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.