The Stacks project

Lemma 29.47.6. Let $A$ be a ring.

  1. The category of ring maps $A \to B$ inducing a universal homeomorphism on spectra has a final object $A \to A^{awn}$.

  2. Given $A \to B$ in the category of (1) the resulting map $B \to A^{awn}$ is an isomorphism if and only if $B$ is absolutely weakly normal.

  3. The category of ring maps $A \to B$ inducing isomorphisms on residue fields and a universal homeomorphism on spectra has a final object $A \to A^{sn}$.

  4. Given $A \to B$ in the category of (3) the resulting map $B \to A^{sn}$ is an isomorphism if and only if $B$ is seminormal.

For any ring map $\varphi : A \to A'$ there are unique maps $\varphi ^{awn} : A^{awn} \to (A')^{awn}$ and $\varphi ^{sn} : A^{sn} \to (A')^{sn}$ compatible with $\varphi $.

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

\[ B = A[x_1, \ldots , x_ n]/J \]

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

\[ B = A[x_1, \ldots , x_ n]/J \quad \text{and}\quad B' = A[x_1, \ldots , x_{n'}]/J' \]

then we can consider

\[ B'' = A[x_1, \ldots , x_{n + n'}]/J'' \]

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

\[ A^{awn} = \mathop{\mathrm{colim}}\nolimits B \]

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 $A^{awn} \to B^{awn}$ is a universal homeomorphism 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. $\square$


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