Normal Schemes

Normality = "not too far from smooth"

Recall that the punctured plane $U= \mathbb{A}^2- \{(0,0)\}$ viewed as an open subscheme of $X = \mathbb{A}^2$ is NOT affine. In proving this, we computed the global section of the punctured plane and found that $O_X(U) = O_X(X)$. This means in particular that every function on $U$ extends to the whole of $X$. This is an analogue of Hartogs's Lemma in complex geometry: we can extend a holomorphic function defined on the complement of a set of codimension at least two on a complex manifold over the missing set.

In algebraic geometry, there is an analogue. We can extend functions over points (in codim at least 2) not only if they are smooth, but also if they are "mildly singular"Hart, i.e. normal. Locally Noetherian normal schemes satisfy Hartogs's Lemma. Consequently rational functions without poles are defined everywhere.

Defn. A scheme $X$ is called normal if all of its stalks are normal, i.e. integral domains that are integrally closed in its field of fraction.

Claim. Normal schemes are reduced.
Proof. A scheme is reduced iff all of its stalks are reduced.
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Recall that if $A$ is integrally closed then so is its nontrivial localization $S^{-1}A$. From this we can deduce the following.

Claim. If $A$ is integrally closed then $\text{Spec} A$ is normal.

Claim. If $X$ is quasi-compact then $X$ is normal iff its normal at every closed point.
Proof. Here we simply use the fact that if a property $P$ is compatible with localization, then to check that $P$ holds for a quasi-compact scheme, it suffices to check it at closed points.  We replicate the proof from previous post here.

Suppose $X$ is normal at every closed point. Let $x \in X$. Then $\overline{\{x\}}$ contains a closed point $y$. Let $U= \text{Spec} A$ be an affine neighborhood of $y$. Then $U$ must contains $X$ as $y$ is in the closure of $x$. Thus we have $\mathbb{V}(\mathfrak{p}_x) \ni \mathfrak{p}_y$ so $\mathfrak{p}_y \supset \mathfrak{p}_x$ so we have a localization map
$$A_{\mathfrak{p}_y} \to A_{\mathfrak{p}_x}.$$
As $A_{\mathfrak{p}_y}$ is normal and normality is compatible with localization, we have $A_{\mathfrak{p}_x}$ is also normal.
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From commutative algebra, we know that for an integral domain $A$, TFAE:

• $A$ is integrally closed;
• $A_{\mathfrak{p}}$ is integrally closed for every $\mathfrak{p}$ prime in $A$;
• $A_{\mathfrak{m}}$ is integrally closed for every $\mathfrak{m}$ max in $A$.

Thus we have.

Proposition. Let $X$ be a scheme. TFAE

1. $X$ is normal;
2. For every affine open $U$ in $X$, $O_X(U)$ is normal; (in particular, $U$ itself is normal)
3. There is a cover of $X$ by open affine $U_{\alpha}$ such that $O_X(U_{\alpha})$ is normal

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Proposition. Let $X$ be a scheme. TFAE

1. $X$ is normal;
2. There exists an open covering $X = \bigcup X_i$ such that each open subscheme $X_i$ is normal;
3. Every open subscheme of $X$ is normal.

Proof.
(1) $\implies $ (2). Suppose $X$ is normal. Then for every open affine $U$ of $X$, $O_X(U)$ is normal. This implies $U$ itself is normal.
(2) $\implies$ (1). Cover $X$ by affine covers of $X_i$. Each of this affine piece is normal so $X$ is normal.
(1) $\implies$ (3). Every open subscheme of $X$ can be covered by affine opens of $X$. 

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Warning: Normal schemes are NOT necessarily integral.

Example. $X = \text{Spec} k \sqcup \text{Spec} k$ is a normal scheme. (Note: disjoint union of normal schemes is normal since normality is a stalk-local condition). However $X = \text{Spec} (k \times k) = \text{Spec} k[x]/(x(x-1))$ so its global section is not an integral domain.
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Claim: Global sections of an irreducible normal scheme is normal.

Proof.  Note a normal scheme is reduced so an irreducible normal scheme is integral. So we can embed all sections of $O_X$ to its function field $K(X)$ and all restriction maps are inclusion. In particular, we have $O_X(X) = \bigcap_{U \subset X, \text{open}}O_X(U) =\bigcap_{U \subset X, \text{affine}}O_X(U) $.

Suppose $s \in K(X)$ is integral over $O_X(X)$. Then it is integral over $O_X(U)$  for $U$ affine, so $s \in O_X(U)$, as it is normal by Proposition above. Thus $s \in \bigcap_{U \subset X, \text{affine}}O_X(U) = O_X(X).$

The Property of being Reduced

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A scheme $X$ is said to be reduced if $O_X(U)$ is reduced for all open $U$. 

Key point

Reduced-ness is stalk-local and affine local. However, it is not an open condition . 

