}{\cancel{b_m}}x^{m+n-m}=a_nx^n+\text{ lower deg polynomial }

Now, $f(x)-q(x)g(x)=\sum{a_ix^i}-a_nx^n+\sum c_ix^i = a_nx^n+...$ So $deg\[f(x)-q(x)g(x)\]\\leq n-1$ By inductive hypothesis $f-q\cdot g=h\cdot g+r$ for $h,r\in F\[x\]$ and either $r=0\text{ or } deg(r)\<deg(g)$ So $f=(h+q)\cdot g+r$ **Example:** Divide $x^7+2x+1$ by $x^2+1$ (with residue) over $\mathbb{R}$

\begin{array}{r} x^5-x^3+x\phantom{} \ x^2+1{\overline{\smash{\big)},x^7+2x+1}} \ \underline{-(x^7+x^5)} \phantom{111}\ -x^5+2x+1\phantom{} \ \underline{-(x^5+x^3)}\phantom{111} \ -x^3+2x+1\ \underline{x^3+x\phantom{1111}}\ r(x)\longrightarrow x+1

\end{array}

$\implies x^7 +2x +1 = (x^5-x^3+x)\underbrace{(x^2+1)}*{\text{deg = 2}} + \underbrace{(x+1)}*{\text{deg = 1}}$ **Theorem:** $F\[x\]$ ($F$ is a field) is a PID **Proof:** Let $0\neq I \rhd F\[x\]$ Pick $0\neq f\in I$ of lowest possible degree Since $f\in I, \braket{\ f\ }\subseteq I$ WTS $I=\braket{\ f\ }$ Let $g\in I,$ then by the divison algorithm, $\exists\ q,r\in F\[x\]\\ st.\ g(x)=q(x)f(x)+r(x)$ Either $r(x)=0$ or $deg(r)\<deg(f)$ $r(x)=\underbrace{g(x)}*{\in I}-q(x)\underbrace{f(x)}*{\in I} \in I$ Then, $r(x)$ must equal 0 because $f$ was of lowest possible degree in $I$ so $deg(r)\<deg(f)$ which is impossible. Then, $g(x)=q(x)\cdot f(x)\in \braket{\ f\ }$ Therefore, $I\subseteq \braket{\ f\ }$ so $I \in \braket{\ f\ }$. In particular, $I$ is principal **Corollary:** ${\text{ Ideals of } F\[x\]\\ } \xleftrightarrow{1:1} {\text{ monic polynomial of } F\[x\]\\ }$ Every ideal is a principal and we can normalize its highest term, multiplying it by a unit scalar $\braket{\ f\ }=\braket{\ g\ },\ f,g \text{ monic } \implies f=g$ $F\[x\]$ is a PID if every ideal is generated by a single element $I \rhd F\[x\]$ $I = \braket{\ f(x)\ }=F\[x\]\\cdot f(x)=\left{\ g(x)\cdot f(x) \mid g(x)\in F\[x\]\\ \right}$ **Example:** $\braket{\ x^2+1\ } \rhd \mathbb{R}\[x\] \to \braket{\ x^2+1\ } = {\ g(x)\cdot (x^2+1)\mid g(x)\in \mathbb{R}\[x\]\\ }$ $\underbrace{x^4-1}\_{=\ q(x)(x^2+1)+r(x)}\in \braket{\ x^2+1\ }=(x^2-1)(x^2+1)$ $\braket{\ f\ }= \braket{\ g\ } \iff f=u\cdot g\mid u\in F\[x\]^x$ $F\[x\]^x=F \backslash {0} \implies p(x)\cdot q(x)=1$ **Corollary:** Every Ideal $I\rhd F\[x\]$ is $I=\braket{\ f(x)\ }$ where $f(x)$ is monic: $f(x)=x^n+a\_{n-1}x^{n-1}+\cdots + a_0$ Moreover, if $f,g$ are monic polynomials then $\braket{\ f\ }=\braket{\ g\ } \iff f=g$ Note: Must assume $f,g$ are monic because $\braket{\ x^2+1\ } = \braket{\ -2x^2-2\ }$ ${\ \text{Ideas in }F\[x\]\\ } \xleftrightarrow{1:1}{\ \text{non-constant monic polynomnial}\ }$ $\braket{\ f\ } \longleftrightarrow f$ **Question:** Why Ideals? * $f:R\to S \implies ker(f)\lhd R$ * $I\lhd R\to \exists\ \underbrace{\pi: R\to R/I}\_{ker(\pi)=I}$ ### Quotient Rings Let $R$ be a ring, $I\lhd R$. Define the quotient ring $R/I$

\boxed{R / I={\ r+I\mid r\in R}}\ r+I:{\ r+\alpha\mid \alpha \in I\ }={\ x\in R\mid x-r\in I\ }

**Example:** $R=\mathbb{Z}$ $I=3\mathbb{Z}$

R/I = \mathbb{Z}/3\mathbb{Z}={\ a+3\mathbb{Z}: a\in \mathbb{Z}\ } \ = 0+3\mathbb{Z}= {\cdots, -6,-3,0,3,6,\cdots } \ = 1+3\mathbb{Z}= {\cdots, -5,-2,1,4,7,\cdots } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots

**Remark:** $r+I=s+I\iff r-s\in I$ $2+3\mathbb{Z}=5+3\mathbb{Z} \iff 5-2 = 3 \in 3\mathbb{Z}$ ### Making R / I a Ring Addition: $\underbrace{(r+I)+(s+I)}*{\text{addition of cosets}}=\underbrace{(r+s)}*{\text{addition in the ring}}+I$ $(2+3\mathbb{Z})+(2+3\mathbb{Z})=(2+2)+3\mathbb{Z} = 4+3\mathbb{Z} = 1+3\mathbb{Z}$ Multiplication: $(r+I)\cdot (s+I)=r\cdot s+I$ $(2+3\mathbb{Z})\cdot (2+3\mathbb{Z})=4+3\mathbb{Z}=1+3\mathbb{Z}$ Well Defined?

(2+3\mathbb{Z})=(-7+3\mathbb{Z})=(14+3\mathbb{Z})\ (-7+3\mathbb{Z})+(14+3\mathbb{Z})=(7+3\mathbb{Z})=(1+3\mathbb{Z})\ (-7+3\mathbb{Z})\cdot (14+3\mathbb{Z})=(-98+3\mathbb{Z})=(1+3\mathbb{Z})

\*Need to show this is always true **Claim:** Addition and Multiplication on $R/I$ are well defined WTS: **Proof:** Suppose that $r+I,\ s+I\in R/I$ Assume that $r+I=r'+I,s+I=s'+I$ WTS: $(r+s)+I=(r'+s')+I\iff (r+s)-(r'+s')\in I$ $(r+s)-(r'+s') = \underbrace{(r-r')}*{\in I}+\underbrace{(s-s')}*{\in I}$ since $r+I=r'+I,\ s+I=s'+I$ Μultiplication: Suppose that $r+I=r'+I, s+I=s'+I\in R/I$ $(r+I)\cdot (s+I)=(r'+I)\cdot (s'+I)\iff rs+I=r's'+I$ WTS: $r\cdot s - r'\cdot s'\in I$ $r'-r=\alpha \in I$ $s'-s=\beta \in I$ $\implies rs-r's'=rs-(r+\alpha)(s+\beta)= \underbrace{-r\beta}*{\in I}-\underbrace{\alpha s}*{\in I}-\underbrace{\alpha\beta}\_{\in I}\in I$ So, multiplication and addition are well-defined ### Zero Element $0+I=I$ $(0+I)+(r+I)=(0+r)+I=r+ I$ $0+I$ is the zero coset $1+ I$ is the unity coset $(1+I)\cdot (r+I)=(1\cdot r)+I=r+I$ Exercise: Show ring axioms hold in $R/I$ There is a surjective ring homomorphism $\pi:R\to R/I\\ \\pi(r)=r+I$ Exercise: Show that $\pi$ is a surjective ring homomorphism and find $ker(\pi)$ (which should be $I$) Given $I\rhd R \to R/I={\ r+I: r\in R\ }$ is a quotient ring, $0\_{R/I}=I$, $1\_{R/I}=1+I$ **Recall:** $r+I=s+I\iff r-s\in I$ $\pi: R\to R/I \\ \\pi(r)=r+I$ $\pi$ is a surjective homomorphism where $ker(\pi)=I$ **Remark:** If $R$ is a finite $R$, then $\underbrace{|I| \mid |R|}\_{\text{Legrange theorem}}$and $|R/I| = \frac{|R|}{|I|}$ **Examples:** 1. $R=\mathbb{Z}, I=3\mathbb{Z}$ , $\bar x$ means coset of $x, x+I$ $R/I=\mathbb{Z}/3\mathbb{Z} = {\ \bar 0, \bar 1, \bar 2\ }$ is a complete set of distinct coset representations

