Commutative Algebra/Modules, submodules and homomorphisms

Basics

Analogously, one can define right $R$-modules with an operation $R\times M\to M,(r,m)\mapsto mr$; the difference is only formal, but it will later help us define bimodules in a user-friendly way.

For the sake of brevity, we will often write module instead of left $R$-module.

• Exercise 5.1.1: Prove that every Abelian monoid $(M,+)$ together with an operation as specified in 1.) - 4.) of definition 5.1 is already a module.

Submodules

The following lemma gives a criterion for a subset of a module being a submodule.

Lemma 5.3:

A subset $N\subseteq M$ is a submodule iff

$\mathrm{\forall }r\in R,n,q\in N:rn-q\in N$.

Proof:

Let $N$ be a submodule. Then since $-q\in N$ since we have an Abelian group and further $rn\in N$ due to closedness under the module operation, also $rn+(-q)=:rn-q\in N$.

If $N$ is such that $\mathrm{\forall }r\in R,n,q\in N:rn-q\in N$, then for any $n,m\in N$ also $n+m=n+(-1_R)(-m)\in N$.

Proof:

Well-definedness: If $m+N=p+N$, then $m-p\in N$, hence $r(m-p)=rm-rp\in N$ and thus $rm+N=rp+N$.

1. $1_R(m+N)=(1_{r}m)+N=m+N$
2. $r(n+N+m+N)=r((m+n)+N)=r(m+n)+N=rm+rn+N=rm+N+rn+N$
3. $(r+s)(m+N)=(r+s)m+N=rm+sm+N=rm+N+rn+N$
4. analogous to 3. (replace $+$ by $\cdot$)$◻$

Sum and intersection of submodules

We shall now ask the question: Given a module $M$ and certain submodules $\{N_{\alpha }{\}}_{\alpha \in A}$, which module is the smallest module containing all the $N_{\alpha }$? And which module is the largest module that is itself contained within all $N_{\alpha }$? The following definitions and theorems answer those questions.

Proof:

1. $\sum _{\alpha \in A}{N}_{\alpha }$ is a submodule:

• It is an Abelian subgroup since if $\sum _{l=1}^k{r}_l{n}_{{\alpha }_l},\sum _{j=1}^m{s}_l{n}_{{\beta }_j}\in \sum _{\alpha \in A}{N}_{\alpha }$, then

$\sum _{l=1}^k{r}_l{n}_{{\alpha }_l}-\sum _{j=1}^m{s}_l{n}_{{\beta }_j}=\sum _{l=1}^k{r}_l{n}_{{\alpha }_l}+\sum _{j=1}^m(-s_l)n_{{\beta }_j}\in \sum _{\alpha \in A}{N}_{\alpha }$.

• It is closed under the module operation, since

$s\left(\sum _{l=1}^k{r}_l{n}_{{\alpha }_l}\right)=\sum _{l=1}^k(s{r}_l)n_{{\alpha }_l}\in \sum _{\alpha \in A}{N}_{\alpha }$.

2. Each $N_{\alpha }$ is contained in $\sum _{\alpha \in A}{N}_{\alpha }$:

This follows since $1_r{n}_{\alpha }\in \sum _{\alpha \in A}{N}_{\alpha }$ for each $\alpha \in A$ and each $n_{\alpha }\in N_{\alpha }$.

3. $\sum _{\alpha \in A}{N}_{\alpha }$ is the smallest submodule containing all the $N_{\alpha }$: If $K\le M$ is another such submodule, then $K$ must contain all the elements

$\sum _{l=1}^k{r}_l{n}_{{\alpha }_l},k\in \mathbb{N},r_l\in R,n_{{\alpha }_l}\in N_{{\alpha }_l}$

due to closedness under addition and submodule operation.$◻$

Proof:

1. It's a submodule: Indeed, if $r\in R,n,p\in \bigcap _{\alpha \in A}{N}_{\alpha }$, then $n,p\in N_{\alpha }$ for each $\alpha$ and thus $n-rp\in N_{\alpha }$ for each $\alpha$, hence $n-rp\in \bigcap _{\alpha \in A}{N}_{\alpha }$.

2. It is contained in all $N_{\alpha }$ by definition of the intersection.

3. Any set that contains all elements from each of the $N_{\alpha }$ is contained within the intersection.$◻$

We have the following rule for computing with intersections and sums:

Proof:

$\subseteq$: Let $l+n\in (L+N)\cap K$. Since $L\subseteq K$, $l\in K$ and hence $n\in K$. Since also $n\in N$ by assumption, $l+n\in L+K\cap N$.