Stalk-local.

Reduced-ness is a stalk-local property: $X$ is reduced iff $(O_X)_x$ is reduced for all $x$.

Paraphrasing this, a scheme is not reduced iff its talk at some point is not reduced. 

proof. (Trivial) Suppose $X$ is not reduced. Then there exists a section nonzero $s \in O_X(U)$ such that $s^n = 0$ for some $n$. As $s$ is not $0$ and $O_X$ is a sheaf, $s_x \neq 0$ for some $x \in U$, but $s_x^n = 0$ in $(O_X)_x$.

On the other hand, suppose $[(s, U)]^n = 0$ in $(O_X)_x$ and $[(s, U)] \neq 0$. Then $s|_W \neq 0$ on some $x \in W \subset U$.

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Affine Local

Affine Communication Lemma.

A property $P$ enjoyed by some affine open sets of a scheme $X$ is called affine-local if

• If $\text{Spec} A \hookrightarrow X$ has property $P$ then so does $\text{Spec} A_f$.
• If $A = (f_1, \ldots, f_n)$ and if $\text{Spec} A_{f_i} \hookrightarrow X$ has property $P$ for all $i$ then so does $\text{Spec} A.$

Suppose $P$ is an affine-local property. Suppose $X = \bigcup_{i} \text{Spec} A_i$ where $\text{Spec} A_i$ all have property $P$. Then every open affine subset of $X$ has property $P$.

Reducedness is affine-local

Suppose $A$ is a ring generated by $f_1, \ldots, f_n$. Then $A$ is reduced iff each $A_i$ is.

$X$ is reduced iff X can be covered by affine open sets $\text{Spec} A$ where $A$ is reduced.

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Example: The scheme $X = \text{Spec} k[x,y]/(y^2, xy)$ is non-reduced. 

The ring $k[x,y]/(y^2, xy)$  contains the element $y \neq 0$ such that $y^2 = 0$. At the stalk of every point $p$, we will still have $y_p^2 = 0$. So $(O_X)_p$ is reduced iff $y_p = 0$. This is possible if we invert $x$, i.e. if $p \in D(x) = $ $x$-axis minus origin.

We show the origin is the only point at which $X$ is non-reduced.

Fuzzy Origin

Proof.  This is equivalent to saying that $\text{Spec} \left(k[x,y]/(y^2, xy)\right)_x$ is reduced, as 

the points of $X$ are prime ideals containing $y^2$ and $xy$, i.e. $(y)$- the $x$-axis.  We think of  $X = \text{Spec} k[x,y]/(y^2, xy)$ then as the $x$ axis with some derivatives at $(0,0)$. 

To say that the origin is the only non-reduced point is equivalent to saying that that on $x$-axis minus the origin, $X$ is reduced. In other words $O_X(D(x))$ is reduced. 

We have $A_x := \left(k[x,y]/(y^2, xy)\right)_x = k[x, y, x^{-1} ]/(y^2, xy)$. As $x$ is a unit in this ring and $xy = 0$, we must have $y = 0$ in this ring. So $A_x = k[x, y, x^{-1}]/(y) = k[x, x^{-1}]$ is reduced. 

At the origin, we have $$A_{(x, y)} = \left(k[x,y]/(y^2, xy)\right)_{(x, y)}  = \frac{k[x,y]_{(x,y)}}{(y^2, xy)}.$$

Thus $y$ is zero in $A_{(x,y)}$ iff the image of $y$ in $k[x,y]_{(x,y)}$ lies in $(y^2, xy)$. Thus for some $h$ outside of $(x, y) we have

$$h y = y^2 f + xy g.$$

Dividing both sides by $y$ gives $h \in (x, y)$ a contradiction. 

More Illuminating Explanation.   Remember that $\text{Spec} k[x,y]/(y^2, xy)$ is just the $x$-axis together with (the span of ) some extra derivative vectors at the origin $\displaystyle \frac{\partial}{\partial_x}, \frac{\partial}{\partial_y}, \frac{\partial^2}{\partial_{x^2}}$ (cf. Visualizing Nilpotents). For $y$ to be $0$ at the stalk of $\mathfrak{p}$ it has to be zero in a neighborhood of $\mathfrak{p}$ (where neighborhood here include the derivative vectors). So for  $y$ to be zero at the stalk of the origin, we not only need $y|_{(0,0)} = 0$ but also that $\frac{\partial}{\partial_x}, \frac{\partial}{\partial_y}$ and $\frac{\partial^2}{\partial_{x^2}}$ applied to $y$ is $0$. The later does not hold as $\frac{\partial}{\partial_y}(y) \neq 0.$  (It gives $y$, the $y$-term)

To say this rigorously, remember that $\frac{\partial}{\partial_y}(y)$ is just the image of $y$ in $k[x, y] \to k[x,y]/(y^2)$ and clearly the image of $y$ is not $0$.