\def\arraystretch{1.5} \begin{array}{c|c|c|c} \cdot & \bar 0 & \bar 1 & \bar 2 \ \hline \bar 0 & \bar 0 & \bar 0 & \bar 0 \ \hline \bar 1 & \bar 0 & \bar 1 & \bar 2 \ \hline \bar 2 & \bar 0 & \bar 2 & \bar 1 \end{array} \phantom{111111111111} \def\arraystretch{1.5} \begin{array}{c|c|c|c}

• & \bar 0 & \bar 1 & \bar 2 \ \hline \bar 0 & \bar 0 & \bar 1 & \bar 2 \ \hline \bar 1 & \bar 1 & \bar 2 & \bar 0\ \hline \bar 2 & \bar 2 & \bar 0 & \bar 1 \end{array}

1. $F\[x\]/\braket{\ f(x)\ }={\ \overline{ h(x)}: deg(h)\<deg(f)\ }$ Claim: this is a complete set of distinct cosets **Proof:** Fix a coset in $F\[x\]/\braket{\ f(x)\ }$ and pick any eleemnt in it, $g(x)$ By the division algorithm $g(x)=q(x)f(x)+r(x)$, where $r(x)=0$ or $deg(r)\<deg(f)$ Now, $\bar g = \overline{q\cdot f +r} = \underbrace{\overline{qf}}\_{=\ 0 \text{ b/c } qf\in \braket{f}} + \overline r \implies \bar g = \bar r$ WTS: They are distinct cosets Suppose $h_1,h_2$ have degree smaller than f If $\overline{h_1}=\overline{h_2}\implies h_1=h_2$ (Claim) But, $deg(h_1(x)-h_2(x)) \< deg(f)$ ← Exercise: prove this However, $h_1(x)-h_2(x)= q(x)f(x)\implies h_1-h_2=0$ Because $deg(LHS)\<deg(RHS)$ So, $h_1(x)=h_2(x)$ $\square$ **Remark:** In $F\[x\],\ deg(f(x)\pm g(x))\leq max{\ deg(f),\ deg(g)\ }$ $deg(f(x)\cdot g(x) ) \geq deg(f)+deg(g)$ Unless $f$ is $0$ or $g$ is $0$ 2a. $F=\mathbb{Z}\_2 = {\ 0, 1\ }$

\begin{rcases} \mathbb{Z}_2[x]/\braket{\ x^2+x+1\ }= {\ \bar 0, \bar 1,\bar x, \overline{x+1}\ } \ \mathbb{Z}_2[x]/\braket{\ x^2+1\ }= {\ \bar 0, \bar 1,\bar x, \overline{x+1}\ } \end{rcases}

Symbolically, they are the same, but cosets are different Exercise: write addition and multiplication tables for these rings. Conclude their additive groups are isomorphic to $\mathbb{Z}\_2 \times \mathbb{Z}\_2$. Conclude that $\mathbb{Z}\_2\[x\]/\braket{\ x^2+x+1\ }$ is a field and $\mathbb{Z}\_2\[x\]/\braket{\ x^2+1\ }$ is not a field How to determine the ring structure of $R/I$ ## 1st Isomorphism Theorem Suppose $f: R\to S$ is a ring homomorphism Then, $R/\underbrace{kerf}*{Ideal of R} \cong \underbrace{Imf}*{\text{Subring of } S}$

\boxed{\text{First Isomorphic thm: } R/kerf\cong Imf}

Proof: Define a homomorphism $g: R/ker(f)\to Im(f)$ $g(r+ker(f))= f(r)\in Im(f)$ First, show g is well defined If $\bar r = \bar s,$ then $g(\bar r ) = f(r)\text{ and } g(\bar s) = f(s)$ $r-s\in ker(f)$ So, $r=s+\alpha$ for $\alpha \in ker(f)$ $f(r)=f(s+\alpha)= f(s)+f(\alpha)=f(s)$ Second, show g is a ring homomorphism ← Exercise Next, show g is surjective

\begin{matrix} R& \rightarrow {R/ker(f)} \xrightarrow{\textasciitilde\text{g}} & Im(f) \subseteq S\ & & \ R & \xrightarrow{f} & Im(f) \subseteq S \end{matrix}

Let $s \in Im(f),$ then $\exists \ r\in R\ s=f(r)$ So $g(\bar r)= f(r) = s$ So $g\in Im(g)$ Finally, we show $g$ is injective. $ker(g) = {\ r+ker(f): g(r+ker(f))=0\ } \\ = {\ r+ker(f): f(r)=0 \ } \\ = {\ r+ker(f): r\in ker(f) \ } =\ {\ ker(f)\ }$ So, $g$ in injective $\square$ **Example:** 1. $\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}\_n$ $f:\mathbb{Z}\rightarrow \mathbb{Z}\_n,$ ← f surjective ring homomorphism $f(a)=a(mod\ n)$ st. $ker(f)=n\mathbb{Z}$ By 1st isomorphism theorem $\mathbb{Z}/\underbrace{ker(f)}*{n\mathbb{Z}} \cong \underbrace{Im(f)}*{\mathbb{Z}\_n}$ 1. $F\[x\]/\braket{\ x- \alpha \ } \cong F$ ← $\alpha \in F$ Find a surjective ring homomorphism $f:F\[x\] \rightarrow F$ S.t. $ker(f) = \braket{\ x-\alpha \ }$ Conclude by 1st isomorphism theorem that $F\[x\]/\braket{\ x-\alpha \ }$ is isomorphic to $F$ **Examples:** $F=\Bbb Q ,\ f(x)=x^2-2$ $\mathbb Q \[x\] / \braket{\ x^2-2\ }={\ \overline{a+bx}: a,b\in \mathbb Q\ }$ How to determine the structure of $R/I$ ? 1st Isomorphism Theorem $F: R\to S\\ R/ker(s)\cong Im(f)$ Claim: $\mathbb Q \[x\] / \braket{\ x^2-1 \ } \cong \mathbb Q \[\\sqrt 2\] \subseteq \mathbb{R}$ Lemma: Let $0\neq I \lhd F\[x\]$ Then, $I = \braket{\ f(x)\ }\iff f(x)\in I$ is of minimum degree $f: \Bbb Q \[x\] \to \Bbb Q \[\\sqrt 2\]$ $f$ is surjective and $ker(f) = \braket{\ x^2 -2\ } \overbrace\implies^{\text{1st Iso Thm}} \Bbb Q \[x\]/\braket{x^2 -2}\cong \Bbb Q \[\\sqrt 2\]$ $f(p(x))=p(\sqrt 2)$ $f(x^2 -2)= (\sqrt2)^2 -2=0$ **Example:** $\mathbb{R}\[x\] / \braket{\ x^2 +1\ }\cong \Bbb C$ $f: \mathbb{R}\[x\]/\braket{\ x^2 +1\ }\to \Bbb C$ $f(p(x))=p(i)$ Ex. $f(-1+2x+3x^2-\frac{1}{\sqrt 2}x^3 )= -1 +2i+3i^2-\frac{1}{\sqrt2 }i^3=-4+(2+\frac{1}{\sqrt2}) i$ $ker(f) =$ (use the lemma!) Exercise: Find ker(f) $\mathbb{R}\[x\]/\braket{\ x^2-1\ }\cong \mathbb{R}\oplus \mathbb{R}$ When is $R/I$ a field? 1. If $R$ is commutative, then so is $R/I$ (Exercise: Give an example of a homomorphism R st. $R/I$ is commutative) 1. When is every element of $R/I$ a unit? ### Maximal Ideal **Definition:** We say $I \rhd R$ is a maximal ideal if there is no proper ideal strictly containing $I$ Equivalently, $I\subseteq J \lhd R\implies I=J$ **Example:** F - Field $\implies {\ 0\ } \lhd f$ is maximal Suppose that ${\ 0\ } \subseteq I \lhd F$ If $\exists\ 0 \neq x \in J$, then $1 = x^{-1}\cdot x \in J \to J$ is improper It follows that ${\ 0\ }$ is maximal **Theorem:** If $R$ is a commutative ring and $I \rhd R$ $R/I$ is a field $\iff I$ is a maximal ideal $I$ is maximal if $\forall \ I \subseteq J \lhd R\implies I = J$ Note: Proof in lecture 18/19 will add later **Example:** $R=\mathbb{Z}$, every ideal takes the form $\underbrace{n\mathbb{Z}}*{n\neq 1}$, $n\in \mathbb{Z}*{\geq 0}$ If $n$ is not prime, then it has a proper factor $1\< m\<n$. Then, $n=k\cdot m,\ k\in \mathbb{Z}$ So $n\mathbb{Z}\subset m\mathbb{Z},$ (proper subset because $m$ is not divisible by $n$ $\implies n\mathbb{Z}$ is not maximal Equivalently, if $n\mathbb{Z} \lhd \mathbb{Z}$ is maximal then $n$ is prime Conversely, if $n=p$ is prime, then $\implies \mathbb{Z}/p\mathbb{Z}\cong \mathbb{Z}\_p \text{ is a field}$ $\implies \mathbb{Z}/p\mathbb{Z}\ \text{ is a field}$ $\implies p\mathbb{Z}\lhd \mathbb{Z} \text{ is maximal}$ In $\mathbb{Z},$ $n\mathbb{Z} \lhd \mathbb{Z} \text{ is maximal } \iff \mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}\_n \text{ is a field }\iff n\text{ is prime}$ **Example:** $R=\mathbb{Z}\_6$ Lattice of proper ideals

