# Module homomorphism

In algebra, a module homomorphism is a function between modules that preserves the module structures. Explicitly, if M and N are left modules over a ring R, then a function $f:M\to N$ is called an R-module homomorphism or an R-linear map if for any x, y in M and r in R,

$f(x+y)=f(x)+f(y),$ $f(rx)=rf(x).$ If M, N are right R-modules, then the second condition is replaced with

$f(xr)=f(x)r.$ The preimage of the zero element under f is called the kernel of f. The set of all module homomorphisms from M to N is denoted by $\operatorname {Hom} _{R}(M,N)$ . It is an abelian group (under pointwise addition) but is not necessarily a module unless R is commutative.

The composition of module homomorphisms is again a module homomorphism, and the identity map on a module is a module homomorphism. Thus, all the (say left) modules together with all the module homomorphisms between them form the category of modules.

## Terminology

A module homomorphism is called a module isomorphism if it admits an inverse homomorphism; in particular, it is a bijection. Conversely, one can show a bijective module homomorphism is an isomorphism; i.e., the inverse is a module homomorphism. In particular, a module homomorphism is an isomorphism if and only if it is an isomorphism between the underlying abelian groups.

The isomorphism theorems hold for module homomorphisms.

A module homomorphism from a module M to itself is called an endomorphism and an isomorphism from M to itself an automorphism. One writes $\operatorname {End} _{R}(M)=\operatorname {Hom} _{R}(M,M)$ for the set of all endomorphisms between a module M. It is not only an abelian group but is also a ring with multiplication given by function composition, called the endomorphism ring of M. The group of units of this ring is the automorphism group of M.

Schur's lemma says that a homomorphism between simple modules (a module with no non-trivial submodules) must be either zero or an isomorphism. In particular, the endomorphism ring of a simple module is a division ring.

In the language of the category theory, an injective homomorphism is also called a monomorphism and a surjective homomorphism an epimorphism.

## Examples

• The zero map MN that maps every element to zero.
• A linear transformation between vector spaces.
• $\operatorname {Hom} _{\mathbb {Z} }(\mathbb {Z} /n,\mathbb {Z} /m)=\mathbb {Z} /\operatorname {gcd} (n,m)$ .
• For a commutative ring R and ideals I, J, there is the canonical identification
$\operatorname {Hom} _{R}(R/I,R/J)=\{r\in R|rI\subset J\}/J$ given by $f\mapsto f(1)$ . In particular, $\operatorname {Hom} _{R}(R/I,R)$ is the annihilator of I.
• Given a ring R and an element r, let $l_{r}:R\to R$ denote the left multiplication by r. Then for any s, t in R,
$l_{r}(st)=rst=l_{r}(s)t$ .
That is, $l_{r}$ is right R-linear.
• For any ring R,
• $\operatorname {End} _{R}(R)=R$ as rings when R is viewed as a right module over itself. Explicitly, this isomorphism is given by the left regular representation $R{\overset {\sim }{\to }}\operatorname {End} _{R}(R),\,r\mapsto l_{r}$ .
• $\operatorname {Hom} _{R}(R,M)=M$ through $f\mapsto f(1)$ for any left module M. (The module structure on Hom here comes from the right R-action on R; see #Module structures on Hom below.)
• $\operatorname {Hom} _{R}(M,R)$ is called the dual module of M; it is a left (resp. right) module if M is a right (resp. left) module over R with the module structure coming from the R-action on R. It is denoted by $M^{*}$ .
• Given a ring homomorphism RS of commutative rings and an S-module M, an R-linear map θ: SM is called a derivation if for any f, g in S, θ(f g) = f θ(g) + θ(f) g.
• If S, T are unital associative algebras over a ring R, then an algebra homomorphism from S to T is a ring homomorphism that is also an R-module homomorphism.

## Module structures on Hom

In short, Hom inherits a ring action that was not used up to form Hom. More precise, let M, N be left R-modules. Suppose M has a right action of a ring S that commutes with the R-action; i.e., M is an (R, S)-module. Then

$\operatorname {Hom} _{R}(M,N)$ has the structure of a left S-module defined by: for s in S and x in M,

$(s\cdot f)(x)=f(xs).$ It is well-defined (i.e., $s\cdot f$ is R-linear) since

$(s\cdot f)(rx)=f(rxs)=rf(xs)=r(s\cdot f)(x),$ and $s\cdot f$ is a ring action since

$(st\cdot f)(x)=f(xst)=(t\cdot f)(xs)=s\cdot (t\cdot f)(x)$ .

Note: the above verification would "fail" if one used the left R-action in place of the right S-action. In this sense, Hom is often said to "use up" the R-action.

Similarly, if M is a left R-module and N is an (R, S)-module, then $\operatorname {Hom} _{R}(M,N)$ is a right S-module by $(f\cdot s)(x)=f(x)s$ .

## A matrix representation

The relationship between matrices and linear transformations in linear algebra generalizes in a natural way to module homomorphisms between free modules. Precisely, given a right R-module U, there is the canonical isomorphism of the abelian groups

$\operatorname {Hom} _{R}(U^{\oplus n},U^{\oplus m}){\overset {f\mapsto [f_{ij}]}{\underset {\sim }{\to }}}M_{m,n}(\operatorname {End} _{R}(U))$ obtained by viewing $U^{\oplus n}$ consisting of column vectors and then writing f as an m × n matrix. In particular, viewing R as a right R-module and using $\operatorname {End} _{R}(R)\simeq R$ , one has

$\operatorname {End} _{R}(R^{n})\simeq M_{n}(R)$ ,

which turns out to be a ring isomorphism (as a composition corresponds to a matrix multiplication).

