Rings and Modules

Chentian Wu

2024-11-08

\newcommand{\Z}{\mathbb Z} \newcommand{\R}{\mathbb R} \newcommand{\1}{\textbf 1} \newcommand{\0}{\textbf 0} \newcommand{\Hom}{\text{Hom}} \newcommand{\Aut}{\text{Aut}} \newcommand{\End}{\text{End}} \newcommand{\Iso}{\cong} \newcommand{\Im}{\text{Im}} \newcommand{\Id}{\text{id}} \newcommand{\GL}{\text{GL}} \newcommand{\End}{\text{End}} \newcommand{\Ker}{\text{ker}} \newcommand{\Norm}{\unlhd} \newcommand{\L}{\mathcal L}

1. Rings

Definiton. Ring
A ring is a triple $(R, +:R × R\to R, \times:R × R\to R)$ , such that:

$$(R, +)$$ is an abelian group.
Associativity. $\forall a,b,c\in R$ $(a\times b)\times c = a\times (b\times c)$
Distribution. $\forall a,b,c\in R$ $(a+b)\times c = (a\times c) +(b\times c) \\ a\times (b+c) = (a\times b)+(a\times c)$

We usually denote the identity of $$(R, +)$$ as $\0$ .

$a\times b$ can be written as $$ab$$

Definiton. Commutative ring
A ring $$R$$ is called a commutative ring if $\forall a,b\in R.a\times b=b\times a$

Example.

$C(\R)$ is the set of all continuous real-valued functions defined on $\R$ . $(C(\R), +, \times)$ is a commutative ring with $\begin{aligned} \0&:=x\mapsto 0\\ \1&:=x\mapsto 1 \end{aligned}$

Definiton. Multiplicative identity
If there is some $\1\in R$ , and $\1\neq \0$ , such that

\forall r \in R. \1\times r =r\times\1 = r

$$R$$ is said to have identity, and $\1$ is the multiplicative identity (or identity)

Some new textbooks require multiplicative identity as a must for a ring.

Definiton. Zero divisor and unit
If a ring $$R$$ has multiplicative identity, an element $$a$$ is called a

zero divisor: if there is a nonzero element $b\in R$ such that $$ab=0$$ or $$ba=0$$
unit: if there is a element $b\in R$ such that $$ab=ba=1$$

Lemma. Properties of rings
If $$R$$ is a ring, then the following properties are true

$\forall a\in R.a\0=\0a=\0$
$\forall a,b\in R. (-a)b=a(-b)=-(ab)$
$\forall a,b\in R.(-a)(-b)=ab$
If $$R$$ has an identity $\1$ , then the identity is unique and $-a=(-\1)a$

Definiton. Ring homomorphism
Let $$(R, +_R, \times_R)$$ and $$(S, +_S , \times_S )$$ be two rings. A map

f : R\to S

is called a ring homomorphism, if $\forall a,b\in R$

\begin{aligned} f (a+_R b) &= f (a) +_S f (b)\\ f (a\times_R b) &= f (a)\times_S f (b) \end{aligned}

The set of all homomorphisms from a ring $$R$$ to a ring $$S$$ is denoted by $\Hom(R,S)$

Just like a group homomorphism, a ring homomorphism retains the operation structure

Lemma. a composition of ring homomorphisms is still a ring homomorphism

Definiton. Ring isomorphism / isomorphic
If $f:R\to S$ is a bijective ring homomorphism, then we say $$f$$ is a ring isomorphism. If there’s a isomorphism between $$R$$ and $$S$$ , we say $$R$$ and $$S$$ are isomorphic, denoted by $R\Iso S$

It’s easy to prove that in this case $f^{-1}$ is also a ring isomorphism

Definiton. Ring automorphism group

The set of all bijective homomorphisms from a ring $$R$$ to itself is (obviously) a group, called the automorphism group of $$R$$ denoted by $\Aut(R)$

Note that $\Aut(R)\leq S_R$

Definiton. Kernel, image of ring homomorphism
$$R$$ and $$S$$ are two rings. $f\in\Hom(R,S)$ , then

\Ker (f):=f^{-1}(\{\0\})

called the kernel of $$f$$ , and

\Im(f):=f(R)

called the image of $$f$$ .

