Hom and the Tensor Product

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

In this post we take a closer look at \(\Hom\) and the tensor product.

Hom functor

Previously we saw that \(\Hom_{\lMod{A}}(-,N)\) and \(\Hom_\lMod{A}(M,-)\) are left exact functors, and we called the modules \(M\) and \(N\) that make them exact functors projective and injective modules, respectively. The following two propositions go in a somewhat different direction, showing that any splitting exact sequence remains exact after applying \(\Hom\).

Proposition 1 Given a splitting exact sequence

\[0 \longrightarrow M \overset{u}{\longrightarrow} L \overset{v}{\longrightarrow} N \rightarrow 0\]

and any \(A\)-module \(K\), the induced sequence

\[0 \rightarrow \Hom_\lMod{A}(N,K) \rightarrow\Hom_\lMod{A}(L,K) \rightarrow\Hom_\lMod{A}(M,K) \rightarrow 0\]

is a splitting exact sequence. Conversely, if the above sequence is exact for every \(K\), then the original exact sequence is a splitting exact sequence.

Proof

That the given exact sequence \(0 \rightarrow M \rightarrow L \rightarrow N \rightarrow 0\) splits is equivalent to the existence of a suitable retraction \(r:L \rightarrow M\). (§Exact Sequences, ⁋Proposition 10) Now consider

\[\Hom_\lMod{A}(r, \id_K):\Hom_\lMod{A}(M,K) \rightarrow\Hom_\lMod{A}(L,K)\]

From the identity \(r\circ u=\id_M\) we know that \(\Hom_\lMod{A}(r,\id_K)\) has a section, and applying §Exact Sequences, ⁋Proposition 10 again shows that the second sequence splits.

For the converse, set \(K=M\) and consider the short exact sequence

\[0 \rightarrow \Hom_\lMod{A}(N,M) \rightarrow \Hom_\lMod{A}(L,M) \rightarrow\Hom_\lMod{A}(M,M) \rightarrow 0\]

Then there exists a suitable \(f\in\Hom_\lMod{A}(L,M)\) with \(f\circ u=\id_M\), so applying §Exact Sequences, ⁋Proposition 10 again completes the proof.

Similarly, the following holds.

Proposition 2 Given a splitting exact sequence

\[0 \longrightarrow M \overset{u}{\longrightarrow} L \overset{v}{\longrightarrow} N \rightarrow 0\]

and any \(A\)-module \(K\), the induced sequence

\[0 \rightarrow \Hom_\lMod{A}(K,M) \rightarrow\Hom_\lMod{A}(K,L) \rightarrow\Hom_\lMod{A}(K,N) \rightarrow 0\]

is a splitting exact sequence. Conversely, if the above sequence is exact for every \(K\), then the original exact sequence is a splitting exact sequence.

Homomorphism \(M^\ast\otimes_AN \rightarrow \Hom_{\rMod{A}}(M,N)\)

For convenience, assume that \(A\) is a commutative ring. Then \(\lMod{A}=\rMod{A}\), so there is no need to distinguish whether an arbitrary \(A\)-module is a left or right \(A\)-module. In this case, for aesthetic reasons when writing subscripts, we fix \(\rMod{A}\) as our notation. Now let three \(A\)-modules \(M,N,L\) be given. Then we can construct the following \(A\)-module homomorphism

\[\nu:\Hom_{\rMod{A}}(M,L)\otimes_A N \rightarrow\Hom_{\rMod{A}}(M,L\otimes_AN)\]

More generally, if \(A,B\) are rings (not necessarily commutative), \(M\) is a left \(A\)-module, \(N\) is a left \(B\)-module, and \(L\) is an \((A,B)\)-bimodule, then we can construct the following \(\mathbb{Z}\)-module homomorphism

\[\Hom_{\lMod{A}}(M,L)\otimes_AN \rightarrow \Hom_{\lMod{A}}(M, L\otimes_BN)\]

([Bou] II.4.2)

First, for any \(u\in\Hom_{\rMod{A}}(M,L)\) and any \(y\in N\), the formula

\[x\mapsto u(x)\otimes_Ay\]

defines an \(A\)-linear map from \(M\) to \(L\otimes_AN\). Then one checks that the above function sending \((u,y)\in\Hom_{\rMod{A}}(M,L)\times N\) to an element of \(\Hom_{\rMod{A}}(M, L\otimes_AB)\) is \(A\)-balanced. Therefore it induces the \(A\)-linear map \(\nu\).

