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Let $R$ be some local $k$-algebra, which is finite dimensional as a vector space. (Eisenbud lists 0-krull dimension as an extra hypothesis in this section, but it seems redundant - I think it follows from Noether normaliation...)

($k$ is a field.)

I am looking for an example of the following:

Some example of a short exact sequence over $R$ on which $Hom(\_, R)$ fails to be exact, and some $R$-module $M$ which is not reflexive. (I mean that $Hom(Hom(M,R),R)$ is not isomorphic to $M$ as an $R$-module.)

(I am reading the opening of Ch. 21 in Eisenbud, and he asserts these as possibilities if $R$ is bad enough. They are believable, but I like concreteness.)

I thought about $S = k[x]/(x^n)$, $n \not = 0$, but I think this ring is self-injective by Baer's criterion, so is not going to provide an example. (At least, $Hom(\_, S)$ exact will follow if $S$ is self-injective. Not sure if this will provide and obstruction to finitely generated modules not being reflexive.)

Maybe I need some more complicated ring, e.g. $k[x,y]/(x^2,xy,y^2)$ (stolen from Wikipedia as an example of a non-Gorenstein ring)? Not really sure what I should be trying to set up here - especially if I want to to find some module which is not reflexive.

Thanks!

(A hint will suffice - I would prefer to work out the computation on my own once I know what to compute...)

Elle Najt
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  • Have you tried $M = k$ the residue field? – Youngsu Mar 26 '16 at 04:09
  • Your guess of taking the ring $R=k[x,y]/(x^2,xy,y^2)$ is excellent and now follow Youngsu and try $M=k$. What is $Hom_R(k,R)$? This should immediately answer your question on reflexivity. For failure of exactness, try $0\to (x,y)\to R\to k\to 0$. – Mohan Mar 26 '16 at 04:14
  • The residue field was the first thing I tried with that second ring, but I must have computed something incorrectly: $Hom_R(k,R)$ is in natural isomorphism with the (x,y) torsion elements of $R$ which is the ideal $(x,y)$, by sending a morphism to the image of $1$. Now $Hom( (x,y), R)$ - ah, it contains more elements due to the symmetry in the variables, for example by switching the role of $x$ and $y$ (sending $ax + by$ to $ay + bx$). I overlooked this the first time. Thank you both! (I guess Baer's criterion is saying saying the short exact sequence example is "universal" in some sense.) – Elle Najt Mar 26 '16 at 04:39

1 Answers1

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Let $R=K[X,Y]/(X,Y)^2$, and $\mathfrak m=(x,y)$.

Both questions reduce to show $$\operatorname{Hom}_R(\mathfrak m,R)\not\simeq R/\mathfrak m.$$ (I don't understand from the comments how you proved this, so let me give a try.)

Since $\mathfrak m^2=(0)$ we deduce that $\mathfrak m$ is also an $R/\mathfrak m=K$-module (vector space). Moreover, $\dim_K\mathfrak m=2$ and hence $\mathfrak m\simeq K^2$. (This is an isomorphism of $K$-vector spaces, and of $R$-modules as well.) Then we have $\operatorname{Hom}_R(\mathfrak m,R)\simeq\operatorname{Hom}_R(K^2,R)\simeq\operatorname{Hom}_R(K,R)^2\simeq\mathfrak m^2$, and this shows us that the length of the $R$-module $\operatorname{Hom}_R(\mathfrak m,R)$ equals $2\ell(\mathfrak m)=4$. On the other side, $\ell(R/\mathfrak m)=1$, so $\operatorname{Hom}_R(\mathfrak m,R)\not\simeq R/\mathfrak m.$

user26857
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  • You are right in that my argument was not precise - I gave some reason to believe that $Hom_R(m,R)$ was "larger" than $R/m$, corresponding to precomposition by the automorphism of $K^2 \to K^2$ that sends $(a,b)$ to $(b,a)$. But of course this doesn't really show the lack of an isomorphism. Thank you for showing me how to do this computation! – Elle Najt Mar 26 '16 at 17:33