Definiton. Field
A ring $$R$$ is called a field if $(R\setminus \{\0\}, \times)$ is an abelian group.

Or we can say, a field is a commutative ring with identity where every non-zero element is a unit.

Definiton. Integral domain (整环)
A ring $$R$$ is called a integral domain if the following three conditions are satisfied

$$R$$ is commutative
$\1\in R$
$\forall a,b\in R.a\neq\0\land b\neq \0\implies a\times b\neq \0$

The concept of integral domain was proposed so that we can better research about divisibility theory with it. That is, in a integral domain, $$ab=ac$$ with $a\neq \0$ , then we have $$b=c$$

Corollary.

Any field is a integral domain where every non-zero element has a multiplicative inverse.
Any finite integral domain is a field.

Pf. Suppose $$R$$ is a finite integral domain. Note that if $a\neq \0$ , the map $f:x\mapsto ax$ should be injective because otherwise

a(x_1-x_2)=\0\implies a=\0

Since $$R$$ is finite, $f:R\to R$ should be a bijection, then $\exists b.ab=\1\in R$ . Since the choice of $$a$$ is arbitrary, we know that $(R\setminus \{0\}, \times)$ is an (abelian) group, so $$R$$ is a field.

Example

$C(\R)$ defined above is not an integral domain because consider $f:=x\mapsto \begin{cases}0 & x \leq 0 \\ x & x >0\end{cases}\\ g:=x\mapsto \begin{cases}x & x \leq 0 \\ 0 & x >0\end{cases}$ but $fg\neq \0:=x\mapsto 0\in C(\R)$ .

Definiton. Subring

A subring of ring $$R$$ is a subgroup of $$(R, +)$$ that is closed under multiplication. i.e. $S \subset R$ is called a subring, if $(S, +) \leq (R, +)$ , and

\forall a, b \in S.a\times b\in S

A subring should be closed under $$+$$ and $\times$

Lemma. If $$R$$ is a ring, $S\subset R$ , TTFE

$$S$$ is a subring
there exists corresponding $+:S^2\to S$ and $\times:S^2\to S$ such that the inclusion map from $$S$$ to $$R$$ is a ring homomorphism

Lemma. Intersections of subrings of a ring is still a subring of the ring.

Easy to prove using just definitions

Definiton. Endomorphism ring
Let $$(A, +)$$ be an abelian group. Define

\begin{aligned} +:\Hom(A, A)\times\Hom(A, A)&\to\Hom(A, A) \\ f + g &\mapsto (x\mapsto f(x) + g(x))\\ \times:\Hom(A, A)\times\Hom(A, A)&\to\Hom(A, A)\\ f\times g&\mapsto f\circ g \end{aligned}

Then $(\Hom(A, A), +, ×)$ is a ring with multiplicative identity, called the endomorphism ring, denoted as $\End(A)$ . $$A$$ is a left $\End(A)$ -module by

$$f \cdot a = f(a) $$

Definiton. Group ring
Let $$R$$ be a ring, $$G$$ a group, the group ring, denoted as $$R[G]$$ , consists of maps from $$G$$ to $$R$$ such that all but finitely many $g\in G$ is sent to $$0$$ . i.e.

R[G]:=\{f\in \text{Map}(G, R): |\{g\in G\mid f(g)\neq 0\}|< \infty\}

Then $(R[G], +, \times)$ is a ring with the following satisfied:

$$(a + b)(g) = a(g) + b(g)$$
$(a \times b)(g)=\displaystyle{\sum_{h\in G, a(h)\neq 0}a(h)b(h^{-1}g)}$

When $$R$$ is a field, a module of $$R[G]$$ is called an $$R$$ -representation.

If $$R$$ has identity, then $$R[G]$$ has identity

\1_{R[G]}=\begin{cases} 1 & a=e\\ 0 & a\neq e \end{cases}

Note, another easy way to understand group ring is as the Foote and Dummit textbook, every element $a\in R[G]$ can be written as $a=\sum_{i=1}^n a_ig_i$ where $$n$$ is the number of elements in $$G$$ , $a_i\in R$ , $g_i\in G$ . Then the addition and multiplication is just similar to polynomial additions and multiplications. Specifically, if $$G$$ is a cyclic group, it’s exactly the same as polynomial operations.