Note the above isomorphism is canonical; no choice is involved. On the other hand, if one is given a module homomorphism between finite-rank free modules, then a choice of an ordered basis corresponds to a choice of an isomorphism $F\simeq R^{n}$ . The above procedure then gives the matrix representation with respect to such choices of the bases. For more general modules, matrix representations may either lack uniqueness or not exist.

## Defining

In practice, one often defines a module homomorphism by specifying its values on a generating set. More precisely, let M and N be left R-modules. Suppose a subset S generates M; i.e., there is a surjection $F\to M$ with a free module F with a basis indexed by S and kernel K (i.e., one has a free presentation). Then to give a module homomorphism $M\to N$ is to give a module homomorphism $F\to N$ that kills K (i.e., maps K to zero).

## Operations

If $f:M\to N$ and $g:M'\to N'$ are module homomorphisms, then their direct sum is

$f\oplus g:M\oplus M'\to N\oplus N',\,(x,y)\mapsto (f(x),g(y))$ and their tensor product is

$f\otimes g:M\otimes M'\to N\otimes N',\,x\otimes y\mapsto f(x)\otimes g(y).$ Let $f:M\to N$ be a module homomorphism between left modules. The graph Γf of f is the submodule of MN given by

$\Gamma _{f}=\{(x,f(x))|x\in M\}$ ,

which is the image of the module homomorphism MMN, x → (x, f(x)).

The transpose of f is

$f^{*}:N^{*}\to M^{*},\,f^{*}(\alpha )=\alpha \circ f.$ If f is an isomorphism, then the transpose of the inverse of f is called the contragredient of f.

## Exact sequences

Consider a sequence of module homomorphisms

$\cdots {\overset {f_{3}}{\longrightarrow }}M_{2}{\overset {f_{2}}{\longrightarrow }}M_{1}{\overset {f_{1}}{\longrightarrow }}M_{0}{\overset {f_{0}}{\longrightarrow }}M_{-1}{\overset {f_{-1}}{\longrightarrow }}\cdots .$ Such a sequence is called a chain complex (or often just complex) if each composition is zero; i.e., $f_{i}\circ f_{i+1}=0$ or equivalently the image of $f_{i+1}$ is contained in the kernel of $f_{i}$ . (If the numbers increase instead of decrease, then it is called a cochain complex; e.g., de Rham complex.) A chain complex is called an exact sequence if $\operatorname {im} (f_{i+1})=\operatorname {ker} (f_{i})$ . A special case of an exact sequence is a short exact sequence:

$0\to A{\overset {f}{\to }}B{\overset {g}{\to }}C\to 0$ where $f$ is injective, the kernel of $g$ is the image of $f$ and $g$ is surjective.

Any module homomorphism $f:M\to N$ defines an exact sequence

$0\to K\to M{\overset {f}{\to }}N\to C\to 0,$ where $K$ is the kernel of $f$ , and $C$ is the cokernel, that is the quotient of $N$ by the image of $f$ .

In the case of modules over a commutative ring, a sequence is exact if and only if it is exact at all the maximal ideals; that is all sequences

$0\to A_{\mathfrak {m}}{\overset {f}{\to }}B_{\mathfrak {m}}{\overset {g}{\to }}C_{\mathfrak {m}}\to 0$ are exact, where the subscript ${\mathfrak {m}}$ means the localization at a maximal ideal ${\mathfrak {m}}$ .

If $f:M\to B,g:N\to B$ are module homomorphisms, then they are said to form a fiber square (or pullback square), denoted by M ×B N, if it fits into

$0\to M\times _{B}N\to M\times N{\overset {\phi }{\to }}B\to 0$ where $\phi (x,y)=f(x)-g(x)$ .

Example: Let $B\subset A$ be commutative rings, and let I be the annihilator of the quotient B-module A/B (which is an ideal of A). Then canonical maps $A\to A/I,B/I\to A/I$ form a fiber square with $B=A\times _{A/I}B/I.$ ## Endomorphisms of finitely generated modules

Let $\phi :M\to M$ be an endomorphism between finitely generated R-modules for a commutative ring R. Then

• $\phi$ is killed by its characteristic polynomial relative to the generators of M; see Nakayama's lemma#Proof.
• If $\phi$ is surjective, then it is injective.

See also: Herbrand quotient (which can be defined for any endomorphism with some finiteness conditions.)

## Variant: additive relations

An additive relation $M\to N$ from a module M to a module N is a submodule of $M\oplus N.$ In other words, it is a "many-valued" homomorphism defined on some submodule of M. The inverse $f^{-1}$ of f is the submodule $\{(y,x)|(x,y)\in f\}$ . Any additive relation f determines a homomorphism from a submodule of M to a quotient of N

$D(f)\to N/\{y|(0,y)\in f\}$ where $D(f)$ consists of all elements x in M such that (x, y) belongs to f for some y in N.

A transgression that arises from a spectral sequence is an example of an additive relation.