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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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    Oh, it is trivial! Thank you so much.2012-09-11

1 Answers 1

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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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    Ok.$I$think I understood (1). Thank you.2012-09-11