A little exercise about free modules

Question

Let $R$ be a non-zero ring with unity. Let $a$ be a proper ideal of $R$. I wish to prove the following statement:

$R/a$ is a free left module over $R$ iff $a=0$

where $R/a$ has the usual scalar multiplication induced by considering the left regular module $_{R}R$

I'm having little trouble proving that if $R/a$ is free over $R$ then one must have $a=0$. Here is what I've done so far:

Suppose that $R/a$ is free over $R$. Let $X$ be a basis of $R/a$ over $R$. Since $a$ is a proper ideal, we have that $R/a \neq 0$ and so $X \neq \emptyset$ as a generating set. Let $x + a \in X$. Since $X$ is linearly independent over $R$, we have in particular that $ \{x+a\}$ must be linearly independent over $R$. This means that $\forall r\in R$, $r(x+a)=a \implies r=0$ or equivalently $\forall r \in R, rx \in a \implies r=0$.

Now if one could take $x=1$ the result would follow immediately. Can anyone help me from the last step forward? Or provide a different way of looking at the problem? Thank you in advice.

Answer 1

If $R$ is commutative, this is true. Noncommutative rings like the ring of endomorphisms of an infinite-dimensional vector space are counterexamples in general. I'll give a relatively inexplicit proof-this could be reduced to manipulating elements.

So, assume $R$ is commutative. If $R/a$ were isomorphic to $R$, then the map $R\to R/a\cong R$ would split, making $R\cong R\oplus M$. We want to show $M=0$, for which it's enough to suppose $R$ is local: $M=0$ if and only if $M_{\mathfrak p}=0$ for every prime ideal $\mathfrak p$. Then if $R$ is local with maximal ideal $\mathfrak m$ and residue field $\kappa,$ we have $\kappa\cong \kappa\oplus M/\mathfrak m M$, so $M/\mathfrak m M=0$. But then by Nakayama's lemma, $M=0$, as was to be shown.

EDIT: A simpler answer which doesn't use commutativity was posted and deleted, so I'll reproduce it here for now. If $R$ is a unital ring, not necessarily commutative, $a$ is two-sided and $f:R\to R/a$, then for $x\in a, xy\in a$ for every $y\in R$. Thus $x[f(1)]=[xf(1)]=[f(x)]=[0]$ in $R/a$, and $f$ cannot be injective.

To be quite clear about how this fails in case $a$ is only a left ideal, take $R$ the endomorphism ring of an infinite-dimensional vector space and $a$ the left ideal of endomorphisms killing the supspace generated by $e_{2k}$ the standard basis vectors with even index. Then $f:R\cong R/a$ as left modules by sending the identity to the endomorphism $T:e_{2k-1}\mapsto 0, e_{2k}\mapsto e_k$. Since $T$ is surjective, $f$ is injective, in contradiction to the two-sided case.

Answer 2

Suppose that $R/a$ is free over $R$. Let $X$ be a basis of $R/a$ over $R$. Since $a$ is properly contained in $R$, we have that $R/a \neq 0$ and so $X \neq \emptyset$ as a generating set. Let $x + a \in X$. Since $X$ is linearly independent over $R$, we have in particular that $ \{x+a\}$ must be linearly independent over $R$. This means that $\forall r\in R$, $r(x+a)=a \implies r=0$ or equivalently $\forall r \in R, rx \in a \implies r=0$. But since $a$ in particular a right ideal, one has that $\forall r \in a, rx \in a$. So for $r \in a$, it follows that $r=0$ by the previous linear independence argument, and thus $a=0$.