# What is the universal property of normalization?

What is the universal property of normalization? I'm looking for an answer something like

If X is a scheme and Y→X is its normalization, then the morphism Y→X has property P and any other morphism Z→X with property P factors uniquely through Y.

I've realized that my answer is wrong. Here's the right answer: if $Z$ is a normal scheme and $f: Z \to X$ is a morphism such that each associated point of $Z$ maps to an associated point of $X$, then $f$ factors through $n$.

A counterexample that shows why what I said previously doesn't work: let $f$ be the inclusion of the node into the nodal curve. There is no unique lift of $f$ to the normalization of the nodal curve.

What's going on here: taking the total ring of fractions is not a functor for arbitrary morphisms of reduced rings. You need a morphism such that no NZD gets mapped to a ZD, which is equivalent (for Noetherian rings) to saying that the preimage of any associated prime is an associated prime.

• I don't completely understand your explanation of what goes wrong in the case of the nodal cubic. What non-zero-divisors are being sent to zero-divisors? Oct 1 '09 at 21:11
• Ok, now I think I understand. We have a map k[x,y]/(y^2-x^3-x^2) --> k given by x,y-->0. But the normalization of the first ring is k[y/x], and since x gets mapped to a zero-divisor, you don't get an induced map on fraction fields. Oct 1 '09 at 22:00
• I have a dumb question -- what is an "associated point"? Oct 9 '09 at 1:58
• @Kevin: An associated prime of an R-module M is a prime which is equal to the annihilator of an element of M. Equivalently, a prime P is an associated prime of M if M contains a submodule isomorphic to R/P. An associated point of Spec(R) is an associated prime of R (regarded as a module over itself). An associated point of a scheme is a point which is an associated point in some affine open. The generic point of a component is always an associated point, and non-generic associated points (like (x,y) in k[x,y]/(x^2,xy)) are called embedded points. Oct 9 '09 at 6:15
• Suppose your scheme is locally Noetherian. Then an associated point is a point $p$ such that on any affine open set about it say $Spec(A)$, $p$ is an associated prime for $A$. (If $X$ is actually Noetherian, then it is quasi-compact, so there are only finitely many associated points).
– Matt
Feb 5 '10 at 4:23

If $X$ is a variety, the normalization $X'\to X$ is the maximal finite birational map to X, and the minimal dominant map from a normal variety to X, where maximal means any other finite birational map $Y\to X$ fits in a unique diagram $X'\to Y\to X$, and minimal means any other dominant map $Z\to X$ from a normal variety $Z$ fits in a unique diagram $Z\to X'\to X$.

In particular a variety is normal if it admits no non trivial finite birational maps. This gives you immediately that nodal curves are not normal, and neither are any varieties obtained by identifying points.

The meaning of normality is explained extremely clearly on page 181, in Mumford's unpublished second volume Algebraic geometry book, revised and edited now by Oda, and available online. http://www.math.upenn.edu/~chai/624_08/math624_08.html

The main geometric point is that normal points are locally irreducible, or "unibranch". The more delicate analytic aspect is the Hartog's extension theorem.

The key is Zariski's main theorem: I quote from Mumford-Oda:

"6. Zariski’s Main Theorem A second major reason why normality is important is that Zariski’s Main Theorem holds for general normal schemes. To understand this in its natural context, first consider the classical case: $k = C$, $X$ a $k$-variety, and $x$ is a closed point of $X$. Then we have the following two sets of properties:

N1) $X$ formally normal at $x$, i.e., $\mathcal{O}_{x,X}$ an integrally closed domain.

N2) $X$ analytically normal at $x$, i.e., $\mathcal{O}_{x,X,an}$, the ring of germs of holomorphic functions at $x$, is an integrally closed domain.

N3) $X$ normal at $x$.

N4) Zariski’s Main Theorem holds at $x$, i.e., $\forall \; f : Z \to X$, $f$ birational and of finite type with $f^{−1}(x)$ finite, then $\exists \; U \subset X$ Zariski-open with $x \in U$ and res $f : f^{−1}U \to U$ an isomorphism.

U1) $X$ formally unibranch at $x$, i.e., Spec $\mathcal{O}_{x,X}$ irreducible.

U2) $X$ analytically unibranch at $x$, i.e., Spec ($\mathcal{O}_{x,X,an}$) irreducible, or equivalently, the germ of analytic space defined by $X$ at $x$ is irreducible.

