# Algebraic vs analytic normality

Let $X$ be a complex algebraic variety. We can ask if $X$ is normal as an algebraic variety, but also, if its analytification is normal as a complex analytic space. Is there a relationship between the two?

Do we have $$\text{algebraic normality} \implies\text{analytic normality}$$ or $$\text{analytic normality} \implies\text{algebraic normality}$$ or both or neither?

Over $\mathbf{C}$, algebraic normalization and analytic normalization are equivalent concepts. See

N. Kuhlmann: Die Normalisierung komplexer Räume, Math. Ann. 144 (1961), 110-125, ZBL0096.27801.

Quoting directly from Satz 4, p. 122 of the aforementioned paper:

Es sei $Y$ eine irreduzible algebraische Varietät über dem Grundkörper $\mathbf{C}$ der komplexen Zahlen, $Y^*$ der zugeordnete komplexe Raum, $Y'$ die Normalisierung von $Y$ (im Sinne der algebraischen Geometrie). Dann ist der $Y'$ zugeordnete komplexe Raum die Normalisierung von $Y^*$ (im analytischen Sinne).

The english translation is straightforward:

Let $Y$ be an irreducible algebraic variety over the field $\mathbb{C}$ of complex numbers, $Y^*$ the associated complex space, $Y'$ the normalization of $Y$ (in the sense of algebraic geometry). Then the complex space associated with $Y'$ is the normalization of $Y^*$ (in the analytic sense).

In particular, $Y$ is normal (in the algebraic sense) if and only if its associated complex space $Y^*$ is normal (in the analytic sense).

• OP asks about $\mathbb{C}$. Is the situation over other fields known? (This may be a big enough follow-up to be a new Question, so I'm fine with a "too big to answer in a comment" response.) Jun 22 '18 at 18:44
• This is a very interesting question, of which I do not know the answer :-) Jun 22 '18 at 20:40
• However, I guess that on an algebraic closed field of characteristic zero nothing changes, by Lefschetz principle. Jun 22 '18 at 20:41
• @EricTowers What do you mean by other fields? This statement is correct as long as an analytic local ring $\mathcal O_{X^{an},x}$ is excellent for any closed point $x$ (also you need to have a reasonable notion of analytic geometry over this field even to make sense of this question). For example, this holds for non-archimedean complete fields of rank 1 and complex numbers. I don't know whether it holds in the real-analytic situation, but I guess it is false since we don't even have any analogue of Oka's theorem as far as I know (I may be wrong).
– gdb
Jun 24 '18 at 4:55

Francesco Polizzi's answer is perfectly fine, but let me try to explain the technique which helps to relate a lot of "local" properties of locally finite type schemes over $\mathbf C$ to their counterparts in the world of complex analytic spaces.

For example, the proof that I will explain below actually shows that a locally finite type scheme $X$ over $\mathbf C$ is reduced/normal/Cohen-Macauley/Gorenstein/regular iff it's analytification $X^{an}$ has the same property. Moreover, exactly the same method works in the non-archimedean situation.

The first difficulty in showing equivalence of analytic and algebraic definitions of normality is that we can't directly relate the algebraic local ring $\mathcal O_{X,x}$ to its analytic analogue $\mathcal O_{X^{an},x}$ for a closed point $x\in X$. The second difficulty is that for a non-closed point $x\in X$ we don't even have any analogue of $\mathcal O_{X,x}$ in the analytic world. So, we should be able to prove that we can test normality (regularity/reducedness/so on) of a locally finite type $\mathbf C$-scheme only on closed points. (This is a subtle question, for instance, this is a celebrated theorem of Serre that we can check regularity on closed points for an arbitrary noetherian scheme, though the locus of regularity is not always open).

What helps us to overcome both difficulties is the notion of an excellent ring. I don't want to recall the definition here, you can read it here (and you can read a self-contained exposition of the main results about excellent rings in Chapter 13 of Matsumura's book "Commutative algebra"). But let me try to explain the main features of excellent rings:

1) If $A$ is a local excellent ring with a maximal ideal $\mathfrak m$, then the completion map $A \to \hat A_{\mathfrak m}$ is regular (meaning that it is flat and all fibers are geometrically regular).

2) The normalization morphism $\operatorname{Spec} A^{norm} \to \operatorname{Spec} A$ is a finite morphism.

3) Any essentially finite type algebra (a direct limit of finite type algebras) over an excellent ring is excellent (This is a real theorem, see Thm. 77 in Matsumura's book for a proof).

4) For any analytic space $X$ and a point $x$, a local ring $\mathcal O_{X,x}$ is excellent. This is a serious theorem, it is rather hard even to prove that this ring is noetherian (see the proof of Oka's theorem in Grauert-Remmert's book "Analytic Coherent Sheaves"). Unfortunately, I don't know any good reference for the excellence of this ring. Let me just emphasize that the proof that I know uses regularity and noetherianness of $\mathcal O_{\mathbf C^n,0}$ as an input for this fact.

