The Stacks project

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

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 $B'$. $\square$

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. $\square$

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 $\square$

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

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

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. $\square$

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. $\square$

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. $\square$

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. $\square$