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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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