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The following two on commutative algebra are true?

Let $S$ be a f.g. algebra over a field $k$. Let $e$ be an integer. Then (1) There is an ideal $I\subset S$ such that if $Q$ is a maximal ideal of $S$ then 

$dim S_Q\ge e$ iff $Q\supset I.$

(2) if $S=S_0\oplus S_1\oplus \cdots$ is a graded algebra, f.g. over $S_0=k$ then

$dim S\ge e.$


EDIT. (1) is done by Matt. I rewite (2).

Is it trivially true when $R$ is a field? If not, how should we modify it?

Theorem 14.8b(Eisenbud CA p316)

Let $S$ be a f.g. algebra over a Noetheian ring  $R$. Let $e$ be an integer. Then

(2) if $S=S_0\oplus S_1\oplus \cdots$ is a graded algebra, f.g. over $S_0=R$ then there exists an ideal $J$ of $R$ such that for any prime ideal $P\subset R$,

$dim K(R/P)\otimes S\ge e$ iff $P\supset J.$

Here $K(\cdot)$ means a quotient field.

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Let me explain my motivation. (2) seems to be wrong for $S=k[x]$. But Eisenbud's book CA p316L-2 says it is true. Ii is in proof for main theorem of elimination theory. –  Tom Sep 11 '12 at 0:52
    
The original one is as follows. Is it trivially true when $R$ is a field? Theorem 14.8 Let $S$ be a f.g. algebra over a Noetheian ring $R$. Let $e$ be an integer. Then (1) There is an ideal $I\subset S$ such that if $Q$ is a maximal ideal of $S$ and$P:=R\cap Q$ then $dim S_Q/PS_Q\ge e$ iff $Q\supset I.$ (2) if $S=S_0\oplus S_1\oplus \cdots$ is a graded algebra, f.g. over $S_0=R$ then there exists an ideal $J$ of $R$ such that for any prime ideal $P\subset R$, $dim S\ge e$ iff $P\supset J.$ –  Tom Sep 11 '12 at 2:15
    
Typo fixed. The original one is as follows. Is it trivially true when $R$ is a field? Theorem 14.8 Let $S$ be a f.g. algebra over a Noetheian ring $R$. Let $e$ be an integer. Then (1) There is an ideal $I\subset S$ such that if $Q$ is a maximal ideal of $S$ and$P:=R\cap Q$ then $dim S_Q/PS_Q\ge e$ iff $Q\supset I.$ (2) if $S=S_0\oplus S_1\oplus \cdots$ is a graded algebra, f.g. over $S_0=R$ then there exists an ideal $J$ of $R$ such that for any prime ideal $P\subset R$, $dim K(R/P)\otimes S\ge e$ iff $P\supset J.$ –  Tom Sep 11 '12 at 2:35
    
Dear Tom, I don't understand (2). You seem to have let $e$ be any integer, so certainly it can't be that $\dim S_0 \geq e$ in general (just take $e$ to be bigger than $\dim S_0$!). If $I$ is not the unit ideal (for a given value of $e$), then (2) is true. Regards, –  Matt E Sep 11 '12 at 3:14
1  
Dear Tom, Reading over your comments, I now see what Thm. 4.8 says. The point is that if $e$ is too large, then $J$ will be the unit ideal of $k$, rather than the $0$ ideal. In other words, in the case when $R = k$ (a field), one takes $J$ to be the zero ideal if $e \leq \dim S$, and one takes $J$ to be the unit ideal otherwise. So in this case, yes, the statement is trivial. Regards, –  Matt E Sep 11 '12 at 4:46

1 Answer 1

up vote 1 down vote accepted

Regarding (1): Look at Spec $S$. It is a union of finitely many irred. components, say $X_i$, of dimension $d_i$. Then the localisation of $S$ at $Q$ has dimension equal to the max of the $d_i$ for those $i$ for which $Q \in X_i$. Thus it has dimension $\geq e$ provided that $Q$ lies in at least one $X_i$ for which $d_i \geq e$. So, if we let $I$ be the ideal that cuts out those components of dimension $\geq e$, then $Q \supset I$ iff $S_Q$ has dimension $\geq e$.

If $I$ is not the unit ideal, i.e. if there actually is a component of dimension $\geq e$, then certainly $\dim S \geq e$. So this proves (a corrected variant of) (2).

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Sorry. My background knowledge is only Eisrnbud book p1-316 without some exercises skipped. So, I know the definition of Spec S only. Irreducible component of dimension $d$? –  Tom Sep 11 '12 at 4:58
    
Oh! There are only finitely many minimal ideals $I$. Let $d_i:=dim R/I.$ –  Tom Sep 11 '12 at 5:03
    
Ok. I think I understood (1). Thank you. –  Tom Sep 11 '12 at 5:05

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