$\supseteq$: Let $l+m\in L+(K\cap N)$. Since $L\subseteq K$, $l\in K$ and since further $m\in K$, $l+m\in K$. Hence, $l+m\in K\cap (L+N)$.$◻$

More abstractly, the properties of the sum and intersection of submodules may be theoretically captured in the following way:

Lattices

There are some special types of lattices:

In fact, it suffices to require conditions 1. and 2. only for sets $S$ with two elements. But as we have shown, in the case that $L$ is the set of all submodules of a given module, we have the "original" conditions satisfied.

Proof:

First, we note that least upper bound and greatest lower bound are unique, since if for example $u,u^{\prime }$ are least upper bounds of $S$, then $u\le u^{\prime }$ and $u^{\prime }\le u$ and hence $u=u^{\prime }$. Thus, the joint and meet operation are well-defined.

The commutative laws follow from $\{a,b\}=\{b,a\}$.

The idempotency laws from clearly $a$ being the least upper bound, as well as the greatest lower bound, of the set $\{a,a\}$.

The first absorption law follows as follows: Let $u$ be the least upper bound of $\{a,b\}$. Then in particular, $u\ge a$. Hence, $a$ is a lower bound of $\{a,u\}$, and any lower bound $l$ satisfies $l\le a$, which is why $a$ is the greatest lower bound of $\{a,u\}$. The second absorption law is proven analogously.

The first associative law follows since if $u$ is the least upper bound of $\{a,b,c\}$ and $v$ is the upper bound of $\{a,b\}$, then $u\ge v$ (as $u$ is an upper bound for $\{a,b\}$) and if $w$ is the least upper bound of $\{v,c\}$, then $w=u$ since $u$ is an upper bound and further, $w\ge v\ge a$ and $w\ge b$. The same argument (with $a$ and $c$ swapped) proves that $u$ is also the least upper bound of the l.u.b. of $\{b,c\}$ and $a$. Again, the second associative law is proven similarly.$◻$

From theorems 5.5-5.7 and 5.10 we note that the submodules of a module form a modular lattice, where the order is given by set inclusion.

Exercises

• Exercise 5.2.1: Let $R$ be a ring. Find a suitable module operation such that $R$ together with its own addition and this module operation is an $R$-module. Make sure you define this operation in the simplest possible way. Prove further, that with respect to this module operation, the submodules of $R$ are exactly the ideals of $R$.

Homomorphisms

We shall now get to know the morphisms within the category of modules over a fixed ring $R$.

Since we are cool, we will often simply write morphisms instead of homomorphisms where it's clear from the context in order to indicate that we have a clue about category theory.

We have the following useful lemma:

Lemma 5.12:

$f:M\to N$ is $R$-linear iff

$\mathrm{\forall }r\in R,m,p\in M:f(rm+p)=rf(m)+f(p)$.

Proof:

Assume first $R$-linearity. Then we have

$f(rm+p)=f(rm)+f(p)=rf(m)+f(p)$.

Assume now the other condition. Then we have for $m,p\in M$

$f(m+p)=f(1_{R}m+p)=1_{R}f(m)+f(p)=f(m)+f(p)$

and

$f(rm)=f(rm+0)=rf(m)+f(0)=rf(m)$

since $f(0)=0$ due to $f(0)=f(0+0)=f(0)+f(0)$; since $M$ is an abelian group, we may add the inverse of $f(0)$ on both sides.$◻$

Lemma 5.13:

If $f:M\to N$ is $R$-linear, then $\mathrm{\forall }m\in M:f(-m)=-f(m)$.

Proof:

This follows from the respective theorem for group homomorphisms, since each morphism of modules is also a morphism of Abelian groups.$◻$

Lemma 5.14:

Let $f:M\to N$ be a morphism. The following are equivalent:

1. $f$ is an isomorphism
2. $\ker f=\{0\}$
3. $f$ has an inverse which is an isomorphism

Proof:

Lemma 5.15:

The kernel and image of morphisms are submodules.

Proof:

1. The kernel:

$f(rn-q)=rf(n)+f(-q)=rf(n)-f(q)=0$

2. The image:

$rf(m)\stackrel{=+f(-p)}{\overbrace{-f(p)}}=f(rm-p)$$◻$

The following four theorems are in complete analogy to group theory.