{\ 0, 3\ }=\begin{matrix} & \mathbb{Z}_6 & \ \phantom{111}\diagup& &\diagdown\phantom{111} \ \braket{\ 3\ } & & \braket{\ 2\ } \ \phantom{111}\diagdown& &\diagup \phantom{111}\ & {\ 0\ }& \end{matrix}= {\ 0, 2,4\ }

$\braket{\ 2\ }, \braket{\ 3\ }\lhd \mathbb{Z}\_6 \text{ are maximal }$ $\begin{rcases} \\mathbb{Z}\_6/\braket{\ 2\ }\cong \mathbb{Z}\_2 \\ \\mathbb{Z}\_6/\braket{\ 3\ }\cong \mathbb{Z}\_3 \\end{rcases} \text{ 1st Isomorphism Theorem}$ **Example:** $R=\mathbb{Z}\_2\oplus \mathbb{Z}\_2= {\ (0,0),\ (1,0),\ (0,1),\ (1,1)\ }$

\text{Lattice of add subgroups}\ \begin{matrix} & \mathbb{Z}_2\oplus \mathbb{Z}_2 & \ \phantom{111}\diagup& |&\diagdown\phantom{111} \ \braket{\ (1,0)\ } & \braket{\ (1,1)\ } & \braket{\ (0,1)\ } \ \phantom{111}\diagdown&| &\diagup \phantom{111}\ & {\ 0\ }& \end{matrix}

\text{Lattice of Proper Ideals}\ \begin{matrix} & \mathbb{Z}_2\oplus \mathbb{Z}_2 & \ \phantom{111}\diagup& &\diagdown\phantom{111} \ \braket{\ (1,0)\ } & & \braket{\ (0,1)\ } \ \phantom{111}\diagdown& &\diagup \phantom{111}\ & {\ 0\ }& \end{matrix}

**Example:** $R=\mathbb{Z}$

\text{Lattice of Proper Ideals}\

 $F\[x\],\ F$ is a field, is a PID. What are the maximal ideals? What are the fields obtained as quotient / homomorphic images of $F\[x\]$? $I\rhd F\[x\],$ then $I=\braket{\ f(x)\ }$ **Remark:** 1. $\braket{\ f(x)\ }= F\[x\]\\iff f(x) \text{ is a unit} \iff f(x)\text{ is a non-zero scalar}$ 1. $\braket{\ f(x)\ }= \braket{\ g(x)\ } \iff f(x)=u\cdot g(x)\text{ for } u\in F\[x\]^x \iff f(x)=\alpha \cdot g(x) \text{ for } \alpha \neq 0 \text{ scalar}$ 1. $\braket{\ f(x)\ } \subseteq \braket{\ g(x)\ }\iff g(x) \mid f(x),\\ \text{ namely }\exists\ q(x) \in F\[x\]:f(x)=q(x)\cdot g(x)$ #  Irreducibility **Definition:** We say that $f(x)\neq \text{ scalar } \in F\[x\]$ is irreducible if for any $g(x)\in F\[x\]$ such that $g(x)\mid f(x),$ either $g(x)=\alpha$ is scalar or $g(x)=\alpha \cdot f(x)$ for a scalar $\alpha \neq 0$ **Definiton:** $g(x) \text{ divides } f(x),\ g(x) \mid f(x) \text{ if } \exists \ q(x)\in F\[x\]\\ st. \ f(x)=g(x)\cdot q(x)$ We say $g(x)$ is a proper divisor of $f(x)$ if $g(x)\mid f(x)$ and $g(x)\neq \text{scalar and }g(x)\neq \alpha \cdot f(x),\ \alpha \in \mathbb{R}$ **Example:** $f(x)=x^2 -1 \in \mathbb{R}\[x\]$ * $\underbrace{(\frac{1}{2}x^2+\frac{1}{2})}\_{g}\underbrace{(-2)}\_h=\underbrace{x^2-1}\_f$ $g,h$ are not proper ideals * $x^2 -1=\underbrace{(x-1)}\_p\underbrace{(x-1)}\_q$ $p,\ q$ are proper divisors **Remark:** $g\mid f$ is a proper divisor $\iff 0\<\\text{deg(g)}\< \text{deg(f)}$ **Proof:** $\impliedby 0 \<\\text{deg(g)} \<\\text{deg(f)}\implies g\neq \alpha \cdot f\text{ as } (\alpha \cdot f)=\text{deg(f)}$ $\implies \text{deg(g)} > 0$ ← $g$ is a proper divisor If $\text{deg(g)=deg(f)},$ then$f(x)=g(x)\cdot q(x)\implies \text{deg(q(x))}=0 \implies q(x)=\alpha \in \mathbb{R}$ This is a contradiction since $g$ is a proper divisor $\square$ **Dictionary:** $I=\braket{\ f(x) \ }\lhd F\[x\],\ J=\braket{\ g(x)\ }\lhd F\[x\]$

\text{Dictionary} \ \def\arraystretch{1.7} \begin{array}{c | c} I\rhd R & F[x] \ \hline I=0 & f(x)=0 \ \hline I=R & f(x) \text{ is a non-zero scalar} \ \hline I\lhd R\text{ (proper) } & \text{ deg( f(x) ) > 0} \ \hline I\subseteq J & g(x) \mid f(x) \ \hline I= J & g(x) =\alpha \cdot f(x),\alpha \in \mathbb{R},\alpha \neq 0 \ \hline I\subset J & g(x) \mid f(x)\text{, but }g(x)\neq \alpha \cdot f(x) \ \hline I\subset J \lhd R \text{ (proper) } & g(x) \mid f(x),g(x)\neq \alpha \cdot f(x), g(x)\text{ not a scalar }\newline &\iff g(x) \text{ is a proper divisor of } f(x) \ \hline I\lhd R \text{ (maximal) } & f(x)\text{ has no proper divisors (irreducible)} \ \end{array}

