The Stacks project

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. $\square$

Proof. Combine Properties, Lemma 28.4.3 and Lemma 29.47.2. $\square$

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. $\square$

Proof. We prove (1) and (2) and we omit the proof of (3) and (4) and the final statement. Consider the category of $A$-algebras of the form

where $J$ is a finitely generated ideal such that $A \to B$ defines a universal homeomorphism on spectra. We claim this category is directed (Categories, Definition 4.19.1). Namely, given

then we can consider

where $J''$ is generated by the elements of $J$ and the elements $f(x_{n + 1}, \ldots , x_{n + n'})$ where $f \in J'$. Then we have $A$-algebra homomorphisms $B \to B''$ and $B' \to B''$ which induce an isomorphism $B \otimes _ A B' \to B''$. It follows from Lemmas 29.45.2 and 29.45.3 that $\mathop{\mathrm{Spec}}(B'') \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism and hence $A \to B''$ is in our category. Finally, given $\varphi , \varphi ' : B \to B'$ in our category with $B$ as displayed above, then we consider the quotient $B''$ of $B'$ by the ideal generated by $\varphi (x_ i) - \varphi '(x_ i)$, $i = 1, \ldots , n$. Since $\mathop{\mathrm{Spec}}(B') = \mathop{\mathrm{Spec}}(B)$ we see that $\mathop{\mathrm{Spec}}(B'') \to \mathop{\mathrm{Spec}}(B')$ is a bijective closed immersion hence a universal homeomorphism. Thus $B''$ is in our category and $\varphi , \varphi '$ are equalized by $B' \to B''$. This completes the proof of our claim. We set

where the colimit is over the category just described. Observe that $A \to A^{awn}$ induces a universal homeomorphism on spectra by Lemma 29.46.2 (this is where we use the category is directed).

Given a ring map $A \to B$ of finite presentation inducing a universal homeomorphism on spectra, we get a canonical map $B \to A^{awn}$ by the very construction of $A^{awn}$. Since every $A \to B$ as in (1) is a filtered colimit of $A \to B$ as in (1) of finite presentation (Lemma 29.46.11), we see that $A \to A^{awn}$ is final in the category (1).

Let $x, y \in A^{awn}$ be elements such that $x^3 = y^2$. Then $A^{awn} \to A^{awn}[t]/(t^2 - x, t^3 - y)$ induces a universal homeomorphism on spectra by Lemma 29.46.10. Thus $A \to A^{awn}[t]/(t^2 - x, t^3 - y)$ is in the category (1) and we obtain a unique $A$-algebra map $A^{awn}[t]/(t^2 - x, t^3 - y) \to A^{awn}$. The image $a \in A^{awn}$ of $t$ is therefore the unique element such that $a^2 = x$ and $a^3 = y$ in $A^{awn}$. In exactly the same manner, given a prime $p$ and $x, y \in A^{awn}$ with $p^ px = y^ p$ we find a unique $a \in A^{awn}$ with $a^ p = x$ and $pq = y$. Thus $A^{awn}$ is absolutely weakly normal by definition.

Finally, let $A \to B$ be in the category (1) with $B$ absolutely weakly normal. Since $A^{awn} \to B^{awn}$ induces a universal homeomorphism on spectra and since $A^{awn}$ is reduced (Lemma 29.47.5) we find $A^{awn} \subset B^{awn}$ (see Algebra, Lemma 10.30.6). If this inclusion is not an equality, then Lemma 29.46.6 implies there is an element $b \in B^{awn}$, $b \not\in A^{awn}$ such that either $b^2, b^3 \in A^{awn}$ or $pb, b^ p \in A^{awn}$ for some prime number $p$. However, by the existence and uniqueness in Definition 29.47.1 this forces $b \in A^{awn}$ and hence we obtain the contradiction that finishes the proof. $\square$

Proof. We will prove (1) and (2) and omit the proof of (3) and (4). Let $h : X' \to X$ be a morphism of schemes. If (1) holds for $X$ and $X'$, then $X' \times _ X X^{awn} \to X'$ is a universal homeomorphism and hence we get a unique morphism $(X')^{awn} \to X' \times _ X X^{awn}$ over $X'$ by the universal property of $(X')^{awn} \to X'$. Composed with the projection $X' \times _ X X^{awn} \to X^{awn}$ we obtain $h^{awn}$. If in addition (2) holds for $X$ and $X'$ and $h$ is an open immersion, then $X' \times _ X X^{awn}$ is absolutely weakly normal (Lemma 29.47.4) and we deduce that $(X')^{awn} \to X' \times _ X X^{awn}$ is an isomorphism.

Recall that any universal homeomorphism is affine, see Lemma 29.45.4. Thus if $X$ is affine then (1) and (2) follow immediately from Lemma 29.47.6. Let $X$ be a scheme and let $\mathcal{B}$ be the set of affine opens of $X$. For each $U \in \mathcal{B}$ we obtain $U^{awn} \to U$ and for $V \subset U$, $V, U \in \mathcal{B}$ we obtain a canonical isomorphism $\rho _{V, U} : V^{awn} \to V \times _ U U^{awn}$ by the discussion in the previous paragraph. Thus by relative glueing (Constructions, Lemma 27.2.1) we obtain a morphism $X^{awn} \to X$ which restricts to $U^{awn}$ over $U$ compatibly with the $\rho _{V, U}$. Next, let $Y \to X$ be a universal homeomorphism. Then $U \times _ X Y \to U$ is a universal homeomorphism for $U \in \mathcal{B}$ and we obtain a unique morphism $g_ U : U^{awn} \to U \times _ X Y$ over $U$. These $g_ U$ are compatible with the morphisms $\rho _{V, U}$; details omitted. Hence there is a unique morphism $g : X^{awn} \to Y$ over $X$ agreeing with $g_ U$ over $U$, see Constructions, Remark 27.2.3. This proves (1) for $X$. Part (2) follows because it holds affine locally. $\square$

Proof. The equivalence of (1) and (2) is clear from Lemma 29.47.7. If (3) holds, then $X^{sn} \to X$ is an isomorphism and we see that (2) holds.

Assume (2) holds and let $\pi : Y \to X$ be a universal homeomorphism inducing isomorphisms on residue fields with $Y$ reduced. Then there exists a factorization $X \to Y \to X$ of $\text{id}_ X$ by Lemma 29.47.7. Then $X \to Y$ is a closed immersion (by Schemes, Lemma 26.21.11 and the fact that $\pi$ is separated for example by Lemma 29.10.3). Since $X \to Y$ is also a bijection on points, the reducedness of $Y$ shows that it has to be an isomorphism. This finishes the proof. $\square$

Proof. This is proved in exactly the same manner as Lemma 29.47.9. $\square$