The Stacks project

Lemma 29.47.2. Being seminormal or being absolutely weakly normal is a local property of rings, see Properties, Definition 28.4.1.

Proof. Suppose that $A$ is seminormal and $f \in A$. Let $x', y' \in A_ f$ with $(x')^3 = (y')^2$. Write $x' = x/f^{2n}$ and $y' = y/f^{3n}$ for some $n \geq 0$ and $x, y \in A$. After replacing $x, y$ by $f^{2m}x, f^{3m}y$ and $n$ by $n + m$, we see that $x^3 = y^2$ in $A$. Then we find a unique $a \in A$ with $x = a^2$ and $y = a^3$. Setting $a' = a/f^ n$ we get $x' = (a')^2$ and $y' = (a')^3$ as desired. Uniqueness of $a'$ follows from uniqueness of $a$. In exactly the same manner the reader shows that if $A$ is absolutely weakly normal, then $A_ f$ is absolutely weakly normal.

Assume $A$ is a ring and $f_1, \ldots , f_ n \in A$ generate the unit ideal. Assume $A_{f_ i}$ is seminormal for each $i$. Let $x, y \in A$ with $x^3 = y^2$. For each $i$ we find a unique $a_ i \in A_{f_ i}$ with $x = a_ i^2$ and $y = a_ i^3$ in $A_{f_ i}$. By the uniqueness and the result of the first paragraph (which tells us that $A_{f_ if_ j}$ is seminormal) we see that $a_ i$ and $a_ j$ map to the same element of $A_{f_ if_ j}$. By Algebra, Lemma 10.24.2 we find a unique $a \in A$ mapping to $a_ i$ in $A_{f_ i}$ for all $i$. Then $x = a^2$ and $y = a^3$ by the same token. Clearly this $a$ is unique. Thus $A$ is seminormal. If we assume $A_{f_ i}$ is absolutely weakly normal, then the exact same argument shows that $A$ is absolutely weakly normal. $\square$


Comments (0)


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Your email address will not be published.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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0EUM. Beware of the difference between the letter 'O' and the digit '0'.