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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.

Pedro A. Matos
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  • Are you sure this is true? If so I need to think a little harder about whether or not I found a counterexample. – Matt Samuel Jan 10 '15 at 00:56
  • Suppose $R\simeq R\oplus R$ as modules (this is true for some rings). Then the submodule $I$ generated by $(1,0)$ in $R\oplus R$ consists of all elements of the form $(b,0)$. When we go back to $R$, this should be an ideal, and $R/I$ is free. – Matt Samuel Jan 10 '15 at 01:01
  • @MattSamuel: Something's fishy here, because I believe my answer but I also don't see anything wrong with your example. – Jim Jan 10 '15 at 01:05
  • Nevermind, my answer assumes that the ideal has a non zero-divisor. – Jim Jan 10 '15 at 01:07
  • @Jim I think the example works, but I'm second guessing it because I'm assuming the OP has a good reason for thinking this is true. – Matt Samuel Jan 10 '15 at 01:10
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    The problem here was that no one said "commutative" yet. – Kevin Carlson Jan 10 '15 at 01:18

2 Answers2

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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.

Kevin Carlson
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  • I think @PVAL 's answer was right. I think for my example the supposed ideal must only be a left ideal, and really the result is true for all two sided ideals. – Matt Samuel Jan 10 '15 at 06:41
  • Yes, PVAL's answer would work for a two-sided ideal, and is much simpler than mine. – Kevin Carlson Jan 10 '15 at 22:12
  • @KevinCarlson I'm sorry for having ignored all this information until now. Your answer gives a conceptual proof, and I like it. But in fact, I just recently found what appears to be a simpler proof, and it actually only depends on $a$ being a proper right ideal (which means it is still true if we take $a$ to be a proper ideal) and includes the non-comutative case. I will post the proof as an answer to my question. – Pedro A. Matos Feb 28 '15 at 21:43
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    @PedroA.Matos I gave a counterexample to the claim when $a$ is only a left ideal, which can be easily modified to a counterexample in the right ideal case. – Kevin Carlson Feb 28 '15 at 22:32
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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 $a=0$.

Pedro A. Matos
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