**Corollary:** TFAE: $(0\neq f(x)\in F\[x\]$ 1. $f(x)$ is irreducible 1. $f(x)$ has no divisors $g(x)\mid f(x)$ of $0\<\\text{deg(g)}\<\\text{deg(f)}$ 1. $\braket{\ f(x)\ }\lhd F\[x\]$ is maximal 1. $F\[x\]/\braket{\ f(x)\ }$ is a field **Example:** Over any field $F$, any linear polynomial $f(x)=ax+b,\ a\neq 0$ is irreducible If $g(x)\mid f(x),\ \deg g \leq \deg f = 1$ so $\deg g = { 0, 1 }$, so it cannot be proper Also, $F\[x\]/\braket{\ ax+b\ }\cong F$ $1^{st}$ Isomorphism Theorem: $\phi: F\[x\]\\to F \\ \phi(f(x))= f(-\frac{b}{a})$ A surjective ring homomorphism with $\ker \phi = \braket{\ ax+b\ }= \braket{\ x- (-\frac{b}{a})}$ **Theorem:** Let $f(x)\in F\[x\], \alpha \in F,$ Then, $f(\alpha )=0\iff x-\alpha \mid f(x)$ **Proof:** $\impliedby x-\alpha \mid f(x), \text{ then } \exists\ q(x) \in F\[x\]\\ st. f(x)=q(x)(x-\alpha)$ So, $f(\alpha)=q(\alpha )(\alpha -\alpha ) =0$ so, $\alpha$ is a root of $f$ $\implies$ Suppose $f(\alpha ) = 0$, $\exists\ q,r\in F\[x\] \ st. \ f(x)=q(x)(x-\alpha )+r(x) \implies \deg(r)\<\\deg(x-\alpha )=1,$ so $r(x)=\beta$ a scalar Substitute $x=\alpha$ $0=f(\alpha)=q(\alpha)(\alpha -\alpha)+\beta \implies\beta = 0 \implies f(x)=q(x)(x-\alpha)$ Thus, $(x-\alpha ) \mid f(x)$ $\square$ **Example:** $f(x)=x^3-2z+1\in \mathbb{R}\[x\]$ $f(1)=0$ $x^3-2x+1=(x^2+x-1)(x-1)$ **Corollary:** If $f(x)$ is a polynomial of degree $> 1$ and $f(x)$ has a root in $F$ then $f(x)$ is not irreducible. Equivalently, if $f$ is irreducible with $\deg(f)>1,$ then $f$ has no roots. Warning! The converse is not true (in general) $f(x)=x^4 +3x^2+2$ has no roots over $\mathbb{R}$ but it is reducible. $f(x)=(x^2+1)(x^2-2)$ **Corollary:** Let $f(x)\in F\[x\]$ st. $\deg f=n$, then $f$ has at most $n$ roots **Proof:** Induction on $n$ Base Case: $n=1$ $f(x) = ax+b,\ a\neq 0$ $ax+b\iff x=-\frac{b}{a}$ so there is $1$ root Inductive step: $\deg f = m+1$ If $f$ has no roots, trivial Suppose $f(\alpha)=0,$ then $f(x)= q(x)(x-\alpha)$ $\deg q = m$ so $q(x)$ has at most $m$ roots. Let $\beta \in F$ be an arbitary root of $f$ $0= f(\beta)=\underbrace{q(\beta)}*{\in \ F} \underbrace{(\beta - \alpha )}*{\in\ F}$ so Either

\begin{cases} \ q(\beta)=0 \ \ \beta - \alpha = 0 \end{cases}

Thus, there are at most $m+1$ roots $\square$ **Lemma:** If $f(x)\in F\[x\] \ st.\ f(x)$ is non constant $\deg f(x)\leq 3$ then $f$ has a root $\iff f$ is not irreducible **Example:** $f(x)=x^2 +x+1\in \mathbb{R}\[x\]$ $deg(f)=2$ $\Delta = b^2 -4ac = -3 \< 0 \implies f$ has no roots $f$ has no roots in $\mathbb{R}$ but it has roots in $\Bbb C$ $F$ - Field $F\[x\] / \braket{\ f(x)\ }$ is a field $\iff f(x)$ is irreducible, $f(x)$ cannot be written as $f(x)=g(x)h(x)$ $x+1=(\frac{1}{2}x+\frac{1}{2})\cdot 2$ in $\mathbb{R}$ is not proper **Recall:** * Linear polynomials are always irreducible * If $f(x)$ of $\deg f > 1$ and $\exists \ \alpha \in F: f(\alpha) = 0\implies f(x) \text{ is reducible }$ * If $\deg f\in {\ 2, 3\ }$, then $f$ irreducible $\iff f$ has no roots in $F$. Not necessarily true for $\deg f \geq 4$ **Remark:** Being irreducible depends on the field. $x^2 +1$ is irreducible in $\mathbb{R}\[x\]$ but is reducible in $\mathbb{C}\[x\]$ **Examples:** 1. $x^2+x+1$ in $\mathbb{Z}\_2\[x\]$ $\deg f=2,\ f(0) = 1, f(2)=1$ so irreducible 1. $x^2 +x+1$ in $\mathbb{Z}\_3\[x\]$ $f(0)=1,\ f(1)=0\implies x\cdot 1 \mid f(x)$

\begin{array}{r} x+2\phantom{1111} \ x-1{\overline{\smash{\big)},x^2+x+1}} \ \underline{-(x^2-x)} \phantom{111}\ 2x+1\phantom{} \ \underline{-(2x-2)}\phantom{} \ 3\ 3_{mod\ 3}=0 \end{array}

So, $x^2 +x+1 = (x+2)^2$ 1. $x^2 +2x+1 \text{ in } \mathbb{Z}\_3$ ← Exercise 1. $f(x)=x^4+x^2 +x+1\text{ in } \mathbb{Z}\_3$ ← Exam Question? Has no roots but $\deg f = 4$ so may still be reducible ← $f=g\cdot h$ $\text{Possible Degrees}\\ \\begin{rcases} \\cancel{(1)(3)} \\ (2)(2) \\ \\cancel{(3)(1)} \\ \\end{rcases}\implies (1)(3),\ (3)(1)$ Impossible because f has no roots It remains to check if $f=g\cdot h$ with $\deg g,h=2$ $x^4+x^2+x+1=(a_1x^2+b_1x+c_1)(a_2x^2+b_2x+c_2)$ $= \underbrace{a_1a_2}\_{=\ 1}x^4+(\cdots)x^3+\cdots$ $= (x^2+b_1x+c_1)(x^2+b_2x+c_2)$ $=x^4+\underbrace{(b_1+b_2)}*{=\ 0}x^3+ \\underbrace{(c_2+b_1b_2+c_1)}*{=\ 1}x^2+ \\underbrace{(b_1c_2+c_1b_2)}*{=\ 1}x+ \\underbrace{c_1c_2}*{=\ 1}$ $\implies b_1=b_2,\ c_2=\frac{1}{c_1}$ $= (x^2+bx+c)(x^2-bx+\frac{1}{c})$ $= x^4+(c-b^2+\frac{1}{c})x^2+(\frac{b}{c}-cb)+1$ $$\begin{cases} c-b^2+\frac{1}{c}=1 \implies \\ \\frac{b}{c}-cb=1\implies \left( b = \cfrac{1}{\cfrac{1}{c}-c}\right),\ c\neq 0,1,2 \\end{cases}$$ So there is no solution. Thus, $f$ is irreducible 1. $x^4+1$ in $\Bbb Q\[x\]$ $f=g\cdot h\implies \text{possible degrees: (1)(3), (2)(2), (3)(1)}$ Check if $f$ has a root $x^2 +1\geq 1$, so $f$ has no real roots So, $\deg g,h =2$ $x^4+1=(x^2+ax+b)(x^2+cx+d)$ $= x^4 + \underbrace{(c+a)}*{=\ 0}x^3+\underbrace{(d+ad+b)}*{=\ 0}x^2 +\underbrace{(ad+bc)}*{=\ 0}x+\underbrace{bd}*{=\ 1}$

\begin{cases} \frac{1}{b}-a^2+b=0 \ a\cdot \frac{1}{b}-ba=0

\end{cases} \implies a=0\implies \frac{1}{b}=0\implies 1+b^2=0 \implies b=\frac{1}{b}\implies \frac{1}{b}-a^2+b=0\implies 2b-a^2=0\implies a^2=2b\implies a^2=-2 \text{ or } a^2 = 2 \implies \text{ Impossible in }\Bbb Q

