Lemma 29.47.5. A seminormal scheme or ring is reduced. A fortiori the same is true for absolutely weakly normal schemes or rings.
Proof. Let $A$ be a ring. If $a \in A$ is nonzero but $a^2 = 0$, then $a^2 = 0^2$ and $a^3 = 0^3$ and hence $A$ is not seminormal. $\square$
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