Definiton. Polynomial ring
$G=(\Z, +)$ , $$R$$ is a ring, then $R[\mathbb Z]$ can be identified with Laurent polynomials with coefficients in $$R$$ (denoted by $R[z, z^{-1}]$ ) via

a\mapsto \sum_{n\in \Z}a(n)z^n

The subring where $$a(n) = 0$$ for all $$n < 0$$ can be seen as the polynomial ring $$R[z]$$ .
It could be shown that if $$R$$ is a integral domain, then $$R[z]$$ is also an integral domain.

We can think of polynomial as subrings of a group ring, then we always use FFT while doing polynomial multiplications to cut down the complexity

Definiton. Matrix ring
Fix an arbitrary ring $$R$$ and $$n$$ be a positive number. Let $$M_n(R)$$ be the set of all $n\times n$ matrices with entries from $$R$$ . Then $$M_n(R)$$ becomes a ring with operations defined as

\begin{aligned} +&:(X,Y)\mapsto (i\mapsto j\mapsto (X_{ij}+Y_{ij}))\\ \times&:(X,Y)\mapsto (i\mapsto j\mapsto \sum_{k=1}^nX_{ik}\times Y_{kj}) \end{aligned}

The additive identity is $i\mapsto j\mapsto \0$ , and the multiplicative identity is $i\mapsto j\mapsto (i\textsf{ == }j)\textsf{ ? }1:0$

Definiton. Field of fractions
If $$R$$ is an integral domain, we call

Q = R × (R\setminus \{\0\})/\sim

the field of fractions of $$R$$ , where $\forall a,b,c,d\in R$

(a, b) \sim (c, d) \implies ac=bd

with definitions of operators

$+_Q: Q \times Q\to Q$ be defined as $([(a, b)], [(c, d)]) \mapsto [(ad + bc, bd)]$
$×_Q : Q \times Q \to Q$ be $([(a, b)], [(c, d)]) \mapsto [(ac, bd)]$

then one can show that $$Q$$ is a field (remember to check $$+_Q$$ , $$\times_Q$$ being well defined first), called the fraction field of $$R$$ .

Remark.
If $$R$$ is a field, then every non-zero element of $$R$$ has a multiplicative inverse. It could then be shown that the map

i:r\mapsto [(r,\1)]

is an bijection from $$R$$ to $$Q$$ , thus $R\Iso Q$

Definiton. Opposite ring
Let $(R, +, \times)$ be a ring, then $(R, +, (a, b)\mapsto(b\times a))$ is called the opposite ring of $$R$$ .

similar to the definition of opposite group

Definiton. Direct product of rings
Let $\{R_i\}_{\alpha\in\Lambda}$ be a sequence of rings, then $R=\prod_{\alpha\in\Lambda}{R_\alpha}$ , called the direct product of this sequence of rings, is a ring with under componentwise addition and multiplication, that is

\begin{aligned} r+s&=\alpha\mapsto (r_\alpha +_\alpha s_\alpha)\\ r\times s&=\alpha\mapsto (r_\alpha \times_\alpha s_\alpha) \end{aligned}

The isomorphism theorem in rings and in modules are super similar to that in groups. To show this theorem, we’ll introduce the concept of ideals.

2. Ideals

Definiton. Left ideal and ideal
Let $$R$$ be a ring. A subring $I\subset R$ satisfying the condition that

\forall a\in I.\forall r\in R. ra\in I

is called a left ideal. If furthermore we have $ar\in I$ , we call it an ideal.