Thus, $x^4 + 1$ is irreducible over $\Bbb Q$ However, $x^4+1$ is reducible over $\mathbb{R}$ $a=\sqrt2, b=1$, $(x^2+\sqrt 2x+1)(x^2-\sqrt 2x+1)=x^4+1$ **Note:** a polynomial has a linear factor $\iff$ it has a root Which polynomials are reducible over $\Bbb C?$ ## Fundamental Theorem of Algebra If $f(x)\in \Bbb C \[x\]$ and $f$ is non-constant, then $f$ has $f(x)$ has a complex root **Corollary:** 1. $f(x)\in \Bbb C\[x\] \text{ is irreducible } \iff f(x)\text{ is linear}$ **Proof:** $\impliedby \text{ always true}$ $\implies \text{By FTA, } \exists\ \alpha \in \Bbb C:f(\alpha ) = 0$ and we know an irreducible polynomial cannot have a root unless it is linear 1. If $f(x)\in \Bbb C\[x\], f$ is non-constant, $\deg f = n$, then: $f(x)=c(x\alpha_1)(x-\alpha_2)\cdots(x-\alpha_n)\text{ for some } c_1\alpha \_1, \cdots c_n\alpha_n\in \Bbb C$ **Example:** $f(x)=x^3-1 \in \mathbb{R}\[x\]$ has a real root but it doesn’t split into 3 linear factors $x^3-1=(x-1)(x^2+x+1)$ **Proof:** Induction on $n$ Base case, $n=1$: $f(x)=ax+b=\underbrace a\_ c(x-(-\frac{b}{\underbrace{c}\_\\alpha}))$ Inductive Step: $\deg f= n+1$ By FTA, $\exists$ root of f, call it $\alpha \_{n+1} \in \Bbb C$ So, $f(\alpha *{n+1})=0\text{ and } x-\alpha*{n+1} \mid f(x)$ Thus, $\exists\ g(x)\in \Bbb C \[x\]\\ st.\ f(x)=g(x)(x-\alpha\_{n+1})$ Then, $\deg g = n$ and by the inductive hypothesis $g(x)=c(x-\alpha_1)\cdots(x-\alpha\_{n+1})$ because $f(x)=c(x-\alpha_1)\cdots(x-\alpha_n)(x-\alpha\_{n+1})$ **Corollary:** A polynomial over $\Bbb C\[x\]$ is irreducible $\iff$ it is linear **Theorem:** Let $f(x)\in \mathbb{R}\[x\]$ and $f$ is non-constant, Then, $f(x)=c(x-\beta_1)\cdots (x-\beta_x)\underbrace{q_1(x)\cdots q_m(x)}\_{\text{polynomials of deg 2}}$ So, irreducible polynomials over $\mathbb{R}$ have $\deg \in {\ 1,2 \ }$ Reminder: For $z\in \Bbb C$, if $z=x+iy,$ then $\bar z=x-iy$ and $\overline{z_1+z_2}= \bar z_1+\bar z_2,\ \overline{z_1z_2}=\bar z_1\bar z_2$ $z=\bar z \iff z=x$ **Proof:** Let $f(x)=a_nx^n+\cdots +a_1x+a_0\in \mathbb{R}\[x\],\ n\geq 1$ In particular, $f(x)\in \Bbb C \[x\]$ By corollary of FTA, $f(x)=c(x-\alpha_1)\cdots(x-\alpha_n)$ $\alpha_1,\cdots,\alpha_n\in \Bbb C, C\in \mathbb{R}$ for any $\alpha_i \in \Bbb C$ $f(\alpha_i)=0=a_n\alpha_i^n+\cdots + a_i\alpha \_i+a_0$ Applying Conjugation, $0=\bar 0= \overline{a_n\alpha_i^n+\cdots +a_i\alpha_i +a_0}\\ =\bar{a_n}\bar \alpha \_i^n+\cdots +\bar a_1\bar \alpha \_i+\bar a_0 \\ =a_n\bar \alpha_i^n+\cdots +a_1\bar \alpha_i+a_0=f(\overline \alpha_i)$ So, if $f(\alpha_i)=0,$ then $f(\bar \alpha_i)=0$ We know $f(x)=a_n(x-\alpha_1)\cdots(x-\alpha_n)\in \Bbb C$ $\forall \ 1\leq i\leq n,$ $\exists\ j\ st.\ \alpha_j=\overline \alpha_j$ Either $i=j \implies a_i\in \mathbb{R}$ or $i\neq j\implies \alpha_i =\overline \alpha \_j$ $f(x)=a_n\underbrace{(x-\beta_1)\cdots(x-\beta_k)}*{\in\ \mathbb{R}}\underbrace{(x-\overline \gamma_1)\cdots(x-\gamma_m)(x-\overline \gamma_m)}*{\in \ \Bbb C}$ For every $1\leq i \leq m$, $(x-\gamma_i)(x-\overline\gamma_i)=x^2-\underbrace{(\gamma_i+\overline\gamma_i)}*{\in\ R}+\underbrace{\gamma_i\overline \gamma_i}*{\in\ \mathbb{R}}=q_i(x) \square$ **Example:** $f(x)=x^4-x^2+1\in \mathbb{R}\[x\]$ Let $t=x^2$ $t^2-t+1\implies t=\frac{1}{2}\pm \frac{\sqrt 3}{2}i$ Over $\Bbb C,$ $f(x)=(x^2-(\frac{1}{2}+\frac{\sqrt3}{2}i))(x^2-(\frac{1}{2}-\frac{\sqrt3}{2}i))$ $\sqrt{\frac{1}{2}+\frac{\sqrt3}{2}i}$ ← Convert to polar using $r e^{i\theta}\ r=1, \ \theta =\frac{\pi}{3}$ $=\pm e^{\frac{\pi}{6}i}$ $\sqrt{\frac{1}{2}-\frac{\sqrt3}{2}}=\pm e^{-\frac{\pi}{6}i}$ $$f(x)=(x-e^{\frac{\pi}{6}i})(x+e^{\frac{\pi}{6}i})(x-e^{-\frac{\pi}{6}i})(x+e^{-\frac{\pi}{6}i})\\ =(x-e^{\frac{\pi}{6}i})(x-e^{-\frac{\pi}{6}i})(x+e^{\frac{\pi}{6}i})(x+e^{-\frac{\pi}{6}i})\\ =(x^2-(e^{\frac{\pi}{6}i}+e^{-\frac{\pi}{6}i})x+1)(x^2+(e^{\frac{\pi}{6}i}+e^{-\frac{\pi}{6}i})x+1)\\ =(x^2-2\cos\frac{\pi}6x+1)(x^2+2\cos\frac{\pi}6x+1)\\ =(x^2-\sqrt 3x+1)(x^2+\sqrt 3x+1)$$ # Field Extensions Over $\Bbb C$, $p(x)=c(x-\alpha_1)\cdots(x-\alpha_n)$ Definition: A field extension is $\phi:F\to K$ where $\phi$ is an injective ring homomorphism and $F\subseteq K$ Exercise: If F is a field and $\phi$ is any ring homomorphism, then $\phi$ is injective **Examples:** 1. $\Bbb Q\subseteq \mathbb{R}$ 1. $\mathbb{R}\subseteq \Bbb C$ 1. $\Bbb Q \subseteq \Bbb Q\[\\sqrt 2\]$ 1. If $F$ is a field and $f(x)$ is an irreducible polynomial then $\underbrace{F\[x\]/f(x)}\_{\text{field}} = {\ \overline{h(x)}: \deg h \< \deg f\ }$ $\phi: F \to F\[x\]/\braket{\ f(x)\ } \\ \\phi(a)=\overline a = a+\braket{\ f(x)\ }$ $\phi$ is an injective ring homomorphism 1. $F=\Bbb Q,\ f(x)=x^2-2$ $\phi:\Bbb Q\to \underbrace{\Bbb Q\[x\]/\braket{\ x^2-2\ }}\_k={\ \overline{a+bx}:\ a,b\in \Bbb Q\ }$ $\phi(a)=\bar a$ Consider $f(\lambda)=\lambda ^2-2$ $f(\lambda)\in \Bbb Q\[\\lambda\],f(\lambda)$ is irreducible But, $f(\lambda)\in K\[\\lambda\]$ irreducible and has a root $\Bbb Q\[x\]/\braket{\ x^2-2\ }=K\cong\Bbb Q\[\\sqrt 2\]$ $\psi:Q\[x\]\\to \Bbb Q\[\\sqrt 2\]$ $\psi(p(x))=p(\sqrt2)$ $(\overline x)^2-\overline 2= \overline x\cdot \overline x-\overline 2 = \overline{x^2-2} = \overline 0$ So, if $f(\lambda)$ is irreducible then in the field $K=F\[x\]/\braket{\ f(x)\ }$ $,\ f(\lambda)$ has a root $\overline x$ $\lambda ^2 -2$ factorization over $K$ $\lambda ^2 -2=(\lambda-\overline x)(\lambda +\overline x) = (\lambda-\sqrt 2)(\lambda +\sqrt x)$ 1. $\Bbb Q\[ \sqrt\[3\]{2}\]={\ a+b\sqrt\[3\]{2} +c\sqrt\[3\]{4}:a,b,c\in \Bbb Q\ }\subseteq \mathbb{R}$ is a field $\Bbb Q\[\\sqrt\[3\]{2}\]\\cong \Bbb Q\[x\]/\braket{\ x^3-2\ }$ **Proof:** Using the $1^{\text{st}}$ Isomorphism Theorem, $\phi:\Bbb Q\[x\]\\to \Bbb Q\[\\sqrt\[3\]{2}\]\\ \\phi(p(x))=p(\sqrt\[3\]{2})$ $x^3-2\in \ker \phi$ $\braket{\ x^3-2\ }\subseteq \ker \phi$ Claim: they are equal Notice that $\sqrt\[3\]{2} \notin \Bbb Q$ so $x^3-2$ has no roots in $\Bbb Q$. It is of $\deg 3$, thus it is irreducible Consequently, $\braket{\ x^3-2\ }$ is maximal Hence, $\braket{\ x^3-2\ } =\ker \phi$ So, $\Bbb Q\[x\]/\braket{\ x^3-2\ } =\Bbb Q\[x\]/\ker \phi \cong Im \phi = \Bbb Q\[\\sqrt\[3\]{2}\]$ $f(\lambda)=\lambda ^3 -2$ irreducible over $\Bbb Q$ but it is reducible over $K$ because it has a root $\implies \lambda - \sqrt\[3\]{2}\mid \lambda ^2-2$ over $K$

