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Let $k$ be a field, $R=k[x_1,\dots,x_r]$ the polynomial ring in $r$ indeterminates and $I_1,I_2\dots,I_d$ homogeneous ideals of $R$ generated by linear forms. Define $J = I_1 I_2 \cdots I_d$ and suppose that $\dim R/J = 0$.

Question: Is it true that $(R/J)_d = 0$? If yes, how can we see that?

user26857
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Manos
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1 Answers1

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Since the $I_i$ are generated by homogeneous linear forms, they are prime ideals. Since $J \subset I_i$, you have that $$\mathrm{ht}~J \leq \mathrm{ht}~I_i$$ for all $i$. Since you claim that $J$ has codimension $r$, it must be the case that each $I_i$ has height $r$, which in turn means it is equal to the homogenous maximal ideal $M =(x_1,\ldots,x_r)$. Now the result follows since $(R/M^d)_d =0$.

Adam Boocher
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  • Yes. For quotients of polynomial rings, it is true that $ht I = dim R - dim(R/I)$. To make the answer clearer, I'll update it. – Adam Boocher May 08 '14 at 22:11
  • My favorite first books are "Computational Algebraic Geometry by Hal Schenck" and "Ideals, Varieties and Algorithms" by Cox Little and O'Shea. – Adam Boocher May 08 '14 at 22:20