11
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ It also seems like a sleight of hand to me. $\endgroup$ – xyzzyz Mar 19 '18 at 1:44
2
+100
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ 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
-1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.