Lemma 10.69.6. If $R$ is reduced then every (affine) blowup algebra of $R$ is reduced.

** Being reduced is invariant under blowup **

**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 $R[\frac{I}{a}]$.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #845 by Johan Commelin on

Comment #3589 by slogan_bot on

There are also: