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I'm searching for an example of a module, that is not projective for $k[X,Y]/(X^3,Y^5)$, but projective for the two subalgebras $k[X]/(X^3)$ and $k[Y]/(Y^5)$. (I don't think it is relevant, but in this case the characteristic of the underlying field should be $5$ or $3$).

I suppose that such an example exists, but don't know for sure. If $X$ and $Y$ have the same order which is the cardinality of the field, then such an example exists of course. Anybody has an idea for this inhomogeneous case?

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The short form of my answer is: (1) no submodule of the free module of rank 1 works, and (2) I think (but am not sure) that it suffices to consider the free module of rank 1. – Jack Schmidt Oct 27 '11 at 16:18
up vote 3 down vote accepted

I now found an example of such a module myself:

Take the module with generators $u,v,w$. The $Y$-action just generates new basis vectors and the $X$-action is defined as follows: $XY^iu=Y^iv$ for all $i$, $XY^iv=Y^{i+1}w$ for all $i$ and $Xw=Y^4u$.

This is projective as a module for $k[X]/X^3$ and $k[Y]/Y^5$, but non-projective (non-free) for $k[X,Y]/X^3,Y^5$, since it is $15$-dimensional, but has two-dimensional socle (generated by $Y^4v$ and $Y^4w$).

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Here is a limited no answer.

All three rings $k[x,y]/(x^3,y^5)$, $k[x]/(x^3)$, and $k[y]/(y^5)$ are local, so projective = free.

The dimension of a module that is projective for both of the small rings, must be a multiple of 15, and I suspect it is clear that the action must also fit together so that the resulting module is free over the original ring.

The regular module contains no proper, nonzero submodule that is projective over both subrings (just by dimension count). Over a group ring this would mean the calculation was finished, I believe, but perhaps one has to be more careful in general. Below I give what I hope is a reduction to this case.

If the two powers are the same, then the least common multiple can be quite a bit smaller. For $k[x,y]/(xx,yy)$ one has the submodule generated by $x+y$ is free over both subrings $k[x]/(xx)$ and $k[y]/(yy)$, but of course is not free over the original (local) ring due to dimension problems.

Perhaps this calculation works can be made to work in general: $$R = k[x,y]/(x^3,y^5) = k[x]/(x^3) \otimes k[y]/(y^5) = X \otimes Y$$ If $M$ is $X$-free, then $X\otimes M \cong M$, and if $M$ is $Y$-free, then $Y \otimes M \cong M$. Hence $$R \otimes M \cong (X \otimes Y) \otimes M \cong X \otimes ( Y \otimes M ) \cong X \otimes M \cong M$$ Something goes wrong (isomorphic over which ring?) since this doesn't use the relatively prime hypothesis, but hopefully it reduces the question to the $R$-module $R$, where it is clear in the relatively prime case.

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