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Modern Algebra: Groups, Rings, Modules, Fields

Sam Macdonald

Section 12.2 Module Homomorphisms

Subsection Homomorphisms and Isomorphisms

“Transformation rarely happens accidentally.”
―A.J. Sheppard

Definition 12.20. \(R\)-Module Homomorphism.

Let \(R\) be a ring and let \(M\) and \(N\) be \(R\)-modules. An \(R\)-module homomorphism from \(M\) to \(N,\) sometimes called an \(R\)-map, is a function \(\varphi: M \to N\) such that for all \(r \in R\) and \(m,n \in M\) we have
  1. Preserves Addition.
    \(\varphi(m+n)=\varphi(m)+\varphi(n)\text{,}\) i.e. \(\varphi\) is an additive group homomorphism, and
  2. Preserves Scaling.
    \(\varphi(rm) = r\varphi(m)\text{.}\)
When \(R\) is a field, we refer to \(\varphi\) as a linear transformation.

Exercise 12.21. \(\varphi(0)=0\).

Let \(\varphi\) be an \(R\)-module homomorphism. Then \(\varphi(0)=0\text{.}\)
Solution.
By definition, \(\varphi\) is a homomorphism of abelian groups. Thus \(\varphi(0)=0\) by Theorem 1.63.

Remark 12.22.

Unlike ring homomorphisms, a \(R\)-module homomorphism does not need to map \(1\) to \(1\text{.}\)

Example 12.23. Multipication Map is \(R\)-Map.

  1. Let \(R\) be a commutative ring and \(M\) be an \(R\)-module. For each \(r \in R\text{,}\) the multiplication map \(\varphi_r : M \to M\) given by \(\varphi_r(m) = rm\) is a homomorphism of \(R\)-modules.
  2. Let \(R\) be a ring with \(1 \ne 0\text{,}\) let \(M\) be an \(R\)-module, and let \(N\) be an \(R\)-submodule of \(M\text{.}\) Then the inclusion map \(i : N \to M\) is an \(R\)-module homomorphism.
Solution.
  1. By the definition of \(R\)-module we have
    \begin{equation*} \varphi_r(m + n) = r(m + n) = rm + rn = \varphi_r(m) + \varphi_r(n), \end{equation*}
    and thus \(\varphi_r\) preserves addition. Also, we have
    \begin{equation*} \varphi_r(sm) = r(sm) = (rs)m = (sr)m = s(rm) = s\varphi_r(m), \end{equation*}
    and thus \(\varphi_r\) preserves scaling as well, making \(\varphi_r\) an \(R\)-module homomorphism.

Definition 12.24. \(R\)-Module Isomorphism.

An \(R\)-module homomorphism \(\varphi: M \to N\) is an \(R\)-module isomorphism if there is another \(R\)-module homomorphism \(\phi: N \to M\) such that \(\varphi \circ \phi = \id_M\) and \(\phi \circ \varphi = \id_N\text{.}\)
To check that an \(R\)-module homomorphism \(\varphi : M \to N\) is an isomorphism, it is sufficient to check that it is bijective.

Proof.

Notice that if \(\varphi\) is an \(R\)-module homomorphism then it is a group homomophism as well. Thus Theorem 1.71 yields the result.

Exercise 12.26.

If \(R\) is the zero ring, then there is (up to isomorphism) only one \(R\)-module, the zero module.
Solution.
If \(M\) is any module over the \(0\) ring, then for each \(m \in M\) we have \(m = 1_R \cdot m = 0_R \cdot m = 0_M\text{.}\)

Definition 12.27. Kernel.

The kernel of an \(R\)-module homomorphism is the set \(\ker(\varphi) := \{m \in M \mid \varphi(m) = 0_N\}\)

Subsection \(\Hom\)

“All good things must come to an \(\End\)”
―Geoffrey Chaucer

Definition 12.29. \(\Hom\).

Let \(R\) be a ring and let \(M\) and \(N\) be \(R\)-modules. Then \(\Hom_R(M, N )\) denotes the set of all \(R\)-module homomorphisms from \(M\) to \(N\text{,}\) and \(\End_R(M)\) denotes the set \(\Hom_R(M, M)\text{.}\)

Proof.

in Eloísa notes

Exercise 12.32.