Now how to formalize  "if $y$ vanishes at the stalk of the origin then $\frac{\partial}{\partial_y}(y) =0$"?  To view $\frac{\partial}{\partial_y}$ as an element of $\text{Spec} k[x,y]/(y^2, xy)$ is to have a morphism $k[x,y]/(y^2, xy) \to k[y]/(y^2)$ given by $x \mapsto 0$ and $y \mapsto y$, i.e every $F$ is mapped to its $y$-term.

 (Two polynomials $f, g \in k[x,y]$ are equal in the first ring iff $\displaystyle \frac{\partial}{\partial_x}, \frac{\partial}{\partial_y}, \frac{\partial^2}{\partial_{x^2}}$ applied to them are equal, whereas in the second ring, functions are identified if they agree under $\frac{\partial}{\partial_y}$. As $x$ has no $y$ term, it becomes $0$ in the latter.)

The map $\text{Spec} k[y]/(y^2) \to X$ induces  a pull-back map on stalks, so we have
$$\left(\frac{k[x,y]}{(y^2, xy)}\right)_{(x,y)} \to \left(\frac{k[y]}{(y^2)}\right)_{(y)}$$  

which is a local homomorphism.
If $y$ is $0$ in $\left(\frac{k[x,y]}{(y^2, xy)}\right)_{(x,y)} $ then its image in $\left(\frac{k[y]}{(y^2)}\right)_{(y)}$ must also be $0$. In other words, as the latter is just $\frac{k[y]_{(y)}}{(y^2)}$, we must have the image of  $y$ in $k[y]_{(y)}$ lie in $y^2$ and thus $yh = y^2 f$ for some $h, f \not \in (y)$. This is impossible by unique factorization in $k[y]$. Another way to see this is to see that if $y$ is $0$ in the ring above then $\frac{k[y]_{(y)}}{(y^2)} =  k$, i.e. $y^2$ is kernel of $k \to k[y]_{(y)}$ which is not possible as the latter is injective. 

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Not an open condition:

Reducedness is in general, not an open condition (see example below). However if $X$ is a locally Noetherian scheme then reducedness is open.

Example.
Let $X$ be $\text{Spec} A$ where

$$A = \text{Spec} \frac{\mathbb{C}[x, y_1, \ldots]}{(y_1^2, y_2^2, y_3^2, \ldots, (x-1)y_1, (x-2)y_2,  (x- 3)y_2, \ldots)}$$

Notice that prime ideals of $A$ are exactly prime ideals $\mathfrak{p}$ of $\mathbb{C}[x, y_1, y_2, \ldots]$ containing $ (y_1^2, y_2^2, y_3^2, \ldots, (x-1)y_1, (x-2)y_2,  (x- 3)y_2, \ldots)$. These are exactly prime ideals containing $(y_1, y_2, \ldots,)$.

So $\text{Spec} A$ as a set is homeomorphic to the $$\text{Spec} \frac{\mathbb{C}[x, y_1, \ldots]}{(y_1, y_2, y_3, \ldots } = \text{Spec} \mathbb{C}[x].$$

Claim: the nonreduced points in $X$ correspond to the set $\{\text{x - m}\mid m \in \mathbb{Z}\}$ in $ \text{Spec} \mathbb{C}[x]$. In particular the set of  reduced points in $\text{Spec} A$ is therefore not open.

Proof:  Indeed similar to the above example, $A_{\mathfrak{p}}$ is not reduced iff one of $y_1, y_2 \ldots$ is not zero in the stalk at $\mathfrak{p}$.

Clearly, at points $\mathfrak{p}$ where $(x-m)$ is invertible in $\mathfrak{p}$, we have $y_m = 0$ in $\mathfrak{p}$, so $y_m$ is not nilpotent. Thus $A$ is reduced at $\mathfrak{p}$ if for all $m$, $x- m$ is invertible in $\mathfrak{p}$. This is true if $x-m$ all lie outside $\mathfrak{p}$, i.e. $\mathfrak{p} \neq x-m$ for any $m$. (As $\mathfrak{p}$ is maximal).

Let $\mathfrak{p}$ correspond to $(x-m)$ for some $m$. Then in $A_{\mathfrak{p}}$, we invert $(x - n)$ for every $n\neq m$, so for all such $n$, $y_n = 0$. Thus
$$A_{\mathfrak{p}} = \text{Spec}\frac{\mathbb{C}[x,y_m]}{(y_m^2, (x-m)y_m)},$$
which reduces the problem to the previous example.

Relationship with quasi-compact schemes: Recall that for every open condition $P$, to check if all points of a  quasi-compact scheme $X$ has $P$, it suffices to check that all closed points do. Though reducedness is not an open condition, it is still possible to check the reducedness of a quasi-compact schemes by checking it at closed points.