\begin{array}{r} \lambda^2+\sqrt[3]2\lambda\phantom{11111111111} \ \lambda-\sqrt[3]2\phantom{1}{\overline{\smash{\big)},\lambda^3-2\phantom{11111111111111}}} \ \underline{-(\lambda^3-\sqrt[3]2\lambda^2)} \phantom{1111111111}\ \sqrt[3]2\lambda^2-2\phantom{1111-1} \ \underline{-(\sqrt[3]2\lambda^2-\sqrt[3]4\lambda)}\phantom{1111} \ \sqrt[3]4\lambda-2\phantom{-11}\ \underline{\sqrt[3]4\lambda-\sqrt[3]4 \sqrt[3]4\phantom{}}\ 0

\end{array}

We have a factorization over $K$ $\lambda ^3-2=(\lambda -\sqrt\[3\]{2})(\lambda^2+\sqrt\[3\]{2}\lambda +\sqrt\[3\]4)$ $\Bbb Q\to \Bbb Q\[x\]/\braket{\ x^3-2\ } \cong K\to L=K\[y\]/\underbrace{\braket{\ y^2+\sqrt\[3\]2+\sqrt\[3\]4\ }}\_{\text{irreducible over }K}$ $(\lambda -\sqrt\[3\]2)(\lambda^2+\sqrt\[3\]2+\sqrt\[3\]4)$ **Theorem:** Let $f$ be a polynomial over $F$, then $\exists$ field $F\to K$ st. over $K,\ f(\lambda)=c(\lambda-\alpha_1)\cdots (\lambda-\alpha_n)$ splits In $L,\ \lambda ^3 -2$ splits --- Field Extension: $\phi \to K,$ $K$ extends F $K=F\to K \\ \phantom{00111}\phi \mapsto \overline \phi$ Recall from Linear Algebra: $F-\text{ Field}$ $V-\text{ Vector Space}$: abelian group with scalar multiples $\underbrace{\alpha}*{\in F} \cdot \underbrace{v}*{\in V} \in V$ An important case: If $F\to K$ is a field extension, then $K$ becomes an $F-\text{Vector Space}$ Given, $\alpha \in F,\ v\in K$ define scalar multiple by $\alpha \cdot v := \phi(\underbrace{\alpha}*{\in\ K})\cdot \underbrace{v}*{\in\ K}\in K$ **Examples:** 1. $F=\mathbb{R},\ k=\Bbb C,\ \mathbb{R}\subset \Bbb C$ is a field extension $\Bbb C$ is a vector space over $\Bbb R,$ $\Bbb C$ has vectors while $\mathbb{R}$ has scalars $1+i\in \Bbb C,\ -\frac{1}{2} \subset \mathbb{R}$ $(\frac{1}2 )(1+i)= -\frac{1}2 -\frac{1}2 i \in \Bbb C$ 1. $F-\text{Field}$, $f(x)\in F\[x\]$ is irreducible $K=F\[x\]/\braket{\ f(x)\ }$ $\phi: F\to K\\ \phantom{111} \alpha \mapsto\alpha +\braket{\ f(x)\ }= \overline \alpha$ $\underbrace{\alpha}*{\in \ F} \cdot \underbrace{\overline{p(x)}}*{\in\ K} = \alpha p(x) \in K$ For instance: $F=\mathbb{Z}\_3,\ f(x)=x^2+1$ $K=\mathbb{Z}\_3\[x\]/\braket{\ x^2+1\ }={\ \bar 0, \bar 1,\bar 2, \bar x,\overline{x+1},\overline{x+2},\overline{2x},\overline{2x+1},\overline{2x+2}\ }$ $\underbrace{2}\_{\in\ \mathbb{Z}*3}\cdot \underbrace{\overline{2x+1}}*{4\equiv 1(\text{mod }3)}=\overline{2(2x+1)}=\overline{x+2}$ **Recall:** If $V$ is a vector space over a field $F$, then it has a basis $B$. ($B$ is linearly independent and it spans $V$). If $|B|\<\\infty,\ \dim_F V=|B|$ $B={\ e_1,\cdots ,e_n\ },$ then every $v\in V$ can be uniquely represented as $v=\alpha \_1e_1+\cdots +\alpha \_n e_n,$ $\alpha \_i\in F$ There is a bijective correspondence $V\longleftrightarrow F\\ v\longleftrightarrow (\alpha_1,\cdots, \alpha_n)$ **Examples:** 1. $\mathbb{R}\subset \Bbb C$ $B={\ 1,i\ }\implies \dim\_\\mathbb{R}\Bbb C=2$ $\Bbb C \xleftrightarrow{1:1}\mathbb{R}^2\\ a+bi\longleftrightarrow (a,b)$ 1. $F-\text{Finite Field, } V-\text{Vector Space over } F$ $\dim_F V=n$ $V\xleftrightarrow{1:1} F^n$ $|V|=|F^n|=|F|^n$ If $|F|=q,$ then $|F^n|=q^n=|V|$ ### Characteristics of a Field $F-\text{Field}$ $1=1_F$ $1,1+1,1+1+1,\cdots$ If all these elements are distinct, then $\text{char}(F)=0$ $\text{char}(\Bbb Q)=0$ $\text{char}(\Bbb R)=0$ $\mathbb{Z}\_3={\ 0,1,2\ }$ $\text{char} (F)=\min\ {\ r\in \Bbb N:1_F+\cdots+1_F=0_F }$ So, $\text{char}(\mathbb{Z}\_p)=p$ $\mathbb{Z}\_3\[x\]/\braket{\ x^2+1\ }=K$ $1_F=\overline 1,\ \overline 1,\ \overline 1+\overline 1=\overline 2,\ \overline 1+\overline 1+\overline 1=\overline 0$, $\text{char}(K)=3$ If $F$ is a field with $\text{char}(F)=p,\ p\neq 0,$ then it is an extension of $\mathbb{Z}\_p$ Namely, there is a homomorphism $\phi: \mathbb{Z}*p\to F\\ \phi(0)=0_F,\phi(1)=1_F,\phi(2)=1_F+1_F, \\ \phi(p-1)=1_F+\underbrace{\cdots}*{p-1}+1_F$ Does not work unless $\text{char}(F)=p,$ $F=\mathbb{R},\ \phi:\mathbb{Z}\_3 \not\to \mathbb{R},\ \phi(0)=0,\ \phi(1)=1,\ \phi(2)=2$ $3=1+2=\phi(1)+\phi(2)=\phi(1+2)=\phi(3)=\phi(0)=0$ Suppose that $F$ is a finite field. $\text{char}(F)\neq 0$ $\underbrace{1_F+\cdots+1_F}*{m}=\underbrace{1_F+\cdots+1_F}*{m+r}\implies \underbrace{1_F+\cdots+1_F}\_{r}=0_F$ So finite fields have characteristic $>0$ and prime **Proposition:** F is a field, either $\text{char}(F)=0 or \text{char}(F)=p,\text{prime}$ **Proof:** Suppose $\text{char}(F)=r>0$, Assume to the contrary that $r$ is composite, $r=m_1\cdot m_2$ $(\underbrace{1_F+\cdots+1_F}*{m_1})\cdot(\underbrace{1_F+\cdots+1_F}*{m_2})= \underbrace{1_F+\cdots+1_F}\_{m_1\cdot m_2}=0_F$ Since $F$ is a field, it has no zero-divisors So either $\underbrace{1_F+\cdots + 1_F}*{m_1}=0_F$ or $\underbrace{1_F+\cdots + 1_F}*{m_2}=0_F$ Contradiction in either case becasue $m_1, m_2 \< r.\ \ \square$ **Summary:** 1. $F\text{ finite}\implies \text{char}(F)\neq 0$ 1. $\text{char}(F)\text{ is } 0 \text{ or }p, \text{ prime }$ 1. If $\text{char}(F)=p,\ \mathbb{Z}\_p\to F$ $F$ finite field $\implies \mathbb{Z}\_p\to F$ for $\text{char}(F)=p$ So, $F$ is a vector space over $\mathbb{Z}\_p$ Since $F$ is finite, $\dim\_{\mathbb{Z}\_p}F=n\<\\infty$ $F\xleftrightarrow{1:1}\mathbb{Z}\_p$ $|F|=|\mathbb{Z}\_p^n|=p^n$ **Corollary:** If $F$ is a finite field, then $|F|=p^n$ for $p$ prime **Reminder:** If F is a finite field, then $|F|=p^n$, $p=\text{ char} (F)$ (prime), $n\in N$ ( Idea: $\mathbb{Z}\_p\to F,\ F$ is a $\mathbb{Z}\_p-$ vector space so $|F|=|\mathbb{Z}\_p^n|=p^n$ $\text{char }(F)= \text{min}{\ r\mid 1_F\ \underbrace{+\cdots +}\_r\ 1_F=0_F}$ Goal: If $p$ - prime, $n\in \mathbb{N}$ then $\exists\ F$ - field, $|F|=p^n$ **Lemma:** If $F$is a field, $\text{char} (F)=p$ (>0) then, $\forall a,b\in F:$

