Suppose $I\oplus K$ is a free module, then "$KI\subseteq K\cap I$" 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.
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.
Hint: look carefully the proof of the existence of $K$, $K$ is the kernel of the surjective morphism $F\rightarrow P$.