I have been staring at this proof in the Proceedings of the AMS, and I don't follow the author's logic.

Here's the setup:

$I$ is an ideal in a ring $R$, and it is projective as a right $R$ module. Therefore it is a summand in a free right $R$ module, $F=I\oplus K$.

Now, the line of reasoning continues (verbatim except for symbol changes):

Suppose $K\neq 0$. Since $F$ is isomorphic to a direct sum of copies of $R$, it has canonical multiplication. Let $\operatorname{Ann}_F(I)$ be the annihilator of $I$ in $F$. Then $KI\subseteq K\cap I=0$, so $\operatorname{Ann}_F(I)\neq 0$.

Now if $K$ was a right ideal of $R$, then $KI\subseteq K\cap I$ would be easy to understand. The only set $K$ and $I$ are comparable in $F$, so $(I\oplus 0)\cap( 0\oplus K)=0\oplus 0$ in the direct sum. But the left side is apparently multiplying $K$ through the $F$ module action, so that we're actually talking about $(0\oplus K)I$. Why would one say that's a subset of $I\oplus 0$?

Sure, every element of $(0\oplus K)I$, when expressed as a tuple in $F$ has entries in $I$, but as far as I can see, that doesn't mean anything about its membership in $I\oplus 0$.

I should also say that the ring $R$ is right self-injective and the ideal $I$ has zero left annihilator in $R$, but I'm not sure that it makes a difference. The author appeals to neither property in the line of thought above. In fact, that $I$ has zero left annihilator immediately allows you to say that $Ann_F(I)=0$, but since the whole point of this is to derive a contradiction, I need to see if there's any merit in what the author has claimed.

I haven't managed to cook up a counterexample yet, mostly because I have a hard time realizing projective ideals as summands in free modules. Am I missing some observation or is my doubt justified? I have a sneaking suspicion that a cognitive error occurred on the author's part.

  • $\begingroup$ It also seems like a sleight of hand to me. $\endgroup$ – xyzzyz Mar 19 '18 at 1:44

I think your doubt is justified.

Here’s a very simple example (where the ring is right self-injective and the ideal $I$ has zero left annihilator in $R$).

Let $R=k$, a field, and let $I=k$.

$I$ can be regarded as the (right $k$-module) summand of $F=k\oplus k$ spanned by $(1,1)$, with complement $K$ spanned by $(0,1)$.

Then the annihilator of $I$ in $F$ with the “canonical multiplication” (which I presume means coordinate-wise) is zero.

Of course, in this example you could make a different choice of the embedding of $I$ into $F$ so that it had non-zero annihilator. But the “canonical” multiplication, and hence the annihilator, depend on the choices you make.

  • $\begingroup$ In the second to last paragraph when you say coordinatewise, you mean you think it means $\{f\in F\mid f\cdot i=0 \forall i\in I\subset F \}$ where it's the coordinatewise product in $F$ as a ring without unity? I still have a hard time believing this, but since it's the second time it's been mentioned I'll have to give it more thought. $\endgroup$ – rschwieb Mar 28 '18 at 15:29
  • $\begingroup$ @rschwieb That’s what I thought. I may be wrong (I found the paper you’re referring to, and it’s not very clear), but I couldn’t think of an alternative meaning. Do you have an alternative in mind? $\endgroup$ – Jeremy Rickard Mar 28 '18 at 15:56
  • $\begingroup$ My gut insists that he was just referring to the normal module action of $R$ on $\oplus R$. The proof has at least one other probable mistake, and another weakness (which I think is patchable.) I'm just trying to make sure that I'm not shortchanging this person. $\endgroup$ – rschwieb Mar 28 '18 at 20:23
  • $\begingroup$ Well, the nice part about this proposed example (or at least the one I sketched out just now that's similar to it) is that it rules out both interpretations. The claim that $KI\subseteq I\cap K$ just doesn't hold up either way. $\endgroup$ – rschwieb Mar 28 '18 at 20:48
  • $\begingroup$ What I guess I'm really looking for is a group algebra with a projective augmentation ideal which is not a summand in $R[G]$ itself but is a summand in, say, $R[G]\times R[G]$, so I can really illustrate something is wrong. The example just using vector spaces isn't quite the same. I managed to catch, in one place or another, that having a projective augmentation ideal is equivalent to something about the orders of finite subgroups being regular or maybe units, but I haven't gotten the text yet. $\endgroup$ – rschwieb Mar 29 '18 at 15:03

Answer is abandoned. Only still present to support a future comment.

Aren't $K$ and $I$ ideals in the ring $F$?

(What's its multiplication? Back up two sentences.)

  • $\begingroup$ I'm not really convinced you're on to anything here. It seems farfetched that he intended to talk about the (possibly nonunital) ring $F$, but I'll consider it for a little bit. I interpreted "canonical multiplication" as the ordinary module multiplication given to free modules. $\endgroup$ – rschwieb Mar 26 '18 at 14:34
  • $\begingroup$ Here's why $K$ and $I$ are probably not "ideals in $F$": look, suppose $(\ldots, x_i,\ldots)\in K$. We know for sure that $(\ldots, x_ir,\ldots)\in K$ too for each $r\in R$, but what you're suggesting is that for any $(\ldots, r_i,\ldots)\in F$ we have $(\ldots, x_ir_i,\ldots)\in K$, which does not follow from the definition of $K$ as an $R$ module. $\endgroup$ – rschwieb Mar 26 '18 at 14:44
  • $\begingroup$ @rschwieb : Well, it was a pre-coffee "shot in the dark". I've had the hardest time recalling (or googling) the intended "canonical multiplication"; could you suggest a reference? $\endgroup$ – Eric Towers Mar 26 '18 at 17:07
  • $\begingroup$ That's OK! I'm glad for any eyes on the problem. Certainly doesn't hurt to rule out some possibilities. Frankly I'm not 100% sure what the author meant by "canonical multiplication," and like I said, I only interpreted it as the coordinate-wise action of $R$ on the free module. The author unfortunately does the same thing elsewhere, talking about "a canonical map $\oplus R\to R$ and a canonical map $R\to \oplus R$". Personally I would call the collection of injections and projections canonical, but I'm not aware of a single pair of maps labeled thusly. $\endgroup$ – rschwieb Mar 26 '18 at 17:51
  • $\begingroup$ @rschwieb : I'm used to "canonical map" meaning the maps from the relevant universal property commutative diagram. So for the two direct sum-related maps you mention, see this. I can't think of a relevant diagram for "canonical multiplication". Other than elementwise multiplication, I'm stumped by the phrase. This might be an appropriate question over at mathoverflow. $\endgroup$ – Eric Towers Mar 26 '18 at 20:15

Hint: look carefully the proof of the existence of $K$, $K$ is the kernel of the surjective morphism $F\rightarrow P$.


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