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Let $(R,m)$ be a local ring. Let $x_1,...,x_d$ be a maximal $R$-sequence. Is $\operatorname{Hom}_R(R/m,R/(x_1,...,x_d))$ isomorphic to $R/m$?

user26857
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user177523
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If $(R,m)$ is a local ring and $M$ is an $R$-module, it is true that $\Phi(M) = Hom_R(R/m,M) \cong \left\{x \in M: \, mx =0\right\}$. $\Phi(M)$ is called as the "socle" of $M$ by some authors, e.g. by Bruns & Herzog. Note that the socle is always an $R/m$-vector space.

I don't believe the statement of the question to be true, since otherwise, the socle $\Phi(R)=\left\{x \in R : \, m x =0 \right\}$ of any Artinian local ring $(R,m)$ would be $1$-dimensional: the only maximal $R$-sequence of an Artinian local ring is the empty sequence and the statement would imply that $\Phi(R) \cong R/m$.

Here is a counterexample: Let $R = k[x,y] / (x^2,xy,y^2)$. Then $R$ is artinian local with maximal ideal $m=(\bar{x},\bar{y})$. Its socle is the homogeneous part of $R$ of degree $1$ and this has dimension $2$ as a $k$-vector space.

Manos
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    For a CM local ring this means that the ring is Gorenstein. For a counterexample take any local CM ring which is not Gorenstein. – user26857 Feb 07 '15 at 18:26
  • But $<x,y>^2$ is not maximal R sequence – user177523 Feb 08 '15 at 06:29
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    @SelvaRaja: Correct, it is not. But my point is that i am applying the statement directly to $R$: since it is Artinian, the only $R$-sequence is the empty sequence, so the statement, if it were true, it should apply directly on $R$, and that is not the case. Does that make sense? – Manos Feb 08 '15 at 07:11