Show that for every nonzero integers \(m\) and \(n\) there is a \(\Z\)-module isomorphism \(\Hom_\Z(\Z/n,\Z/m) \cong \Z / (\gcd(m,n))\text{.}\)

Subsection Quotient Modules

“The best way to solve a problem is to remove its cause.”
―Martin Luther King Jr.

Definition 12.33. Quotient Module.

Let \(R\) be a ring, let \(M\) be an \(R\)-module, and let \(N\) be a submodule of \(M\text{.}\) The quotient module \(M/N\) is the quotient group \(M/N\) under \(+\) (so elements of \(M/N\) are additive cosets of the form \(m + N\) with \(m \in M\)) using the following structure:
  1. Rule for Addition.
    Addition is defined by \((m_1 + N) + (m_2 + N) = (m_1+m_2) + N\)
  2. Rule for Scaling.
    Scaling by \(R\) defined to by \(r \cdot (m + N) = rm + N\) for all \(r \in R\) and \(m + N \in M/N\text{.}\)

Remark 12.34.

When working with groups, it was necessary that we modded out by a normal subgroup in order to ensure that our quotient group was well defined. However, since all modules underlying groups are abelian, this condition is satisfied automatically.

Definition 12.36. Module Quotient Map.

Let \(R\) be a ring, let \(M\) be an \(R\)-module, and let \(N\) be a submodule of \(M\text{.}\) Define the canonical quotient map \(\pi: M \to M/N\) by \(\pi(m) = m + N\text{.}\)

Proof.

Among the many things to check here, we will only check a couple.
We need to prove the rule for scaling by \(R\) on \(M/N\) is well-defined: If \(m+N=m'+N\) then \(m-m'\in N\) so \(r(m-m')\in N\) by the definition of submodule. This gives that \(rm-rm'\in N\text{,}\) hence \(rm+N=rm'+N\text{.}\) The module axioms are then pretty straightforward. We already know from 817 CITEX that \(M/N\) is an abelian group under \(+\text{.}\)
Let us check one of the four axioms involving scaling. We have
\begin{equation*} r( (m_1 +N) + (m_2 +N)) = r((m_1 + m_2) + N) = (r(m_1+m_2)) + N = (rm_1 + rm_2) + N = (rm_1 + N) + (rm_2 + N), \end{equation*}
which gives the third such axiom. The other three are also straightforward.
The fact that \(\pi\) is an \(R\)-module homomorphism is also straightforward. Its kernel is \(\{m \in M \mid m + N = N\}\text{,}\) which is equal to \(N\text{.}\)

Example 12.38. \(\Z\)-modules and Quotients.

Recall that \(\Z\)-modules are the same as abelian groups by Theorem 12.10. Submodules and quotient \(\Z\)-modules are the same things as subgroups and quotients of abelian groups.

Definition 12.39. Cokernel.

The cokernel of a map of \(R\)-modules \(M \stackrel{\varphi}{\rightarrow} N\) is the module
\begin{equation*} \text { coker } \varphi:=B / \operatorname{im}(\varphi). \end{equation*}

Proof.

