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Let $k$ be a field, $S=k[x_1,\dots,x_n]$ and $I$ a homogeneous ideal of $S$. Let $f_1,\dots,f_l$ be a minimal generating set of $I$ and let $d$ be the maximal degree among the degrees of the $f_i$. Then $d$ is an invariant of $I$.

Eisenbud in Commutative Algebra with a view toward Algebraic Geometry, page 509, claims that "the reader may easily check" that $d$ is also an invariant of the graded ring $S/I$. Any hints for proving this last statement?

Manos
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1 Answers1

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A minimal generating set of $I$ is a homogeneous $k$-basis of $I/\mathfrak mI$, where $m=(x_1,\dots,x_n)$. This shows that $d=\max\{i:(I/\mathfrak mI)_i\ne0\}$. Now use that $\hbox{Tor}_1^S(k,S/I)\simeq I/\mathfrak mI$.

  • Nice! But this proves that $d$ is an invariant of $I$, correct? I have managed to prove this in an alternative way, but not as elegant as yours. How about my last statement for $d$ being an invariant of $S/I$? This is were i am stuck. – Manos Dec 10 '13 at 21:04
  • Oh wait, i might be able to use what you wrote on the exact sequence $0 \rightarrow I \rightarrow S \rightarrow S/I \rightarrow 0$. – Manos Dec 10 '13 at 21:06
  • For the last isomorphism look here. And yes, that sequence is involved. –  Dec 10 '13 at 21:18
  • Got it! Its not as easy, though, as Eisenbud claimed :) – Manos Dec 10 '13 at 21:37