U3) $X$ unibranch at $x$, i.e., if $X'$ = normalization of $X$ in $R(X)$, $\pi : X' \to X$ the canonical morphism, then $\pi^{−1}(x)$ is a single point.

U4) $X$ topologically unibranch at $x$ — cf. Part I [76, (3.9)].

U5) The Connectedness Theorem holds at $x$, i.e., $\forall f : Z \to X$, $f$ proper, $Z$ integral, $f(\eta Z) = \eta X$ and $\exists U \subset X$ Zariski-open with $f^{−1}(y)$ connected for all $y \in U$, then $f^{−1}(x)$ is connected too.

6.1. I claim: i) all properties N are equivalent,

ii) all properties U are equivalent,

iii) $N \Rightarrow U$."

The reference  is to Mumford's Alg Geom I, Complex projective varieties. Compare also the discussion of ZMT in his red book.

• As implied by the word "formally", the local rings in statements N1) and U1) should be the completions. U4) means there exist arbitrarily small neighborhoods of x that are not disconnected by removing the singular locus of X. Feb 4 '18 at 2:42

Normalization is right adjoint to the inclusion functor from the category of normal schemes into the category of reduced schemes. In other words, if $n:Y\rightarrow X$ is the normalization of $X$ and $f:Z\rightarrow X$ is any morphism where $Z$ is a normal scheme, then $f$ factors uniquely through $n$.

• It seems clear now that you say it. If f:R->S, then you get a unique map R~ -> S~ by sending a/b to f(a)/f(b). I guess my problem is that I don't understand what it means for a scheme to be normal, but Dan Erman tells me that I shouldn't feel bad about that. Oct 1 '09 at 17:14
• By the way, seminormality does satisfy the analog this universal property as originally stated without hypotheses about generic points mapping to generic points. Well, I'm sure it seminormality satisfies it under very moderate geometric hypotheses -- probably locally excellent is enough. May 1 '10 at 2:23

I guess that you may not be worried about this anymore, but Prop. 12.44 in Görtz and Wedhorn's book answers your question (on integral schemes): For every integral and normal scheme Y' and every dominant morphism$f: Y' \to X$ there exists a unique morphism $g: Y' \to Y$ such that $n\circ g = f$.

• Typo: taking into account the notation of the question, the letter $Y$ is being asked to do too much. Jan 21 '16 at 11:25
• Hope now everything is correct, thanks for pointing it out! Jan 21 '16 at 13:04

I think it's okay if you rephrase like this: let X be a scheme with ring of fractions R. Then the normalization of X is the universal example of a normal scheme with ring of fractions R mapping to X.

• But I'd really like to have some universal property that tells me what maps to X look like, even if they don't have the same ring of fractions. Oct 1 '09 at 21:03

While this comes pretty late I wanted to add that normalization has a somewhat stronger universal property than what is mentioned in the other answers:

(N) Let $f: Z \to X$ be a morphism between integral, excellent schemes with $Z$ normal such that the image contains a normal point of $X$. Then $f$ lifts uniquely to the normalization $Y \to X$.

If the cited normal point is the generic point of $X$, one gets back the criterion demanding dominance (ND), as in Giulia's answer. However it also encompasses situations like a normal curve $Z$ on a singular surface $X$ not contained in the singular locus, which also lifts uniquely to the normalization.

To justify (N) one uses (ND) to reduce to $Z$ being the normalization of a closed integral subscheme $R \subset X$ whose generic point $\eta$ is normal in $X$. Denote by $R'$ the reduced closure of $\{\eta\}$ in the normalization of $R \times_X Y$. Because $\eta$ is a normal point of $X$, this is the unique component mapping dominantly to $R$. In the resulting diagram $\require{AMScd}$ $$\begin{CD} R' @>>> R\times_X Y@>>> Y\\ @VVV @VVV @VVV\\ Z @>>> R@>>> X \end{CD}$$ we get the left vertical arrow again by (ND) from $R' \to R$. However, as $X$ is excellent, the map $Y \to X$ is finite, hence $R' \to Z$ is a finite birational map to a normal scheme, so an isomorphism. This gives a lift $Z \to R \times_X Y$ in the above diagram. Any other such lift also induces a section of $R' \to Z$ by (ND) which implies uniqueness.