## 29.55 Weak normalization

We will only define the weak normalization of a scheme when it locally has finitely many irreducible components; similar to the case of normalization.

Lemma 29.55.1. Let $A \to B$ be a ring map inducing a dominant morphism $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ of spectra. There exists an $A$-subalgebra $B' \subset B$ such that

1. $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism,

2. given a factorization $A \to C \to B$ such that $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism, the image of $C \to B$ is contained in $B'$.

Proof. We will use Lemma 29.45.6 without further mention. Consider the commutative diagram

$\xymatrix{ B \ar[r] & B_{red} \\ A \ar[u] \ar[r] & A_{red} \ar[u] }$

For any factorization $A \to C \to B$ of $A \to B$ as in (2), we see that $A_{red} \to C_{red} \to B_{red}$ is a factorization of $A_{red} \to B_{red}$ as in (2). It follows that if the lemma holds for $A_{red} \to B_{red}$ and produces the $A_{red}$-subalgebra $B'_{red} \subset B_{red}$, then setting $B' \subset B$ equal to the inverse image of $B'_{red}$ solves the lemma for $A \to B$. This reduces us to the case discussed in the next paragraph.

Assume $A$ and $B$ are reduced. In this case $A \subset B$ by Algebra, Lemma 10.30.6. Let $A \to C \to B$ be a factorization as in (2). Then we may apply Proposition 29.46.8 to $A \subset C$ to see that every element of $C$ is contained in an extension $A[c_1, \ldots , c_ n] \subset C$ such that for $i = 1, \ldots , n$ we have

1. $c_ i^2, c_ i^3 \in A[c_1, \ldots , c_{i - 1}]$, or

2. there exists a prime number $p$ with $pc_ i, c_ i^ p \in A[c_1, \ldots , c_{i - 1}]$.

Thus property (2) holds if we define $B' \subset B$ to be the subset of elements $b \in B$ which are contained in an extension $A[b_1, \ldots , b_ n] \subset B$ such that (*) holds: for $i = 1, \ldots , n$ we have

1. $b_ i^2, b_ i^3 \in A[b_1, \ldots , b_{i - 1}]$, or

2. there exists a prime number $p$ with $pb_ i, b_ i^ p \in A[b_1, \ldots , b_{i - 1}]$.

There are only two things to check: (a) $B'$ is an $A$-subalgebra, and (b) $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism. Part (a) follows because given $n \geq 0$ and $b_1, \ldots , b_ n \in B$ satisfying (*) and $m \geq 0$ and $b'_1, \ldots , b'_ m \in B$ satisfying (*), the integer $n + m$ and $b_1, \ldots , b_ n, b'_1, \ldots , b'_ m \in B$ also satisfies (*). Finally, part (b) holds by Proposition 29.46.8 and our construction of $B'$. $\square$

Lemma 29.55.2. Let $A \to B$ be a ring map inducing a dominant morphism $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ of spectra. Formation of the $A$-subalgebra $B' \subset B$ in Lemma 29.55.1 commutes with localization (see proof for explanation).

Proof. Let $S \subset A$ be a multiplicative subset. Then $S^{-1}A \to S^{-1}B$ is a ring map which induces a dominant morphism $\mathop{\mathrm{Spec}}(S^{-1}B) \to \mathop{\mathrm{Spec}}(S^{-1}A)$ as well (see Lemmas 29.8.4 and 29.25.9). Hence Lemma 29.55.1 produces an $S^{-1}A$-subalgebra $(S^{-1}B)' \subset S^{-1}B$. The statement means that $S^{-1}B' = (S^{-1}B)'$ as $S^{-1}A$-subalgebras of $S^{-1}B$.

To see this is true, we will use the construction of $B'$ and $(S^{-1}B)'$ in the proof of Lemma 29.55.1. In the first step, we see that $B'$ is the inverse image of the $A_{red}$-subalgebra $B'_{red} \subset B_{red}$ constructed for the ring map $A_{red} \to B_{red}$ and similarly for $(S^{-1}B)'$. Noting that $S^{-1}B_{red} = (S^{-1}B)_{red}$ this reduces us to the case discussed in the next paragraph.

