Can ⨁_I A be isomorphic to ∏_I A for infinite I?

Question

Suppose $A$ is a non-zero ring (say commutative unital) and $I$ is an infinite set. Can it happen that there is an isomorphism of $A$-modules $\bigoplus_{i\in I}A\cong \prod_{i\in I}A$?

The obvious morphism $\bigoplus_{i\in I}A\to \prod_{i\in I}A$ is obviously not an isomorphism, but could there be another one?

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If $A$ is a field, then no.

In the case when $A$ is a field, I think the following argument does the trick. I'd be happy to know if there's a simpler one. If you're like me and my friends, you'll think this is obvious, but you won't be able to prove it.

Case 1: $A$ is finite or countable. Then $\bigoplus_{i\in I}A$ has cardinality $|I|$ and $\prod_{i\in I}A$ has cardinality $|A|^{|I|} = 2^{|I|}>|I|$. There is no bijection between the two sets, so there's no $A$-module isomorphism.

Case 2: $A$ has arbitrary size. Let $k\subseteq A$ be the prime field of $A$ (either $\mathbb F_p$ or $\mathbb Q$). There is a natural inclusion $\prod_{i\in I}k\hookrightarrow \prod_{i\in I}A$. Suppose $S\subseteq \prod_{i\in I}k$ is linearly independent with $|S|>|I|$. Then I claim the image of $S$ under this inclusion is $A$-linearly independent. This would show that $\dim_A (\prod_{i\in I} A)>|I|=\dim_A(\bigoplus_{i\in I}A)$, so the two cannot be isomorphic.

To show the claim, suppose there is a non-trivial $A$-linear relation on $S$. This can be used to construct a $k$-linear map (the "coefficients map" of the relation) $\phi:\prod_{i\in I}k\to A$ so that $\phi(s_1+\cdots +s_n)=0$, but $\phi(s_r)$ is not zero for some $r\in \{1,\dots, n\}$. Expressing $A$ as $k^{\bigoplus J}$ for some $J$, we get $J$ many $k$-linear compositions $$ \phi_j:\prod_{i\in I}k \xrightarrow{\phi} A\cong k^{\bigoplus J} \xrightarrow{p_j}k $$

Since $\phi(s_r)$ is not zero for some $r\in \{1,\dots, n\}$, there is some $j$ so that $\phi_j(s_r)$ is not zero. But $\phi_j(s_1+\cdots +s_n)=0$, so this gives a non-trivial $k$-linear relation on $S$, contradicting the assumption that $S$ is $k$-linearly independent.

If I'm not mistaken, better bookkeeping in the above argument actually shows that $\dim_A (\prod_{i\in I}A)=2^{|I|}$.

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The obvious way to try to settle the question for a general (commutative unital) ring $A$ is to tensor with $A/\mathfrak m$, where $\mathfrak m$ is a maximal ideal of $A$. So I may as well pose the additional question

If $A$ is a (commutative unital) ring and $I$ is an infinite set, is it necessarily true that $(A/\mathfrak m) \otimes_A (\prod_{i\in I} A) \cong \prod_{i\in I} (A/\mathfrak m)$ as $(A/\mathfrak m)$-modules?

Answer 1

The answer to the first question is no. Here is a proof.

Let $A$ be a commutative ring and $\mathfrak{m}$ be a maximal ideal of $A$. Let $k=A/\mathfrak{m}$. Suppose we have an isomorphism of $A$-modules $A^I \cong A^{(I)}$, where $A^{(I)} := \bigoplus_{i \in I} A$. Tensoring with $k$ over $A$ yields an isomorphism of $k$-vector spaces

\begin{equation} \frac{A^I}{\mathfrak{m} \cdot A^I} \cong k^{(I)} \end{equation} where $\mathfrak{m} \cdot A^I$ denotes the $A$-submodule generated by the $m\cdot x$ with $m \in \mathfrak{m}$ and $x \in A^I$. Since $\mathfrak{m} \cdot A^I \subset \mathfrak{m}^I$, we have a surjective $k$-linear map

\begin{equation} \frac{A^I}{\mathfrak{m} \cdot A^I} \to k^I \end{equation} So we get a contradiction by considering the dimension over $k$ (and using the case you first settled).

EDIT : The answer to the second question is also negative in general. Let me prove that for $I=\mathbf{N}$, there exist a couple $(A,\mathfrak{m})$ such that the statement is false. The idea is to take $A=k[X_j, j \in J]$ and $\mathfrak{m}=\langle X_j, j \in J \rangle$ with a sufficiently big set $J$ (the argument should also work if $\mathfrak{m}/\mathfrak{m}^2$ is a sufficiently big $k$-vector space). We want to prove that

\begin{equation} \dim \frac{A^{\mathbf{N}}}{\mathfrak{m} \cdot A^{\mathbf{N}}} > \dim k^{\mathbf{N}}. \end{equation} It is sufficient to show that $\dim \frac{\mathfrak{m}^{\mathbf{N}}}{\mathfrak{m} \cdot A^{\mathbf{N}}} > \dim k^{\mathbf{N}}$. We have

\begin{equation} \dim \frac{\mathfrak{m}^{\mathbf{N}}}{\mathfrak{m} \cdot A^{\mathbf{N}}} \geq \dim \frac{\mathfrak{m}^{\mathbf{N}}}{\mathfrak{m} \cdot A^{\mathbf{N}} + (\mathfrak{m}^2)^{\mathbf{N}}} \end{equation} The latter space can be identified with the cokernel of the natural (injective) map $k^{\mathbf{N}} \otimes V \to V^{\mathbf{N}}$, where $V=\mathfrak{m}/\mathfrak{m}^2$.

Lemma The cokernel of $f : k^{\mathbf{N}} \otimes V \to V^{\mathbf{N}}$ has dimension $\geq \dim V$.

Proof. A basis of $V$ is given by the family $e_j = \overline{X_j}$ indexed by $j \in J$. Write $J$ as a disjoint union $J = \sqcup_{s \in S} J_s$ where each $J_s$ is countable. We have $\operatorname{card} S = \operatorname{card J}$. Fix isomorphisms $\phi_s : \mathbf{N} \to J_s$. For any $s \in S$, let

\begin{equation} u^{(s)} = (e_{\phi_s(n)})_{n \in \mathbf{N}} \in V^{\mathbf{N}}. \end{equation}

Let us say that a sequence $u \in V^{\mathbf{N}}$ has finite rank if the vector space generated by the $u_n$ is finite dimensional. Then the image of $f$ is the subspace of finite rank sequences in $V^{\mathbf{N}}$. It is not difficult to see that a finite linear combination $\sum_s \lambda_s u^{(s)}$ with all $\lambda_s \neq 0$ cannot have finite rank, whence the lemma.

It remains to take $J$ sufficiently big, for example we can take $J=2^{k^{\mathbf{N}}}$.