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. $\square$

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