If $A$ and $B$ are reduced, we have constructed $B'$ as the union of the subalgebras $A[b_1, \ldots , b_ n]$ such that for $i = 1, \ldots , n$ we have

1. $b_ i^2, b_ i^3 \in A[b_1, \ldots , b_{i - 1}]$, or

2. there exists a prime number $p$ with $pb_ i, b_ i^ p \in A[b_1, \ldots , b_{i - 1}]$.

Similarly for $(S^{-1}B)' \subset S^{-1}B$. Thus it is clear that the image of $B' \to B \to S^{-1}B$ is contained in $(S^{-1}B)'$. To show that the corresponding map $S^{-1}B' \to (S^{-1}B)'$ is surjective, one uses Lemma 29.46.3 to clear denominators successively; we omit the details. $\square$

Lemma 29.55.3. Let $A \to B$ be a ring map inducing a dominant morphism $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ of spectra. There exists an $A$-subalgebra $B' \subset B$ such that

1. $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism inducing isomorphisms on residue fields,

2. given a factorization $A \to C \to B$ such that $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism inducing isomorphisms on residue fields, the image of $C \to B$ is contained in $B'$.

Proof. This proof is exactly the same as the proof of Lemma 29.55.1 except we use Proposition 29.46.7 in stead of Proposition 29.46.8 $\square$

Lemma 29.55.4. Let $A \to B$ be a ring map inducing a dominant morphism $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ of spectra. Formation of the $A$-subalgebra $B' \subset B$ in Lemma 29.55.3 commutes with localization (see proof for explanation).

Proof. The proof is the same as the proof of Lemma 29.55.2. $\square$

Lemma 29.55.5. Let $f : Y \to X$ be a quasi-compact, quasi-separated, and dominant morphism of schemes.

1. The category of factorizations $Y \to X' \to X$ where $X' \to X$ is a universal homeomorphism has an initial object $Y \to X^{Y/wn} \to X$.

2. The category of factorizations $Y \to X' \to X$ where $X' \to X$ is a universal homeomorphism inducing isomorphisms on residue fields has an initial object $Y \to X^{Y/sn} \to X$.

Moreover, formation of the factorization $Y \to X^{Y/wn} \to X$ and $Y \to X^{Y/sn} \to X$ commutes with base change to open subschemes of $X$.

Proof. We will prove (1) and omit the proof of (2); also the final assertion will follow from the construction of the factorization. We will use Lemma 29.45.5 without further mention. First, let $Y \to X^{Y/n} \to X$ be the normalization of $X$ in $Y$, see Definition 29.53.3. For $Y \to X' \to X$ as in (1), we obtain a unique morphism $X^{Y/n} \to X'$ compatible with the given morphisms, see Lemma 29.53.4. Thus it suffices to prove the lemma with $f$ replaced by $X^{Y/n} \to X$. This reduces us to the case studied in the next paragraph.

Assume $f$ is integral (the rest of the proof works more generally if $f$ is affine). Let $U = \mathop{\mathrm{Spec}}(A)$ be an affine open of $X$ and let $V = f^{-1}(U) = \mathop{\mathrm{Spec}}(B)$ be the inverse image in $Y$. Then $A \to B$ is a ring map which induces a dominant morphism on spectra. By Lemma 29.55.1 we obtain an $A$-subalgebra $B' \subset B$ such that setting $U^{V/wn} = \mathop{\mathrm{Spec}}(B')$ the factorization $V \to U^{V/wn} \to U$ is initial in the category of factorizations $V \to U' \to U$ where $U' \to U$ is a universal homeomorphism.

