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Let $I=(f_1,\dots,f_r)\subset k[x_1,\dots,x_n]$ be an ideal with generators homogeneous linear polynomials and furthermore we assume that $f_1,\dots,f_r$ form a minimal generating set of $I$. Is the information above enough to tell that the height of this ideal is $r$?

If the direct proof is hard and a nontrivial theorem is needed, please let me know the source. Thanks!

No One
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1 Answers1

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Let $H_i$ be the vanishing set of $f_i$, $H_i$ is a linear hyperplane of $k^n$. The assumption on the $f_i$ means that no $H_i$ contains the intersection of the others, ie that the $f_i$ are linearly independent linear forms.

Now, $R=k[x_1,\ldots,x_n]/I$ is clearly the space of polynomial functions on $H:= \cap_{i=1}^r{H_i}$, (if $k$ is infinite; I think the argument can be refined with bases for finite fields but let’s ignore it for the sake of simplicity) which is a vector space of dimension $n-r$. So it is an integral domain of dimension $n-r$, therefore $I$ is prime and its height is at most $r$.

But better, actually, for the same reason $(0)$, $(f_1), \ldots , (f_1,\ldots,f_{r-1})$ is an increasing chain of prime ideals contained in $I$. So $I$ has height at least $r$, QED.

Aphelli
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  • Could you show me the source of the theorem(s) you are invoking about the relation between dimension of vector space and the height? Also what is the dimension of an integral domain? – No One Dec 12 '19 at 16:52
  • It’s the Krull dimension (the maximum height over all proper ideals). I suppose you can check out, eg Qing Liu, Algebraic Geometry and Arithmetic Curves, Section 2.5. – Aphelli Dec 12 '19 at 16:54
  • I just checked though Qing Liu's book sec 2.5 and didn't find the relation between the dimension of the vector space and the Krull dimension of the ring. I think this is the key part of the proof--- the association of the dimension of the space cut out by the linear polynomials and the Krull dimension/height. Do you think you can give a direct proof for this? – No One Dec 12 '19 at 17:18
  • You mean: let $V$ be a vector space with dimension $d$ and $A$ the ring of polynomial functions on $V$, then $A$ has Krull dimension $d$? But, given a basis $e_1,\ldots,e_d$ of $V$, we get an isomorphism $A \cong k[X_1,\ldots,X_d]$. – Aphelli Dec 12 '19 at 18:15