\begin{cases} (a\cdot b)^{p^i}= a^{p^i}\cdot b^{p^i} \ (a+ b)^{p^i}= a^{p^i}+ b^{p^i} \end{cases}

In other words, $\phi:F\to F \\ \phi(a)=a^p$ is a ring homomorphism. (Frobenius endomorphism **Proof:** 1. EZ 1. $i=1: (a\cdot b)^{p^i}=^? a^{p^i}\cdot b^{p^i}$

(a\cdot b)^{p^i}= \sum_{k=0}^p \begin{pmatrix}p\k\end{pmatrix}a^kb^{p-k}

Claim: If $0\<k\<p$ then $\begin{pmatrix}p\\k\end{pmatrix}=0\_{mod\ p}$ (divisible by p) **Example:** Proof of claim: $\begin{pmatrix}p\\k\end{pmatrix}= \frac{p!}{k!(p-k)!},\ k=0,p\ : k,p-k\<p$, So, $p$ does not factor the denominator! (p is prime and $k!(p-k)!$ is a product of numbers $\< p$. But $p\mid p'$, so $p\mid \frac{p!}{k!(p-k)!}$ Proof of Lemma:

(a\cdot b)^{p}= \sum_{k=0}^p \begin{pmatrix}p\k\end{pmatrix}a^kb^{p-k}=\begin{pmatrix}p\0\end{pmatrix}a^0b^{p-0}+\begin{pmatrix}p\p\end{pmatrix} a^pb^{p-p}=a^p+b

More completely, we proved that $(a+b)^p=a^p+b^p+ \underbrace{p\cdot f(a,b)}\_{f(a,b)\cdot (1_F+\cdots + 1_F)}$ ← a polynomial in $a,b$ General $i$: induction, $i+1: (a+b)^{p^{i+1}}= ((a+b)^{p^i})^p= (a^{p^i}+b^{p^i})^p= A^p+B^p=(a^{p^i})^p+(b^{p^i})^p= a^{p^{i+1}}+b^{p^{i+1}}$ Proposition: Let $K$ be a field, $\text{char(K)}=p$ Let $n\geq 1$ then ${\ x\in K \mid x^{p^n}=x\ }\subseteq K$ is a subring. Moreover, it is a field **Exercise:** Prove that this is a subring Pick $x\in S, x\neq 0$ We know that $x^{p^n}= x\implies x^{p^n}-x=0\implies x\cdot (x^{p^n-1}-1)=0$ But K is a field $\implies$ no zero divisors s.t. $x\neq 0\implies p^{p^n-1}=1 \implies \underbrace{x^{p^n-2}}\_{\text{inverse of x}}\cdot x = x^{p^n-1}= 1\ \square$ Strategy: $p^n$ * $\mathbb{Z}\_p,\ f(x)=x^{p^n}-x$ * $\exists\ \mathbb{Z}*p \to \underbrace{K}*{\text{Field}}$ : $f(x) = (x-\alpha_n)\cdot (x-\alpha\_{p^n})$ (aka splitting field) * $S={\ x\in k\mid x^{p^n}=x\ }\subseteq K$. By the proposition, $S$ is a field. Aim, $|S|=p^n$ 1. $|S|\leq p^n$, notice that the elements of S are roots of $f(x)=x^{p^n}=x$ in the field $K$ Therefore, $f(x)$ has at most $\underbrace{p^n}\_{\text{deg(f)}}$ roots, so $|S|\leq p^n$ 1. $|S|\geq p^n$, $S={\ \alpha_n, \cdots , \alpha \_{p^n}\ }$ - roots of $f(x)=(x-\alpha_1)\cdots (x-\alpha \_n)$ in K, but we need to show that they are all distinct. **Definition:** (derivative)

\boxed {f(x)=a_nx^n+\cdots+a_nx+a_0\implies f’(x)=na_nx^{n-1} + \cdots+2a_2x+a_1}

**Example:** $F=\mathbb{Z}\_3$ $f(x)=2x^5+x^3+2x\implies f'(x)=\underbrace{10x^5}\_1+\underbrace{3x^2}\_0+2=\underline{x^5+2}$ ### Leibniz Rule $(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)$ This is still true over any field with the formal definition ### Multiple Factors and Derivatives If $g(x)^2 \mid f(x)$ then $g(x)\mid f'(x),\ (f,g\in F\[x\])$ Proof: Write $f(x)=g(x)^2\cdot h(x)$ for some $h\in F\[x\],$ so by Leibniz rule,

f’(x)=(g(x)^2\cdot h(x))’=\underbrace{(g(x)^2)}‘h(x)+g(x)^2\cdot h’(x) = g(x)(2g’(x)h(x)+g(x)h’(x)

