Every module over a field is free. Is every commutative ring whose modules are all free a field?

Let $A$ denote a commutative ring. Then if $A$ is a field, we may deduce that every $A$-module is free. Does the converse hold? i.e. If every $A$-module is free, can we deduce that $A$ is a field?

• Here the question was asked for the more general non-commutative case. – Bruno Stonek Mar 19 '14 at 13:25

I would think so.

Let $0\neq x\in A$ a non-invertible element. Then the ideal $Ax$ is proper. Now consider the quotient $A/xA$ as an $A$-module. Since $x\cdot\bar1=\bar 0$, it is torsion, thus not free.

If $I$ is a proper ideal of $A$ then $A/I$ is free by assumption, but any element in a basis will be killed by anything in $I$, so we must have $I = (0)$, and thus $A$ must be a field.

Tobias's and Andrea's answers are pretty much optimal for commutative rings. I'd just like to share a strategy that works for noncommutative rings as well.

For any ring (with identity) $R$, all right $R$ modules are free iff $R$ is a division ring.

Proof: Let $S$ be a simple right $R$ module. Then $S$ is free. Since it's simple, it cannot have more than one copy of $R$ in its decomposition into a sum of copies of $R$. Thus $S\cong R$ as right $R$ modules, and this shows $R$ is a simple right $R$ module. That immediately implies $R$ is a division ring.

• Note that actually my answer also works for noncommutative rings, as long as one just lets $I$ be a left ideal. – Tobias Kildetoft Dec 16 '15 at 8:33

Assume that $R$ is a commutative ring with $1$ whose every module is free. Let $I\subseteq R$ be an ideal. Note that $R/I$ is an $R$-module and therefore free. If $I\neq\{0\}$, then every element of $R/I$ is a torsion element (since we can act by any non-zero element of I to obtain $0$ in $R/I$). Since $R/I$ is also free, $R/I=\{0\}$. Therefore, $I=R$. So, the only ideals are $\{0\}$ and $R$, meaning $R$ is a field.

Assuming $R$ is a domain, then here is another approach:

1. Firstly, $R$ must be a PID, because an ideal that is not principal is not a free $R$-module.

2. Now consider $K$, the quotient ring of $R$, as an $R$-module, and note that $1_R \in K$ is a primitive element (ie. if $1 = ay$, then $a \in R^{\ast}$) and so there is a basis of $K$ containing $1_R$ - call it $\{1_R, v_1, \ldots, v_k\}$. Consider $v_1 = a_1/b_1$ with $b_1\neq 0$ in $R$, then $$a_11_R - b_1v_1 = 0$$ which implies that $a_1 = 0$ whence $v_1 = 0$ - this is impossible.

So it must follow that $K$ is one-dimensional, and hence $K = R$.