Proposition 3 For \(\nu\) defined as above, the following hold.

1. If \(N\) is a projective \(A\)-module, then \(\nu\) is injective. Moreover, if \(N\) is finitely generated projective, then \(\nu\) is bijective.
2. If \(M\) is a finitely generated projective \(A\)-module, then \(\nu\) is bijective.

In particular, if \(L=A\), we obtain the following corollary.

Corollary 4 Let two arbitrary \(A\)-modules \(M,N\) be given. If either \(M\) or \(N\) is a finitely generated projective \(A\)-module, then there is an isomorphism

\[M^\ast \otimes_AN\cong \Hom_{\rMod{A}}(M,N)\]

Also, setting \(N=\Hom_{\rMod{A}}(M', L')\), we obtain the following corollary.

Corollary 5 There exists the following \(A\)-linear map

\[\Hom_\rMod{A}(M,L)\otimes_A\Hom_\rMod{A}(M',L') \rightarrow \Hom_\rMod{A}(M\otimes M', L\otimes L')\]

and this is an isomorphism if one of the following pairs

\[(M,M'),\quad (M,L),\quad (M',L')\]

is a finitely generated projective \(A\)-module.

Trace

We still assume that \(A\) is a commutative ring. Then for any \(A\)-module \(M\),

\[M^\ast\times M \rightarrow A;\qquad (\xi,x) \mapsto \langle x, \xi\rangle\]

is \(A\)-balanced. Therefore this function satisfies the following \(A\)-linear map

\[\tau: M^\ast\otimes_A M \rightarrow A\]

Now if \(M\) is a finitely generated projective \(A\)-module, then by Corollary 4 we can identify the left-hand side with \(\End_\rMod{A}(M)=\Hom_\rMod{A}(M,M)\), and thus a unique \(A\)-linear map from \(\End_\rMod{A}(M)\) to \(A\) is defined.

Definition 6 The map defined above is called the trace map and is denoted by \(\tr\).

Let an arbitrary \(u\in\End_\rMod{A}(M)\) be given. After identifying \(\End_\rMod{A}(M)\) with \(M^\ast\otimes_AM\), we can choose finitely many \(\xi_i\in M^\ast, x_i\in M\) and write it as the sum

\[\sum_i \xi_i\otimes_A x_i\tag{1}\]

and the fact that this element corresponds to \(u\) by Corollary 4 means that the formula

\[u(x)=\sum_i\langle x,\xi_i\rangle x_i\qquad\text{for all $x\in M$}\tag{2}\]

holds. Then by the definition of the trace map,

\[\tr(u)=\sum_i\langle x_i,\xi_i\rangle\]

In general there may be infinitely many pairs \(\xi_i\in M^\ast, x_i\in M\) satisfying (2), but this is because there are infinitely many ways to represent an element of \(M^\ast\otimes_AM\) as in (1), and in any case \(\tau:M^\ast\otimes_AM \rightarrow A\) is well-defined, so \(\tr\) does not depend on this choice.

Moreover, one can check that \(\tr\) is an \(A\)-linear map, and furthermore the following holds.

Proposition 7 For any two finitely generated projective \(A\)-modules \(M,N\) and \(A\)-linear maps \(u:M \rightarrow N\), \(v:N \rightarrow M\) between them,

\[\tr(u\circ v)=\tr(v\circ u)\]