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Let $(R,m,k)$ be a regular local ring of dimension $n$. Let $b_1,\dots,b_n$ be a maximal $R$-sequence and define $J=(b_1,\dots,b_n)$. Let $y_1,\dots,y_n$ be a regular system of parameters of $R$ and write $b_i = \sum b_{ji} y_j$. Then the $\operatorname{Hom}_R (k, R/J) \cong \det(b_{ji}) R/J$ (Corollary 2.3.10 in Bruns and Herzog, CMR).

Question: Why is it true that $\det(b_{ji})$ is inside every ideal $J'$ of $R$ such that $J' \supset J$?

Reference: Bruns and Herzog, CMR, second paragraph of proof of Theorem 2.3.16.

Stefan4024
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Manos
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  • Hi: just out of curiosity, what's the relation of the title to the body question? It escapes me at the moment :) I'm not familiar with regular local rings. Thanks! – rschwieb Feb 18 '14 at 18:04
  • Hi @rschwieb. For a local ring $(R,m,k)$ and an $R$-module $M$, the "socle" of $M$ is defined to be the $R$-module $(0:m)_M \cong Hom_R(k,M)$. – Manos Feb 18 '14 at 18:35
  • Ah ok thanks for letting me know this! I'm used to another definition of socle of a module: the sum of all the module's simple submodules. Maybe they're related or ultimately the same. The maximal ideal $m$ certainly annihilates the socle in my definition, so it's at least a submodule of your definition. – rschwieb Feb 18 '14 at 18:50
  • @rschwieb: Thanks for letting me know too, i was unaware of that definition. I suppose it would be a good question to ask whether the two coincide in the case of a local ring. – Manos Feb 18 '14 at 19:35
  • It's easy to show that both definitions of the socle coincide for modules over commutative local rings. – user26857 Feb 18 '14 at 21:30

1 Answers1

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Let $(S,\mathfrak m)$ be a local artinian ring with socle $(0:\mathfrak m)=aR$, $a\ne 0$. Then $aR\subseteq I$ for any ideal $I\ne (0)$.

If there is an $x\in I$, $x\ne 0$, such that $x\in aR$, then $x=au$ with $u$ invertible (otherwise $u\in\mathfrak m$ and $x\in a\mathfrak m=0$), so $a\in I$.

If for any $x\in I$, $x\ne 0$, we have $x\notin aR$ then for any $x\in I$, $x\ne 0$, there is $m\in\mathfrak m$ such that $xm\ne 0$. For a fixed such $x$ choose $m_1\in\mathfrak m$ such that $m_1x\ne 0$. Repeat the argument for $m_1x$ and find $m_2\in\mathfrak m$ such that $m_2m_1x\ne 0$, and so on. On the other side, $\mathfrak m$ is nilpotent, a contradiction!

user26857
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