Ideal is like the ring version of normal subgroup

Definiton. Equivalent but (seems) weaker definition of ideals
Let $$R$$ be ring, $I\subset R$ is called an ideal if

$(I,+)\leq (R,+)$
$\forall a\in A, r\in R, ra\in A,ar\in A$

Lemma. ideals of a ring
Let $$I$$ and $$J$$ be two ideals of a ring $$R$$ , then

$I\cap J$ is an ideal of $$R$$
If $I\cup J$ is an ideal of $$R$$ then either $I\subset J$ or $J\subset I$ (proof). If $I\subset J$ , we say $$J$$ divides $$I$$ , denoted by $J\mid I$ (note the order)

Definition. Ideal generated by a set
Let $$A$$ be any subset of the ring $$R$$ , $$(A)$$ denotes the smallest ideal of $$R$$ containing $$A$$ called the ideal generated by $$A$$ . Since the intersection of a non-empty collection of ideals of $$R$$ is still an ideal (by the lemma above) and $$A$$ is always contained in at least one ideal, we have

(A)=\bigcap\{I\mid I\text{ is an ideal}, A\subset I\}

Definition. Sums of ideals
If $$I$$ and $$J$$ are ideals of a same ring $$R$$ , $I+J:=\{i+j\mid i\in I, j\in J\}$ is called the sum of $$I$$ and $$J$$ . It could be shown that $$I+J$$ is an ideal of $$R$$ , and is the smallest ideal containing $I\cup J$ .

Definition. Product of ideals
If $$I$$ and $$J$$ are ideals of a same ring $$R$$ , $$IJ:=$$ all finite sums of elements of the form $$ij$$ where $i\in I$ and $j\in J$ is called the product of $$I$$ and $$J$$ . It could be proved that $IJ\subset I\cap J$ .

! Note that $IJ \neq P:=\{ij\mid i\in I,j\in J\}$ . It could be proved that generally $$P$$ is not necessarily an ideal of $$R$$ . Consider $R=\Z[x]$ , $$I=(x^2, 2)$$ , $$J=(x^3, 2)$$ , then we know $x^5+4\not\in P$ but $x^5\in P$ and $4\in P$ , so $$P$$ is not even a subring of $$R$$ .
In fact, $$IJ$$ is the smallest ring containing $$P$$ .

Lemma. If $$R$$ is a commutative ring with $$1$$ , and $$I+J=R$$ , then $IJ=I\cap J$
Pf. let $$a+b=1$$ , for any $x\in I\cap J$ , $x(a+b)=xa+xb=ax+xb\in IJ$ . so $I\cap J\subset IJ$ . qed.

Lemma. The number of ideals of a ring is restricted by its expressive power
If $$R$$ is a ring and $$I$$ is an ideal of $$R$$ , then the following are true

$$I=R$$ iff $$I$$ contains a unit
Assume $$R$$ is commutative, then $$R$$ is a field iff its only ideals are $\0$ and $$R$$ .

Definiton. Maximal ideal
if $$I$$ is an ideal of $$R$$ , and $$R/I$$ is a field, we call $$I$$ an maximal ideal

Note. Another definition of maximal ideal
A more common definition of a maximal ideal $$I$$ of $$R$$ is that,

$$I$$ is an ideal of $$R$$ , and
If $$J$$ is an ideal of $$R$$ and $I\subset J$ , then either $$J=I$$ or $$J=R$$ .
This is way $$I$$ is called maximal. The proof that these two definitions are equivalent could be found here

Definiton. Prime ideal
if $$I$$ is an ideal of $$R$$ , and $$R/I$$ is a integral domain, we call $$I$$ a prime ideal

By definition a maximal ideal is prime

Lemma. Another equivalent definition of prime ideal (link)
Let $$P$$ be a proper ideal of a ring $$R$$ . We say $$P$$ is a prime ideal of $$R$$ if $P\mid IJ\implies P\mid I\lor P\mid J$ , where $$I$$ and $$J$$ are ideals of $$R$$ .

Example.

The singleton $\{0\}$ is a prime ideal of $\Z$ but not a maximal ideal because $\Z/\{0\}$ is not a field.
The set $3\Z$ is a prime ideal of $\Z$ and a maximal ideal of $\Z$

Theorem. Prime ideal test
In general $n\in \Z$ , $n\Z$ is a prime ideal iff $$n$$ is a prime number.

