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The Stacks project

Being reduced is invariant under blowup

Lemma 10.70.9. If R is reduced then every (affine) blowup algebra of R is reduced.

Proof. Let I \subset R be an ideal and a \in I. Suppose x/a^ n with x \in I^ n is a nilpotent element of R[\frac{I}{a}]. Then (x/a^ n)^ m = 0. Hence a^ N x^ m = 0 in R for some N \geq 0. After increasing N if necessary we may assume N = me for some e \geq 0. Then (a^ e x)^ m = 0 and since R is reduced we find a^ e x = 0. This means that x/a^ n = 0 in R[\frac{I}{a}]. \square


Comments (2)

Comment #845 by on

Suggested slogan: Every (affine) blowup algebra of a reduced ring is reduced.

Comment #3589 by slogan_bot on

Suggested slogan: Being reduced is invariant under blowup

(Also: for Noetherian rings one could disect this into both (R) and (S) being invariant individually, if this is not already in the Stacks Project.)

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  • 1 comment(s) on Section 10.70: Blow up algebras

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