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 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 (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 035Q. Beware of the difference between the letter 'O' and the digit '0'.