If $U_1 \subset U_2 \subset X$ are affine opens, then setting $V_ i = f^{-1}(U_ i)$ we obtain a canonical morphism

$\rho _{U_1}^{U_2} : U_1^{V_1/wn} \to U_1 \times _{U_2} U_2^{V_2/wn}$

over $U_1$ by the universal property of $U_1^{V_1/wn}$. These morphisms satisfy a natural functoriality which we leave to the reader to formulate and prove. Furthermore, the morphism $\rho _{U_1}^{U_2}$ is an isomorphism; this follows from Lemma 29.55.2 provided that $U_1 \subset U_2$ is a standard open and in the general case can be reduced to this case by the functorial nature of these maps and Schemes, Lemma 26.11.5 (details omitted). Thus by relative glueing (Constructions, Lemma 27.2.1) we obtain a morphism $X^{Y/wn} \to X$ which restricts to $U^{V/wn} \to U$ over $U$ compatibly with the $\rho _{U_1}^{U_2}$. Of course, the morphisms $V \to U^{V/wn}$ glue to a morphism $Y \to X^{Y/wn}$ (see Constructions, Remark 27.2.3) and we get our factorization $Y \to X^{Y/wn} \to X$ where the second morphism is a universal homeomorphism.

Finally, let $Y \to X' \to X$ be a factorization as in (1). With $V \to U^{V/wn} \to U \subset X$ as above, we obtain a factorization $V \to U \times _ X X' \to U$ where the second arrow is a universal homeomorphism and we obtain a unique morphism $g_ U : U^{V/wn} \to U \times _ X X'$ over $U$ by the universal property of $U^{V/wn}$. These $g_ U$ are compatible with the morphisms $\rho _{U_1}^{U_2}$; details omitted. Hence there is a unique morphism $g : X^{Y/wn} \to X'$ over $X$ agreeing with $g_ U$ over $U$, see Constructions, Remark 27.2.3. This proves that $Y \to X^{Y/wn} \to X$ is initial in our category and the proof is complete. $\square$

Definition 29.55.6. Let $f : Y \to X$ be a quasi-compact, quasi-separated, and dominant morphism of schemes.

1. The factorization $Y \to X^{Y/sn} \to X$ constructed in Lemma 29.55.5 part (2) is the seminormalization of $X$ in $Y$.

2. The factorization $Y \to X^{Y/wn} \to X$ constructed in Lemma 29.55.5 part (1) is the weak normalization of $X$ in $Y$.

Here is a way to reinterpret the seminormalization of a scheme which locally has finitely many irreducible components.

Lemma 29.55.7. Let $X$ be a scheme such that every quasi-compact open has finitely many irreducible components. Let $\nu : X^\nu \to X$ be the normalization of $X$. Then the seminormalization of $X$ in $X^\nu$ is is the seminormalization of $X$. In a formula: $X^{sn} = X^{X^\nu /sn}$.

Proof. Let $f : Y \to X$ be as in (29.54.0.1) so that $X^\nu$ is the normalization of $X$ in $Y$. The seminormalization $X^{sn} \to X$ of $X$ is the initial object in the category of universal homeomorphisms $X' \to X$ inducing isomorphisms on residue fields. Since $Y$ is the disjoint union of the spectra of the residue fields at the generic points of irreducible components of $X$, we see that for any $X' \to X$ in this category we obtain a canonical lift $f' : Y \to X'$ of $f$. Then by Lemma 29.53.4 we obtain a canonical morphism $X^\nu \to X'$. Whence in turn a canonical morphism $X^{X^\nu /sn} \to X'$ by the universal property of $X^{X^\nu /sn}$. In this way we see that $X^{X^\nu /sn}$ satisfies the same universal property that $X^{sn}$ has and we conclude. $\square$

Lemma 29.55.7 motivates the following definition. Since we have only constructed the normalization in case $X$ locally has finitely many irreducible components, we will also restrict ourselves to that case for the weak normalization.

Definition 29.55.8. Let $X$ be a scheme such that every quasi-compact open has finitely many irreducible components. We define the weak normalization of $X$ as the weak normalization

$X^\nu \longrightarrow X^{wn} \longrightarrow X$

of $X$ in the normalization $X^\nu$ of $X$ (Definition 29.54.1). In a formula: $X^{wn} = X^{X^\nu /wn}$.

