## 29.47 Absolute weak normalization and seminormalization

Motivated by the results proved in the previous section we give the following definition.

Definition 29.47.1. Let $A$ be a ring.

We say $A$ is *seminormal* if for all $x, y \in A$ with $x^3 = y^2$ there is a unique $a \in A$ with $x = a^2$ and $y = a^3$.

We say $A$ is *absolutely weakly normal* if (a) $A$ is seminormal and (b) for any prime number $p$ and $x, y \in A$ with $p^ px = y^ p$ there is a unique $a \in A$ with $x = a^ p$ and $y = pa$.

An amusing observation, see [Costa], is that in the definition of seminormal rings it suffices^{1} to assume the existence of $a$. Absolutely weakly normal schemes were defined in [Appendix B, rydh_descent].

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.
$\square$

Next we define seminormal schemes and absolutely weakly normal schemes.

Definition 29.47.3. Let $X$ be a scheme.

We say $X$ is *seminormal* if every $x \in X$ has an affine open neighbourhood $\mathop{\mathrm{Spec}}(R) = U \subset X$ such that the ring $R$ is seminormal.

We say $X$ is *absolutely weakly normal* if every $x \in X$ has an affine open neighbourhood $\mathop{\mathrm{Spec}}(R) = U \subset X$ such that the ring $R$ is absolutely weakly normal.

Here is the obligatory lemma.

Lemma 29.47.4. Let $X$ be a scheme. The following are equivalent:

The scheme $X$ is seminormal.

For every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is seminormal.

There exists an affine open covering $X = \bigcup U_ i$ such that each $\mathcal{O}_ X(U_ i)$ is seminormal.

There exists an open covering $X = \bigcup X_ j$ such that each open subscheme $X_ j$ is seminormal.

Moreover, if $X$ is seminormal then every open subscheme is seminormal. The same statements are true with “seminormal” replaced by “absolutely weakly normal”.

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

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.
$\square$

Lemma 29.47.6. Let $A$ be a ring.

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

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.

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

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 $A^{awn} \to B^{awn}$ 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.
$\square$

Lemma 29.47.7. Let $X$ be a scheme.

The category of universal homeomorphisms $Y \to X$ has an initial object $X^{awn} \to X$.

Given $Y \to X$ in the category of (1) the resulting morphism $X^{awn} \to Y$ is an isomorphism if and only if $Y$ is absolutely weakly normal.

The category of universal homeomorphisms $Y \to X$ which induce ismomorphisms on residue fields has an initial object $X^{sn} \to X$.

Given $Y \to X$ in the category of (3) the resulting morphism $X^{sn} \to Y$ is an isomorphism if and only if $Y$ is seminormal.

For any morphism $h : X' \to X$ of schemes there are unique morphisms $h^{awn} : (X')^{awn} \to X^{awn}$ and $h^{sn} : (X')^{sn} \to X^{sn}$ compatible with $h$.

**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.
$\square$

Definition 29.47.8. Let $X$ be a scheme.

The morphism $X^{sn} \to X$ constructed in Lemma 29.47.7 is the *seminormalization* of $X$.

The morphism $X^{awn} \to X$ constructed in Lemma 29.47.7 is the *absolute weak normalization* of $X$.

To be sure, the seminormalization $X^{sn}$ of $X$ is a seminormal scheme and the absolute weak normalization $X^{awn}$ is an absolutely weakly normal scheme. Moreover, for any morphism $h : Y \to X$ of schemes we obtain a canonical commutative diagram

\[ \xymatrix{ Y^{awn} \ar[d]^{h^{awn}} \ar[r] & Y^{sn} \ar[d]^{h^{sn}} \ar[r] & Y \ar[d]^ h \\ X^{awn} \ar[r] & X^{sn} \ar[r] & X } \]

of schemes; the arrows $h^{sn}$ and $h^{awn}$ are the unique ones compatible with $h$.

Lemma 29.47.9. Let $X$ be a scheme. The following are equivalent

$X$ is seminormal,

$X$ is equal to its own seminormalization, i.e., the morphism $X^{sn} \to X$ is an isomorphism,

if $\pi : Y \to X$ is a universal homomorphism inducing isomorphisms on residue fields with $Y$ reduced, then $\pi $ is an isomorphism.

**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 homomorphism 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.
$\square$

Lemma 29.47.10. Let $X$ be a scheme. The following are equivalent

$X$ is absolutely weakly normal,

$X$ is equal to its own absolute weak normalization, i.e., the morphism $X^{awn} \to X$ is an isomorphism,

if $\pi : Y \to X$ is a universal homomorphism with $Y$ reduced, then $\pi $ is an isomorphism.

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

## Comments (1)

Comment #8600 by Fanjun Meng on