Theorem. First isomorphism theorem in rings
If $$R$$ is a ring, $S\subset R$ is a subring, then TTFE

$$S$$ is a ideal of $$R$$
$(R/S, +, \times)$ is a ring with operations defined as

\begin{aligned} (a+S)+(b+S)&=(a+b)+S\\ (a+S)\times (b+S)&=(a\times b)+S \end{aligned}

There is a ring homomorphism $f\in\Hom(R, R')$ , s.t.

\Ker(f)=S

This is almost the same as that in the version of groups

Definiton. Quotient ring
If $$I$$ is an ideal of a ring $(R, +, \times)$ , then the set of left cosets $R/I:=\{r+I\mid r\in R\}$ is a ring, called the quotient ring of $$R$$ by $$I$$ . The ring operations are defined as

\begin{aligned} (r+I)+(s+I)&=(r+s)+I\\ (r+I)\times (s+I)&=(r\times s)+I \end{aligned}

Remark.
the subring $(\Z/6\Z, +, \times)$ is not an integral domain because $(3+6Z)\times (4+6\Z)=0+6\Z$

3. Modules

Definiton. $$R$$ -module
Let $$(R, +, \times)$$ be a ring. A (left) $$R$$ -module is a triple

(M, + : M × M\to M, · : R × M\to M )

such that:

$$(M, +)$$ is an abelian group
(Associativity) $\forall a, b\in R, x \in M$ ,

$$(a \times b) \cdot x = a \cdot (b \cdot x) $$

(Distribution) $\forall a, b\in R, x, y\in M$ ,

\begin{aligned} (a + b) · x &= a · x + b · x\\ a · (x + y) &= a · x + a · y \end{aligned}

Lemma. Properties of modules
If $$R$$ is a ring and $$M$$ is a $$R$$ -module. For any $a, b\in R, x\in M$ ,

\begin{aligned} \0_R\times a &=a\times \0_R=\0_R\\ \0_R\cdot x&=a\cdot \0_M=\0_M\\ (−a) × b &= a × (−b) = −(a × b)\\ (−a) · x &= a · (−x) = −(a · x)\\ (−a) × (−b) &= a × b\\ (−a) · (−x) &= a · x \end{aligned}

Definiton. Vector space
if $$V$$ is a $$F$$ -module where $$F$$ is a field such that $\forall v\in V.\1 v=v$ , then we call $$V$$ a vector space over $$F$$ .

Definiton. Submodule
A subset $N \subset M$ is called a submodule, if $$(N, +) \leq (M, +)$$ , and $$N$$ is indeed a module of $$R$$ , i.e.

\forall a \in R, x\in N , a · x \in N

Definiton. $$R$$ -module homomorphism
Let $$(M, +_M , \cdot_M)$$ and $$(N, +_N , \cdot_N )$$ be two $$R$$ -modules. A map $f: M\to N$ is called a $$R$$ -module homomorphism if $\forall x,y\in M, a\in R$

\begin{aligned} f(x +_M y) &= f(x) +_N f (y)\\ f(ax) &= a f (x) \end{aligned}

The set of all $$R$$ -module homomorphisms between $$M$$ and $$N$$ is denoted as $\Hom_R(M, N)$ .

Lemma. Intersections of submodules of a $$R$$ -module is still a submodule of that $$R$$ -module.

Lemma. Composition of module homomorphisms is still a module homomorphism

Lemma. homomorphisms of a commutative ring is also a module
Let $$R$$ be a commutative ring, and $$M,N$$ are $$R$$ -modules. The set $\Hom_R(M,N)$ of $$R$$ -homomorphisms from $$M$$ to $$N$$ is also a $$R$$ -module with operations defined as

\begin{aligned} \phi+\psi&:=x\mapsto \phi(x)+\psi(x)\\ r\cdot \phi&:=x\mapsto r\cdot(\phi(x)) \end{aligned}

A direct application of this lemma is, the set of all linear transformations from two vector spaces $$V$$ and $$W$$ over a field $$F$$ , denoted by $\L(V,W)$ , is also a vector space over $$F$$ , with operations defined above.

