Examples of maps $\phi: \operatorname{Hom}(\prod_s N_s, M) \to \prod_s \operatorname{Hom}(N_s, M)$

abstract-algebracommutative-algebrareference-request

Let $R$ be a commutative ring, $\{N_s\}_{s \in S}$ is a family of modules and $M$ is a module over $R$. There is canonical map of modules
$$
\phi: \operatorname{Hom} \bigg(\!\prod_s N_s, M \bigg) \to \prod_s \operatorname{Hom}(N_s, M),
$$
that sends map $f \colon \prod_s N_s \to M$ to the "infinite vector" of maps $(f\circ \iota_k)_s$, where $\iota_k\colon N_k \to \prod_s N_s$ is the canonical injection. It is well-known that this map is isomorphism if $\prod_s N_s$ is replaced by $\bigoplus_s N_s$. But I want to know examples when product version fails to be injective or surjective.

I'm interested in the following examples in order of preferences

  1. Ring $R$ is a field and $\phi$ is neither injective nor surjective.

  2. Ring $R$ is a field and $\phi$ is not injective (surjective).

  3. Map $\phi$ is neither injective nor surjective.

  4. Map $\phi$ is not injective (surjective).

If there are theorems that forbid examples of any of the above types, it is also very interesting.

Best Answer

Equivalently, you are asking about the restriction map $\operatorname{Hom}(P, M) \to \operatorname{Hom}(Q, M)$ induced by the inclusion map $Q\to P$ where $P=\prod_s N_s$ and $Q=\bigoplus_s N_s$. This map is part of an exact sequence $$0\to \operatorname{Hom}(P/Q,M)\to\operatorname{Hom}(P,M)\stackrel{\phi}\to\operatorname{Hom}(Q,M)\to \operatorname{Ext}^1(P/Q,M)\to \operatorname{Ext}^1(P,M).$$ So, $\phi$ is injective iff there are no nontrivial homomorphisms $P/Q\to M$. In particular, for instance, this means that $\phi$ is never injective when $R$ is a field as long as infinitely many of the $N_s$ are nontrivial and $M$ is nontrivial. Or, for an arbitrary family where infinitely many of the $N_s$ are nontrivial, if $M=P/Q$ then $\phi$ will not be injective. On the other hand, if $M$ is an injective module (so in particular, if $R$ is a field, or more generally if $R$ is a semisimple ring), then $\phi$ is always surjective.

For an example where $\phi$ is not surjective, you could take $R=\mathbb{Z}$ and consider the direct product $P=\prod_p\mathbb{Z}/(p)$ where $p$ ranges over the primes. Note that then the quotient $P/Q$ is divisible: given an element $x\in P$ and a nonzero integer $n$, you can divide $x$ by $n$ mod $Q$ by just ignoring the coordinates corresponding to the finitely many primes dividing $n$. But $P$ has no nontrivial divisible submodules, since if $x\in P$ is nonzero on its $p$th coordinate then $x$ is not divisible by $p$. So in particular, $P/Q$ is not isomorphic to a submodule of $P$ so the inclusion map $Q\to P$ does not split. Taking $M=Q$, this says exactly that $\phi$ fails to be surjective since its image does not contain the identity map $Q\to Q$.

To get an example where $\phi$ is neither surjective nor injective, you can just take a direct sum of examples of each type. For instance, with the same product as in the previous example, you could take $M=Q\oplus P/Q$.

Another well-known example where $\phi$ is not surjective is if you take $R=\mathbb{Z}$ and $P$ to be a product of countably infinitely many copies of $\mathbb{Z}$. Then it is a nontrivial theorem that every homomorphism $P\to\mathbb{Z}$ is a linear combination of the projection maps. In particular, there are only countably many such maps, whereas there are uncountably many homomorphisms $Q\to\mathbb{Z}$ (since $Q$ is free on infinitely many generators), so the restriction map $\phi:\operatorname{Hom}(P,\mathbb{Z})\to\operatorname{Hom}(Q,\mathbb{Z})$ is very far from surjective. (On the other hand, it follows from the theorem that $\phi$ is injective in this case, which is rather surprising.)

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