With all these said, let me move to the real arguments. First of all, let me address the second issue from above. The definition of normality is local on both sides, so wlog we can assume that $X=\operatorname{Spec} A$. Then using (2) one can easily check that the locus where $X$ is normal is open in $X$ (this locus is exactly the locus where the natural morphism $X^{norm}\to X$ is an isomorphism). Also, you may look at EGA IV$_3$ 7.8.6 for a generalization of this result. Now to finish the argument we need to note that $X$ is a locally finite type scheme over complex numbers, so if an open subset of $X$ contains all closed points it must be equal to the whole scheme $X$. This solves the second problem and allows to restrict our attention only to closed points on the algebraic side.

Now, the key idea comes into play. We want to relate normality of $\mathcal O_{X,x}$ to normality of $\mathcal O_{X^{an}, x}$ for a closed point $x\in X$. As I pointed out before, we can't do this directly, but what we can do is to relate normality of $\mathcal O_{X,x}$ (resp. $\mathcal O_{X^{an},x}$) to normality of the completion $\hat{\mathcal O_{X,x}}$ (resp. $\hat{\mathcal O_{X^{an},x}}$) and then prove that $\hat{\mathcal O_{X,x}}$ is canonically isomorphic to $\hat{\mathcal O_{X^{an},x}}$. This step is not formal or trivial at all, it uses rather deep results from commutative algebra (and it crucially relies on the fact that $\mathcal O_{X,x}$ and $\mathcal O_{X^{an},x}$ are excellent). Let me formulate the main ingredients that we are still missing to complete the proof.

Theorem 1 (Serre's Criterion of Normality): Let $A$ be a noetherian local ring, then $A$ is normal iff it has properties (R_1) and (S_2).

Theorem 2 (Theorem 23.9 from Matsumura's book "Commutative Ring Theory"): Let $A\to B$ be a flat local homomorphisms of local noetherian rings, then

(i) If $B$ satisfies (R_i)(resp. (S_i)), so does A.

(ii) If both $A$ and the fibre ring $B\otimes_A k(\mathfrak p)$ over any prime ideal $\mathfrak p$ of satisfy (R_i)(resp. (S_i)), so does B.

Remark: For the definition of properties (R_i) and (S_i) you can see this section on Stack's Project. Informally, Serre's criterion says that a noetherian scheme, which is regular in codimension one and such that every function from a codimension 2 open subset can be extended to the whole scheme, is normal.

Now, let's apply the above theorems to the morphisms $\mathcal O_{X,x} \to \hat{\mathcal O_{X,x}}$ and $\mathcal O_{X^{an},x} \to \hat{\mathcal O_{X^{an},x}}$. We know that both rings $\mathcal O_{X,x}$ and $\mathcal O_{X^{an},x}$ are excellent as we discussed above. Thus all fibers of these two morphisms are geometrically regular by the property (1) from above! So, they automatically satisfy properties (R_i) and (S_i) for any i! Hence, Theorem 1 and Theorem 2 together imply that $\mathcal O_{X,x}$ (resp. $\mathcal O_{X^{an},x}$) is normal if and only if $\hat{\mathcal O_{X,x}}$ (resp. $\hat{\mathcal O_{X^{an},x}}$) is. Aha, so the last part is to prove that $\hat{\mathcal O_{X,x}}$ is canonically isomorphic to $\hat{\mathcal O_{X^{an},x}}$. This will imply that normality of $\mathcal O_{X,x}$ is equivalent to normality of $\mathcal O_{X^{an},x}$.

The last step is to actually show that $\hat{\mathcal O_{X,x}}$ is isomorphic to $\hat{\mathcal O_{X^{an},x}}$. This is rather trivial and basically follows from the definition of analytification. Nevertheless, let me explain it (or at least the main idea) in a clear way. From the very definition of the analytification functor, we have a canonical morphism (in the category of locally ringed topological spaces) $$i:X^{an} \to X.$$ This gives us a morphism $\mathcal O_X \to i_*\mathcal O_{X^{an}}$. Consider the stalk of this morphism at x to get a morphism $$\mathcal O_{X,x} \to \mathcal O_{X^{an},x}.$$ Both rings are local and the morphism is local. Hence, it is continuous in $\mathfrak m_x$-adic topology. Thus it induces a morphism on the completions (in a sense of commutative algebra)

$$\hat{\mathcal O_{X,x}} \to \hat{\mathcal O_{X^{an},x}}.$$ In order to check that it is an isomorphism it suffices to prove that $\mathcal O_{X,x}/\mathfrak m_x^n \to \mathcal O_{X^{an},x}/(\mathfrak {m}^{an}_x)^n$ is an isomorphism for any integer $n$. The latter fact follows basically from the very definition of the analytification functor (it is also a good exercise).