Lemma. Submodule criterion
If $$R$$ is a ring and $$M$$ is a $$R$$ -module. A subset $N\subset M$ is a submodule of $$M$$ iff. there exists corresponding $+_M : M ×M \to M$ and $·_M : R×M \times M$ such that $$(M, +_M , \cdot_M )$$ is a left $$R$$ -module and the inclusion map is an $$R$$ -module homomorphism.

Lemma. Another submodule criterion
If $$R$$ is a ring and $$M$$ is a $$R$$ -module. A subset $N\subset M$ is a submodule of $$M$$ iff. $N\neq \emptyset$ and

\forall x,y\in N.\forall r\in R.x+ry\in N

Pf. If $$N$$ is a submodule, then $0\in N$ so $N\neq\emptyset$ . Since $$N$$ is closed under addition and scalar multiplication, so $x+ry\in N$ also holds. If these two conditions hold. Let $$r=-1$$ we know $$N$$ is a (normal) subgroup (under addition) of $$M$$ , and let $$x=0$$ we then know $$N$$ is closed under scalar multiplication. So $$N$$ is a submodule of $$M$$ .

Definiton. Sum of submodules
Let $$A$$ and $$B$$ be submodules of the $$R$$ -module $$M$$ , the sum of $$A$$ and $$B$$ is the set

A+B:=\{a+b\mid a\in A,b\in B\}

It could be checked that $$A+B$$ is still a submodule of $$M$$ and is the smallest submodule containing both $$A$$ and $$B$$

Definiton. Kernel and image
$$M$$ and $$N$$ are two $$R$$ -modules where $$R$$ is a ring. $f\in\Hom_R(M,N)$ , then

\Ker (f):=f^{-1}(\{\0\})

called the kernel of $$f$$ , and

\Im(f):=f(M)

called the image of $$f$$ .

Definiton. $$R$$ -module isomorphism
A bijective $$R$$ -module homomorphism is called a $$R$$ -module isomorphism. If there exists a $$R$$ -module isomorphism between two $$R$$ -module $$M$$ and $$N$$ , then we say $$M$$ and $$N$$ are isomorphic,
denoted by $M\Iso N$

Definiton. $$R$$ -module automorphism group
Let $$M$$ be a left $$R$$ -module. The set of all bijections from $$M$$ to itself forms a subgroup of $$S_M$$ , called the $$R$$ -module automorphism group of $$M$$ , denoted by $\Aut_R(M)$

Example. $\Aut(\R)\neq \Aut_\R(\R)$

See $\R$ is a left $\R$ -module (1-dimensional vector space), then by definition of module automorphism, we have

\Aut_\R(\R)=\{x\mapsto ax\mid a\neq 0\}

See $\R$ simply as a ring, by definition of ring automorphism, we have

\begin{aligned} f(x)=f(1\cdot x)=f(1)f(x)&\implies f(1)=1\\ f(n)=nf(1)&\implies f(n)=n\\ a\geq 0&\implies f(a)=f((\sqrt a)^2)\geq 0\\ a\geq b\implies a-b\geq 0&\implies f(a)\geq f(b) \end{aligned}

The last property we can prove $$f$$ is continuous on $\R$ by definition, and further we can prove the only automorphism on $\R$ is $\Id_\R$

Example. $\Aut_\R(\R^2)$ , by definition, is the same as the set of all invertible linear transformations, that is

\Aut_\R(\R^2)=\GL(2,\R):=\{M\in\R^2\mid \det(M)\neq 0\}

Theorem. First isomorphism theorem in modules
If $$R$$ is a ring. $$M$$ is a $$R$$ -module, $N\leq M$ is a submodule. then $$M/N$$ has a $$R$$ -module structure

(M/N, +, \cdot)

with operations defined as

$$(x+N)+(y+N)=(x+y)+N$$
$r\cdot(x+N)=r\cdot x+N$
And ket $p:M\to M/N$ be the quotient map

p:x\mapsto x+N

then we have

p\in\Hom_R(M,M/N)\quad\quad \Ker(p)=N

We know that $N\Norm M$ (abelian), so the addition is well-defined. Now we only need to check if the definition of scalar multiplication is well-defined, which actually is.

The proof is in fact easier than the first isomorphic theorem of groups