Ignoring the rules for scaling by \(R\text{,}\) we know each of the first four results holds for abelian groups (and the maps are the same). So, we merely need to prove that the rules for scaling are respected in each case. In more detail:
  1. UMP.
    From the UMP for Quotient Groups, we already know that \(\ov{f}\) is a well-defined homomorphism of groups under \(+\) and that it is the unique one such that \(\ov{f} \circ \pi = f\text{.}\) It remains only to show \(\ov{f}\) preserves scaling:
    \begin{equation*} \ov{f}(r (m +N)) = \ov{f} (rm + N) = f(rm) = r f(m) = r \ov{f}(m + N). \end{equation*}
    where the third equation uses that \(f\) preserves scaling.
  2. First Isomorphism Theorem.
    From First Isomorphism Theorem for Groups we already know that there is an isomorphism of abelian groups under \(+\text{,}\)
    \begin{equation*} \ov{\varphi}: M/\ker(\varphi) \xra{\cong} \im(\varphi), \end{equation*}
    given by \(\ov{\varphi}(m + \ker(\varphi)) = \varphi(m)\text{,}\) and it remains only to show this map preserves scaling. This is a special case of what we proved in part (0).
  3. Second Isomorphism Theorem.
    First, note from Exercise 12.9 that \(A+B\) and \(A \cap B\) are indeed submodules. The Second Isomorphism Theorem for Groups yields an isomorphism of abelian groups \(\varphi: A/(A \cap B) \xra{\cong} (A+B)/B\) given by \(\varphi(a + (A\cap B)) = a + B\text{.}\) It remains only to show \(\varphi\) preserves scaling:
    \begin{equation*} \begin{aligned}\varphi(r(a+(A \cap B))) &= \varphi(ra + A \cap B)\\ &=ra + B \\ &=r(a +B) \\ &=r\varphi(a + (A \cap B)) \end{aligned} \end{equation*}
  4. Third Isomorphism Theorem.
    For the third, we already know (from 817) that \(B/A\) is a subgroup of \(M/A\) under \(+\text{.}\) Given \(r \in R\) and \(b +A \in B/A\) we have \(r(b+A) = rb + A\) which belongs to \(B/A\) since \(rb \in B\text{.}\) This proves \(B/A\) is a submodule of \(M/A\text{.}\) Also from 817 we know there is an isomorphism of abelian groups \(\varphi: (M/A)/(B/A) \to M/B\) given by \(\varphi((m+A) + B/A) = m + B\) and it remains only to show it is \(R\)-linear:
    \begin{equation*} \begin{aligned}\varphi(r((m+A) + B/A)) &= \varphi(r(m+A) + B/A)\\ &=\varphi((rm + A) + B/A) \\ &=rm + B \\ &=r(m +B) \\ &=r \varphi((m+A) + B/A) \end{aligned} \end{equation*}
  5. Lattice Isomorphism Theorem.
    The Lattice Theorem is the most complicaed to gerenlize. From 817 we know thre is a bijection between the set of subgroups of \(M\) and that contain \(N\) and subgroups of the quotient group \(M/N\text{,}\) and the maps are the same as given in the statment. We just need to prove that these maps send submodules to submodules. If \(K\) is a submodule of \(M\) containing \(N\text{,}\) then by part (3) we know \(K/N\) is a submodule of \(M/N\text{.}\)
    If \(T\) is a submodule of \(M/N\text{,}\) then \(\pi^{-1}(T)\) is an abelian group. For \(r \in R\) adn \(m \in \pi^{-1}(T)\) we have \(\pi(m) \in T\) and hence \(\pi(rm) = r\pi(m) \in T\) too, since \(T\) is a submodule. This proves \(\pi^{-1}(T)\) is a submodule.

Exploration 12.2. Simple Modules.

Let \(R\) be a commutative ring with \(1 \neq 0\text{.}\) An \(R\)-module \(M\) is simple if it has no nontrivial submodules. Show that \(M \neq 0\) is simple if and only if there exists a maximal ideal \(\fm\) of \(R\) such that \(M \cong R/\fm\text{.}\)
Solution.
As \(M\neq 0\text{,}\) there exists a non-zero element \(m\in M\text{.}\) Let \(f:R\to M\) be defined by \(f(r)=rm\text{.}\) Since \(1\in R\) and \(1m=m\in Rm\text{,}\) we have \(Rm=M\text{.}\) By the First Isomorphism Theorem for Modules we see that \(R/\ker(f)\cong M\text{,}\) making \(\ker(f)\) an ideal in \(R\text{,}\) which we shall conspicuously denote as \(I\) henceforth. By the Lattice Isomorphism Theorem for Rings the only two ideals of \(R/I\) are \(0\) and \(R/I\text{,}\) making \(R/I\) a field by Proposition 9.8 and \(I\) a maximal ideal in \(R\) by Theorem 9.36.

Qual Watch.

Proving Exploration 12.2 was [cross-reference to target(s) "may-2018-5" missing or not unique] on the [cross-reference to target(s) "may-2018" missing or not unique] qualifying exam.
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