2 fixed error

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?

## 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 is can be used to construct a $k$-linear map (the data "coefficients map" of an $A$-linear functional the relation) $\phi:\prod_{i\in I}A\to I}k\to A$ so that $\phi(s_1+\cdots +s_n)=0$, but $\phi(s_r)$ is not zero for some $r\in \{1,\dots, n\}$. Since $\phi$ is $A$-linear, it is $k$-linear. Expressing $A$ as $k^{\bigoplus J}$ for some $J$, we get $J$ many $k$-linear compositions $$\phi_j:\prod_{i\in I}k\hookrightarrow \prod_{i\in I}A I}k \xrightarrow{\phi} A\cong k^{\bigoplus J} \xrightarrow{p_j}k$$

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 $\dim_A (\prod_{i\in I}A)=2^{|I|}$.

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?

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

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?

## 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 is the data of an $A$-linear functional $\phi:\prod_{i\in I}A\to A$ so that $\phi(s_1+\cdots +s_n)=0$, but $\phi(s_r)$ is not zero for some $r\in \{1,\dots, n\}$. Since $\phi$ is $A$-linear, it is $k$-linear. Expressing $A$ as $k^{\bigoplus J}$ for some $J$, we get $J$ many $k$-linear compositions $$\phi_j:\prod_{i\in I}k\hookrightarrow \prod_{i\in I}A \xrightarrow{\phi} A\cong k^{\bigoplus J} \xrightarrow{p_j}k$$

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 $\dim_A (\prod_{i\in I}A)=2^{|I|}$.

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?