Combined with Lemma 29.55.7 we see that for a scheme $X$ which locally has finitely many irreducible components there are canonical morphisms

$X^\nu \to X^{wn} \to X^{sn} \to X$

Having made this definition, we can say what it means for a scheme to be weakly normal (provided it has locally finitely many irreducible components).

Definition 29.55.9. Let $X$ be a scheme such that every quasi-compact open has finitely many irreducible components. We say $X$ is weakly normal if the weak normalization $X^{wn} \to X$ is an isomorphism (Definition 29.55.8).

It follows immediately from the definitions that for a scheme $X$ such that every quasi-compact open has finitely many irreducible components we have

$X\text{ normal} \Rightarrow X\text{ weakly normal} \Rightarrow X\text{ seminormal}$

We can work out the meaning of weak normality in the affine case as follows.

Lemma 29.55.10. Let $X = \mathop{\mathrm{Spec}}(A)$ be an affine scheme which has finitely many irreducible components. Then $X$ is weakly normal if and only if

1. $A$ is seminormal (Definition 29.47.1),

2. for a prime number $p$ and $z, w \in A$ such that (a) $z$ is a nonzerodivisor, (b) $w^ p$ is divisible by $z^ p$, and (c) $pw$ is divisible by $z$, then $w$ is divisible by $z$.

Proof. Assume $X$ is weakly normal. Since a weakly normal scheme is seminormal, we see that (1) holds (by our definition of weakly normal schemes). In particular $A$ is reduced. Let $p, z, w$ be as in (2). Choose $x, y \in A$ such that $z^ p x = w^ p$ and $zy = pw$. Then $p^ p x = y^ p$. The ring map $A \to C = A[t]/(t^ p - x, pt - y)$ induces a universal homeomorphism on spectra. The normalization $X^\nu$ of $X$ is the spectrum of the integral closure $A'$ of $A$ in the total ring of fractions of $A$, see Lemma 29.54.3. Note that $a = w/z \in A'$ because $a^ p = x$. Hence we have an $A$-algebra homomorphism $A \to C \to A'$ sending $t$ to $a$. At this point the defining property $X = X^{wn} = X^{X^\nu /wn}$ of being weakly normal tells us that $C \to A'$ maps into $A$. Thus we find $a \in A$ as desired.

Conversely, assume (1) and (2). Let $A'$ be as in the previous paragraph. We have to show that $X^{X^\nu /wn} = X$. By construction in the proof of Lemma 29.55.1, the scheme $X^{X^\nu /wn}$ is the spectrum of the subring of $A'$ which is the union of the subrings $A[a_1, \ldots , a_ n] \subset A'$ such that for $i = 1, \ldots , n$ we have

1. $a_ i^2, a_ i^3 \in A[a_1, \ldots , a_{i - 1}]$, or

2. there exists a prime number $p$ with $pa_ i, a_ i^ p \in A[a_1, \ldots , a_{i - 1}]$.

Then we can use (1) and (2) to inductively see that $a_1, \ldots , a_ n \in A$; we omit the details. Consequently, we have $X = X^{X^\nu /wn}$ and hence $X$ is weakly normal. $\square$

Here is the obligatory lemma.

Lemma 29.55.11. Let $X$ be a scheme such that every quasi-compact open has finitely many irreducible components. The following are equivalent:

1. The scheme $X$ is weakly normal.

2. For every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ satisfies conditions (1) and (2) of Lemma 29.55.10.

3. There exists an affine open covering $X = \bigcup U_ i$ such that each ring $\mathcal{O}_ X(U_ i)$ satisfies conditions (1) and (2) of Lemma 29.55.10.

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

Moreover, if $X$ is weakly normal then every open subscheme is weakly normal.

Proof. The condition to $X$ be weakly normal is that the morphism $X^{wn} = X^{X^\nu /wn} \to X$ is an isomorphism. Since the construction of $X^\nu \to X$ commutes with base change to open subschemes and since the construction of $X^{X^\nu /wn}$ commutes with base change to open subschemes of $X$ (Lemma 29.55.5) the lemma is clear. $\square$

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