Let $A$ be a non-zero commutative ring with unit, $I$ a infinite set.

Can $\prod_{i\in I}A$ be free as an $A$-module?

I found when $A$ is a field or is isomophic to $\mathbb{Z}/m\mathbb{Z}$, then it is free.

But even when $A=\mathbb{Z}$, it is not free. (Baer Specker group)

It seems $\prod_{i\in I}A$ is always not free when $A$ is a domain with dim$\geq1$?

But I've found it difficult to prove. So I want to prove that when $A$ is PID, $\prod_{i\in I}A$ is always not free. It is suffice to prove that when $I=\mathbb{N}$, $\prod_{i\in \mathbb{N}}A$ is not free.

we have already checked that when $A$ is not DVR,then $\prod_{i\in \mathbb{N}}A$ is not free. (similar to $\mathbb{Z}$)

we remain DVR need to check.

My question is :

1. Is any other $A$ such that $\prod_{i\in I}A$ is free as an $A$-module ?

2. Is $\prod_{i\in I}A$ always NOT free when $A$ is an integral domain with dim$\geq1$?

EDIT: I have already believed if $A$ is not Artinian ring the answer is NOT !(that is $\prod_{i\in I} A$ is not free) If $A$ is Artinian, the cardinality of $\prod_{i\in I}A/m$ are all equal for each maximal ideal $m$, then $\prod_{i\in I} A$ is free, other case $\prod_{i\in I} A$ is NOT free.

Thanks Tom.

Thanks everyone here!

Let $A$ be a non-zero commutative ring with unit, $I$ a infinite set.

Can $\prod_{i\in I}A$ be free as an $A$-module?

I found when $A$ is a field or is isomophic to $\mathbb{Z}/m\mathbb{Z}$, then it is free.

But even when $A=\mathbb{Z}$, it is not free. (Baer Specker group)

It seems $\prod_{i\in I}A$ is always not free when $A$ is a domain with dim$\geq1$?

But I've found it difficult to prove. So I want to prove that when $A$ is PID, $\prod_{i\in I}A$ is always not free. It is suffice to prove that when $I=\mathbb{N}$, $\prod_{i\in \mathbb{N}}A$ is not free.

we have already checked that when $A$ is not DVR,then $\prod_{i\in \mathbb{N}}A$ is not free. (similar to $\mathbb{Z}$)

we remain DVR need to check.

My question is :

1. Is any other $A$ such that $\prod_{i\in I}A$ is free as an $A$-module ?

2. Is $\prod_{i\in I}A$ always NOT free when $A$ is an integral domain with dim$\geq1$?

Let $A$ be a non-zero commutative ring with unit, $I$ a infinite set.

Can $\prod_{i\in I}A$ be free as an $A$-module?

I found when $A$ is a field or is isomophic to $\mathbb{Z}/m\mathbb{Z}$, then it is free.

But even when $A=\mathbb{Z}$, it is not free. (Baer Specker group)

It seems $\prod_{i\in I}A$ is always not free when $A$ is a domain with dim$\geq1$?

But I've found it difficult to prove. So I want to prove that when $A$ is PID, $\prod_{i\in I}A$ is always not free. It is suffice to prove that when $I=\mathbb{N}$, $\prod_{i\in \mathbb{N}}A$ is not free.

we have already checked that when $A$ is not DVR,then $\prod_{i\in \mathbb{N}}A$ is not free. (similar to $\mathbb{Z}$)

we remain DVR need to check.

My question is :

1. Is any other $A$ such that $\prod_{i\in I}A$ is free as an $A$-module ?

2. Is $\prod_{i\in I}A$ always NOT free when $A$ is an integral domain with dim$\geq1$?

Let $A$ be a non-zero commutative ring with unit, $I$ a infinite set.

Can $\prod_{i\in I}A$ be free as an $A$-module?

I found when $A$ is a field or is isomophic to $\mathbb{Z}/m\mathbb{Z}$, then it is free.

But even when $A=\mathbb{Z}$, it is not free. (Baer Specker group)

It seems $\prod_{i\in I}A$ is always not free when $A$ is a domain with dim$\geq1$?

But I've found it difficult to prove. So I want to prove that when $A$ is PID, $\prod_{i\in I}A$ is always not free. It is suffice to prove that when $I=\mathbb{N}$, $\prod_{i\in \mathbb{N}}A$ is not free.

we have already checked that when $A$ is not DVR,then $\prod_{i\in \mathbb{N}}A$ is not free. (similar to $\mathbb{Z}$)

we remain DVR need to check.

My question is :

1. Is any other $A$ such that $\prod_{i\in I}A$ is free as an $A$-module ?

2. Is $\prod_{i\in I}A$ always NOT free when $A$ is an integral domain with dim$\geq1$?

EDIT: I have already believed if $A$ is not Artinian ring the answer is NOT !(that is $\prod_{i\in I} A$ is not free) If $A$ is Artinian, the cardinality of $\prod_{i\in I}A/m$ are all equal for each maximal ideal $m$, then $\prod_{i\in I} A$ is free, other case $\prod_{i\in I} A$ is NOT free.

Thanks Tom.

Thanks everyone here!