Proof:

We define $\bar{\phi }(m+N):=\phi (m)$. This is well-defined since $\ker \phi \subseteq N$. Furthermore, this definition is already enforced by $\bar{\phi }\circ \pi =\phi$. Further, $\bar{\phi }(m+N)=0\iff m\in \ker \phi$.$◻$

Proof:

We set $N=\ker f$ and obtain a homomorphism $\bar{f}:M/\ker f\to K$ with kernel $N/N$ by theorem 5.11. From lemma 5.16 follows the claim.$◻$

Proof:

Since $L\le N$ and $N\le M$ also $L\le M$ by definition. We define the function

$\phi :M/L\to M/N,m+L\mapsto m+N$.

This is well-defined since

$m+L=p+L\iff m-p\in L\Rightarrow m-p\in N\iff m+N=p+N$.

Furthermore,

$m+L\in \ker \phi \iff m+N=0+N\iff m\in N$

and hence $\ker \phi =N/L$. Hence, by theorem 5.17 our claim is proven.$◻$

Proof:

Consider the isomorphism

$\phi :L\to (L+N)/N,\phi (l):=l+N$.

Then $\phi (l)=0\iff l\in N$, which is why the kernel of that homomorphism is given by $L\cap N$. Hence, the theorem follows by the first isomorphism theorem.$◻$

And now for something completely different:

Proof:

Let $a,b\in {\phi }^{-1}(L)$. Then $\phi (a+b)=\phi (a)+\phi (b)\in L$ and hence $a+b\in {\phi }^{-1}(L)$. Let further $r\in R$. Then $\phi (ra)=r\phi (a)\in L$.$◻$

Similarly:

Proof: Let $a,b\in \phi (K)$. Then $a=\phi (i),b=\phi (j)$ and $a+b=\phi (i+j)\in \phi (K)$. Let further $r\in R$. Then $ra=\phi (ri)\in \phi (K)$.$◻$

Exercises

• Exercise 5.3.1: Let $R,S$ be rings regarded as modules over themselves as in exercise 5.2.1. Prove that the ring homomorphisms $\phi :R\to S$ are exactly the module homomorphisms $R\to S$; that is, every ring hom. is a module hom. and vice versa.

The projection morphism

The following two fundamental equations for ${\pi }_N({\pi }_N^{-1}(S))$ and ${\pi }_N^{-1}({\pi }_N(K))$ shall gain supreme importance in later chapters, $S\subseteq M/N$, $K\le M$.

Proof:

Let first $m+N\in S$. Then $m\in {\pi }^{-1}(S)$, since ${\pi }_N(m)=m+N$. Hence, $m+N\in {\pi }_N({\pi }^{-1}(S))$. Let then $m+N\in {\pi }_N({\pi }_N^{-1}(S))$. Then there exists $m^{\prime }\in {\pi }_N^{-1}(S)$ such that ${\pi }_N(m^{\prime })=m+N$, that is $m^{\prime }+N=m+N$. Now $m^{\prime }\in {\pi }_N^{-1}(S)$ means that $\pi (m^{\prime })=m^{\prime }+N\in S$. Hence, $m+N=m^{\prime }+N\in S$.

Let first $m\in K+N$, that is, $m=k+n$ for suitable $k\in K$, $n\in N$. Then ${\pi }_N(m)=k+n+N=k+N={\pi }_N(k)\in {\pi }_N(K)$, which is why by definition $m\in {\pi }_N^{-1}({\pi }_N(K))$. Let then $m\in {\pi }_N^{-1}({\pi }_N(K))$. Then ${\pi }_N(m)=m+N\in {\pi }_N(K)$, that is $m+N=k+N$ with $k\in K$, that is $m=k+n$ for a suitable $n\in N$, that is $m\in K+N$.$◻$

The following lemma from elementary set theory have relevance for the projection morphism and we will need it several times:

Lemma 5.24:

Let $f:S\to T$ be a function, where $S,T$ are completely arbitrary sets. Then $f$ induces a function $2^S\to 2^T$ via $A\mapsto f(A)$, the image of $A$, where $A\subseteq S$. This function preserves inclusion. Further, the function $2^T\to 2^S,B\mapsto f^{-1}(B)$, also preserves inclusion.

Proof:

If $A^{\prime }\subseteq A$, let $y^{\prime }\in f(A^{\prime })$. Then $y^{\prime }=f(x^{\prime })$ for an $x^{\prime }\in A^{\prime }\subseteq A$. Similarly for $f^{-1}$.$◻$