Abstract Algebra 1

Math 103A

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Monoid

A monoid has two properties, $3x^2$

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

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
• $(\mathbb{Z},+)$ - Abelian Group
• $(\mathbb{R},\cdot )$ - Monoid
• $(M_n(\mathbb{R}),+)$ - Abelian Group
  • 2 & 0\\ 1 & -1 \\end{pmatrix} + \\begin{pmatrix} 3 & 1\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} 5 & 1\\ 1 & 0 \\end{pmatrix}$$
  • Identity Element is the 0 matrix
  • Inverse - make each element in matrix negative
• $(M_n(\mathbb{R}),\cdot )$ - Monoid
  • 2 & 0\\ 1 & -1 \\end{pmatrix} \cdot \\begin{pmatrix} 3 & 1\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} 6 & 2\\ 3 & 0 \\end{pmatrix}$$
  • Identity Element is the identity matrix
• $(f:\mathbb{R}⟹\mathbb{R},+)$ - Abelian Group
  • $(f+g)(x)=f(x)+g(x)$
• $(f:\mathbb{R}⟹\mathbb{R},\cdot )$ - Monoid
  • $(f\cdot g)(x)=f(x)g(x)$
  • Identity Element: $f(x)=1$
  • $f(x)=0$ is not invertible
  • Invertible elements have no roots (do not intersect x axis)
• $(f:\mathbb{Z}⟹\mathbb{R},\circ )$ - 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 $(x,y\in S⟹x\cdot y\in S)$
• $1\in S$

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)

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

• & \cdots & \cdots & * \ 0 & \ddots & \ddots & \vdots\ \vdots & \ddots & \ddots & \vdots\ 0 & \cdots & 0 & * \end{pmatrix} \subset M_n(\mathbb{R})$$ is a subring

2. B(\mathbb{R}) {f: \mathbb{R} \mathbb{R}ightarrow \mathbb{R} bounded continuous functions}= {}\subset C(\mathbb{R}) = {f: \mathbb{R} \mathbb{R}ightarrow \mathbb{R} continous 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 communative 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} \mathbb{R}ightarrow \mathbb{Z}\[x\] \mathbb{R}ightarrow (\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 which 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 \\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. $R=\mathbb{R},\mathbb{C},\mathbb{Q}$ $R^x=R\backslash 0$ \\mathbb{C}^x=\mathbb{C}\backslash{0} \\ \\mathbb{Q}^x=\mathbb{Q}\backslash{0} \\ $$
2. $R=\mathbb{Z}$ (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: $a+bi\mathbb{R}ightarrowa-bi$ 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 $R^x=R\backslash 0$

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:

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 $M_2(\mathbb{R}),$

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

Units are never zero divisors

Remark: In general, there might be elements which 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

Properties of Domains

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

Proof

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!