So, $g(x) \mid f'(x)$ **Corollary:** If $(x-\alpha )^2 \mid f(x)$ then $x-\alpha \mid f'(x)$ $(((x-\alpha)^2)'=2x-2\alpha = 2(x-\alpha )$ We aimed to prove $f(x)= x^{p^n}-x= (x-\alpha_n)\cdots (x-\alpha \_{p^n})$ has $p^n$ distinct roots in K, that is, ${\ \alpha_n,\cdots , \alpha \_{p^n}\ }$ are distinct. If $\alpha \_i=\alpha \_j\ (i\neq j)$ then, $(x-\alpha \_i)^2 \mid f(x)$ By the corollary, $x-\alpha \_i \mid f'(x)$ $f(x)= x^{p^n}-x \implies f'(x)=\underbrace{p^n}\_{0 \text{ b/c char(K)=p }}x^{p^n-1}-1 = -1$ So, $x-\alpha_i \mid -1$ is impossible So, all roots are distinct, Consequently, $S={\ x\in K \mid x^{p^n}-x\ } \subseteq K$ is a field of size $p^n$ $1_F+\cdots+1_F=0_F$ $x+\cdots+x=x\cdot (1_F+\cdots+ 1_F) = x\cdot 0_F = 0_F$ ### Properties of Finite Fields Let $F$ be a field 1. $|F|=p^n,\ p=\text{char}(F),\ p \text{ prime}$ 1. $\forall\ p \text{ prime},\ n\geq 1, \exists\ \text{field} \ |F|=p^n$ 1. If $F$ and $K$ are fields $st.\ |F|=|K|=p,\ F\cong K$ 1. $|F^x|=p^n-1,$ abelian group 1. $F^x$ is a cyclic group $F^x\cong \mathbb{Z}\_{p^n-1}$ 1. $|F|=p^n\ \forall\ x\in F^x,\ x^{p^n-1}=1$ By Lagrange Theorem, $\text{ord}(x) \mid |F^x|$ so, $x^{p^n-1}= x^{|F^x|}=1$ 1. $\forall\ x\in F,\ x^{p^n}=x$ 1. $F\subseteq\ K, F$ is a subfield of $K$ $|F|=p^n,\ |K|=p^m,\text{ and } n\mid m$ $K$ is a vector space over $F$ There is no subfield of size $8$ over a field of size $16$ because $8=2^3,\ 16=2^4,\ 3 \not| \ 4$ 1. $|K|=p^m,\ n\mid m,$ then $F={\ x\in K:x^{p^n}=x\ }$ is a subfield of $K,\ |F|=p^n$ ### Fermat’s Little Theorem $\forall\ a\in \mathbb{Z},\ a^p=a(\text{mod } p), 7\implies \text{theorem}$ **Examples:** How many roots does $x^7-1$ have in $F:$ 1. $|F|=27\implies |F^x|=26$ $\text{ord}\_{F^x}(x)\in {\ 1,7\ }, \ 7\not |\ 26,$ so $x=1$, only $1$ root 1. $|F|=8\implies |F^x|=7$ $7\mid 7 ,$ so we have $7$ roots # Applications ## **Diffie–Hellman Protocol** $F- \text{Finite Field}$ $g\in F^x$ $n=\text{ord}\_{F^x}(g)$

\begin{matrix} \text{Alice} & \xrightleftharpoons[g^Y]{g^X} & \text{Bob} \ \ 1\leq X\leq n-1 && 1\leq Y\leq n-1

\end{matrix}

### Subgroup Attack For $a\in \mathbb{N}$

\begin{matrix} \text{Alice} & \xrightleftharpoons[(g^Y)^a]{g^X} &\text{Eve} &\xrightleftharpoons[g^Y]{(g^X)^a} & \text{Bob} \ \Downarrow&&&&\Downarrow\ K=g^{axy}&&&&K=g^{axy} \end{matrix}

Take $a\in \mathbb{N}:\ a\mid n,\ \frac{n}a$ is “small” Now, $K=K=g^{axy}=(g^a)^{xy}$ $\text{ord}*{F^x}(g^a)=\frac{\text{ord}*{F^x}(g)^n}{\gcd(\text{ord}\_{F^x}a)}= \frac{n}{\gcd(n,m)}=\frac{n}{a}$← small number Therefore, $|\braket{\ g^n\ }|= \frac{n}{a}$ and $K\in \braket{\ g^n\ }$ So, Eve can apply brute force to recover $K$ **Solution:** Take $g\in F^x\ st.\ \text{ord}(g)=n$ is prime **Examples:** $\mathbb{Z}*{73}=F\implies |F|^x=72,\ F^x\cong \mathbb{Z}*{72},$ $\ n=72,\ a=36,24,\cdots,\ \frac{n}a = 2,3,\cdots$ $F=\mathbb{Z}\_{83}$ $|F^x|=83-1=82= 2\cdot 41$ $g\in F^x$ where the order is $41$ ## Error-Correcting Codes Assume we are sending a message $s=(s_0,\cdots , s\_{n-1}) \in \mathbb{Z}\_2^n$ Possible Solution: Send the message multiple times This solution is highly inefficient because we are sending n bits of information multiple times. A more efficient solution:

\begin{matrix} s\in \mathbb{Z}_2^n &&s’+\text{error} \ \Downarrow && \Downarrow\ s’\in \mathbb{Z}_2^{n+r}&&s \end{matrix}

### Errors and Hamming Distance $x,y\in \mathbb{Z}\_2^n$ $d(x,y)=$ # of entries which are distinct on $x$ vs $y$ Example:

x=\begin{pmatrix} 1\1\0\1\0 \end{pmatrix},\ y=\begin{pmatrix} 1\0\0\1\1 \end{pmatrix}\implies d(x,y)=2

The hamming distance between $x$ and $y$ is 2 because they differ in two locations in the second position and the last position **Code:** $C\subseteq \mathbb{Z}\_2^n$, a subspace We say that $C$ s an $\[n,k\]$ - code $\implies k=\dim\_{\mathbb{Z}\_2}C$ $|C|=2^n$ **Encoding:**

\underbrace{\mathbb{Z}2^k}{\text{original message}}\to C\subseteq \underbrace{\mathbb{Z}2^n}{\text{encoded message}}

$M\in M\_{n,k}(\mathbb{Z}\_2)$ Encode: $s\mapsto M_s$ * If $\forall\ x\neq y\in C,\ d(x,y) \geq d,$ then the code can correct at least $\frac{d}2$ errors Cycle Code: $C\leq \mathbb{Z}\_2^n$ a code such that if some code in C, then any cyclic shift is also in C

\begin{pmatrix} c_0\\vdots\ c_{n-1} \end{pmatrix}\in C\implies\begin{pmatrix} c_i\c_{i+1}\\vdots\ c_{n-1}\c_0\ \vdots\ c_{i-1} \end{pmatrix}\in C

**Question:** How can we create cyclic codes? Consider the polynomial ring: $\mathbb{Z}*2\[x\]/\braket{\ x^n-1\ }={ \overline{c_0+c_1x \cdots +c*{n-1}x^{n-1}} \mid c_0,\cdots,c\_{n-1}\in \mathbb{Z}\_2} \longleftrightarrow \mathbb{Z}\_2^n$

\mathbb{Z}2[x]/\braket{\ x^n-1\ }={\ \overline{c_0+c_1x \cdots +c{n-1}x^{n-1}} \mid c_0,\cdots,c_{n-1}\in \mathbb{Z}_2\ } \longleftrightarrow \mathbb{Z}_2

\overline{c_0+c_1x+\cdots +c_{n-1}}\xleftrightarrow{1:1}\begin{pmatrix}c_0\ \vdots\ c_{n-1}\end{pmatrix}

**Theorem:** $C\leq \mathbb{Z}\_2^n$ is a cyclic code $\iff \braket{\ g(x)\ }\lhd \mathbb{Z}\_2\[x\]/\braket{\ x^n-1\ }$ We call $g(x)$ the generating polynomial of $C$ **Question:** How do we find, $g(x)\in \mathbb{Z}\_2\[x\]$ such that $C$ is a “good” code? Firstly, notice that the whole process can be formed when $\mathbb{Z}\_2$ is replaced by any field $K$ $\mathbb{Z}\_2\to K,\ |K|=2^r$ $K-\text{finite field}$ $\gamma = K^x$ $\implies \text{ord}\_{k^x}(\gamma)=q-1$ **Example:** $K=\mathbb{Z}\_2\[x\]/\braket{\ x^3+x+1\ }$ $\overline x \in K$ $\overline {x^{\phantom{1}}}, \overline {x^2}, \overline {x^3} = \overline{x+1}$ $\overline {x^4}=\overline{x^2+x},\ \overline {x^5}=\overline{x^2+x+1}$ $\overline {x ^6}=\overline{x^2+1},\ \overline {x^7}=\overline1$ ### **Bose–Chaudhuri–Hocquenghem (1960)** **Theorem:** Suppose that $K$ is a finite field, $|K|=q$ If $\gamma \in K^x$ of $\text{ord}*{K^x}(\gamma)=n,\ \gcd(\underbrace{n}*{\text{odd}},q)=1$$\text{ord}\_{K^x}(\gamma)=n,\ \gcd(n,q)=1$ $|K|=2^r$ $q(x)\in K\[x\]$ such that $g(\gamma)=g(\gamma^2)=\cdots=g(\gamma^{d-1})=0$ Then, $\braket{\ g(x)\ }\subseteq K\[x\]/\braket{\ x^n-1\ }$ is a cyclic code which corrects at least $\frac{d}2$ errors Note: For a code over $\mathbb{Z}\_2,$ replace each $\alpha \in K$ by a length $r$ vector over $\mathbb{Z}\_2$