The Stacks project

Lemma 29.54.5. Let $X$ be a scheme such that every quasi-compact open has finitely many irreducible components.

  1. The normalization $X^\nu $ is a disjoint union of integral normal schemes.

  2. The morphism $\nu : X^\nu \to X$ is integral, surjective, and induces a bijection on irreducible components.

  3. For any integral morphism $\alpha : X' \to X$ such that for $U \subset X$ quasi-compact open the inverse image $\alpha ^{-1}(U)$ has finitely many irreducible components and $\alpha |_{\alpha ^{-1}(U)} : \alpha ^{-1}(U) \to U$ is birational1 there exists a factorization $X^\nu \to X' \to X$ and $X^\nu \to X'$ is the normalization of $X'$.

  4. For any morphism $Z \to X$ with $Z$ a normal scheme such that each irreducible component of $Z$ dominates an irreducible component of $X$ there exists a unique factorization $Z \to X^\nu \to X$.

Proof. Let $f : Y \to X$ be as in ( The scheme $X^\nu $ is a disjoint union of normal integral schemes because $Y$ is normal and every affine open of $Y$ has finitely many irreducible components, see Lemma 29.53.13. This proves (1). Alternatively one can deduce (1) from Lemmas 29.54.2 and 29.54.3.

The morphism $\nu $ is integral by Lemma 29.53.4. By Lemma 29.53.13 the morphism $Y \to X^\nu $ induces a bijection on irreducible components, and by construction of $Y$ this implies that $X^\nu \to X$ induces a bijection on irreducible components. By construction $f : Y \to X$ is dominant, hence also $\nu $ is dominant. Since an integral morphism is closed (Lemma 29.44.7) this implies that $\nu $ is surjective. This proves (2).

Suppose that $\alpha : X' \to X$ is as in (3). It is clear that $X'$ satisfies the assumptions under which the normalization is defined. Let $f' : Y' \to X'$ be the morphism ( constructed starting with $X'$. As $\alpha $ is locally birational it is clear that $Y' = Y$ and $f = \alpha \circ f'$. Hence the factorization $X^\nu \to X' \to X$ exists and $X^\nu \to X'$ is the normalization of $X'$ by Lemma 29.53.4. This proves (3).

Let $g : Z \to X$ be a morphism whose domain is a normal scheme and such that every irreducible component dominates an irreducible component of $X$. By Lemma 29.54.2 we have $X^\nu = X_{red}^\nu $ and by Schemes, Lemma 26.12.7 $Z \to X$ factors through $X_{red}$. Hence we may replace $X$ by $X_{red}$ and assume $X$ is reduced. Moreover, as the factorization is unique it suffices to construct it locally on $Z$. Let $W \subset Z$ and $U \subset X$ be affine opens such that $g(W) \subset U$. Write $U = \mathop{\mathrm{Spec}}(A)$ and $W = \mathop{\mathrm{Spec}}(B)$, with $g|_ W$ given by $\varphi : A \to B$. We will use the results of Lemma 29.54.3 freely. Let $\mathfrak p_1, \ldots , \mathfrak p_ t$ be the minimal primes of $A$. As $Z$ is normal, we see that $B$ is a normal ring, in particular reduced. Moreover, by assumption any minimal prime $\mathfrak q \subset B$ we have that $\varphi ^{-1}(\mathfrak q)$ is a minimal prime of $A$. Hence if $x \in A$ is a nonzerodivisor, i.e., $x \not\in \bigcup \mathfrak p_ i$, then $\varphi (x)$ is a nonzerodivisor in $B$. Thus we obtain a canonical ring map $Q(A) \to Q(B)$. As $B$ is normal it is equal to its integral closure in $Q(B)$ (see Algebra, Lemma 10.37.12). Hence we see that the integral closure $A' \subset Q(A)$ of $A$ maps into $B$ via the canonical map $Q(A) \to Q(B)$. Since $\nu ^{-1}(U) = \mathop{\mathrm{Spec}}(A')$ this gives the canonical factorization $W \to \nu ^{-1}(U) \to U$ of $\nu |_ W$. We omit the verification that it is unique. $\square$

[1] This awkward formulation is necessary as we've only defined what it means for a morphism to be birational if the source and target have finitely many irreducible components. It suffices if $X'_{red} \to X_{red}$ satisfies the condition.

Comments (4)

Comment #8524 by on

The statement of (3) can be written in a slightly more general way without changing the proof: restate "(3) For any birational integral morphism such that has locally finitely many irreducible components [i.e., every quasi-compact open of has finitely many irreducible components] there exists a factorization and is the normalization of ."

Regarding the proof of (4): I don't understand the sentence "as the factorization is unique it suffices to construct it locally on ." When we construct the morphism , shouldn't we say something about the compatibility of these morphisms? (So they can be glued to give rise to a morphism .) The following is the argument I devised: For , let be open affine, and let be open affine. We want to see that the morphisms agree on . Pick an open affine and an open affine . We have canonically defined ring morphisms and , where , are, respectively, the integral closures of , in , . We want to see that the diagram commutes. This can be done by studying this bigger diagram. In this latter diagram:

  1. The horizontal squares commute by construction.

  2. The back squares commute by functoriality of the total ring of fractions (with respect to ring morphisms which preserve nonzerodivisors). In particular, the open immersions , preserve generic points of irreducible components, so the maps on global sections—which are morphisms of reduced rings—preserve nonzerodivisors.

  3. The side squares commute because the map (resp., the map ) sends an integral dependence equation with coefficients in (resp., with coefficients in ) to an integral dependence equation with coefficients in (resp., in ).

  4. The front squares commute because the rest of the diagram commutes and using that is monic.

Comment #8530 by on

Did you read the footnote? It explains why the formulation in (3) is as given; your alternative formulation, although OK in spirit, does not work with the current definition of birational.

Uniqueness in (4) guarantees that the morphisms constructed will glue. Namely, if and are the morphisms constructed for open, then by uniqueness . This type of argument is used in a number of places in the Stacks project without further explanation.

Comment #8535 by on

@#8530 Thank you for the explanation. What got me confused with respect to the birationality issue is that in the proof, third paragraph, it is said "as is birational..." Nonetheless, none of or need to have a finite amount of irreducible components (so cannot be birational under the current definition of birationality, 29.50.1).

The same abuse of terminology happens in the proof of 29.54.6, when it is said " is integral and birational."

Comment #8536 by on

Yes, you are right that when we use the term birational in the proof here and in 29.54.6 we need to restrict to quasi-compact opens. I fixed it here by replacing "birational" by "locally birational". I think "locally birational" is sufficiently clear, but if multiple people disagree I will change it.

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