## 33.6 Geometrically reduced schemes

If $X$ is a reduced scheme over a field, then it can happen that $X$ becomes nonreduced after extending the ground field. This does not happen for geometrically reduced schemes.

Definition 33.6.1. Let $k$ be a field. Let $X$ be a scheme over $k$.

1. Let $x \in X$ be a point. We say $X$ is geometrically reduced at $x$ if for any field extension $k'/k$ and any point $x' \in X_{k'}$ lying over $x$ the local ring $\mathcal{O}_{X_{k'}, x'}$ is reduced.

2. We say $X$ is geometrically reduced over $k$ if $X$ is geometrically reduced at every point of $X$.

This may seem a little mysterious at first, but it is really the same thing as the notion discussed in the algebra chapter. Here are some basic results explaining the connection.

Lemma 33.6.2. Let $k$ be a field. Let $X$ be a scheme over $k$. Let $x \in X$. The following are equivalent

1. $X$ is geometrically reduced at $x$, and

2. the ring $\mathcal{O}_{X, x}$ is geometrically reduced over $k$ (see Algebra, Definition 10.43.1).

Proof. Assume (1). This in particular implies that $\mathcal{O}_{X, x}$ is reduced. Let $k'/k$ be a finite purely inseparable field extension. Consider the ring $\mathcal{O}_{X, x} \otimes _ k k'$. By Algebra, Lemma 10.46.7 its spectrum is the same as the spectrum of $\mathcal{O}_{X, x}$. Hence it is a local ring also (Algebra, Lemma 10.18.2). Therefore there is a unique point $x' \in X_{k'}$ lying over $x$ and $\mathcal{O}_{X_{k'}, x'} \cong \mathcal{O}_{X, x} \otimes _ k k'$. By assumption this is a reduced ring. Hence we deduce (2) by Algebra, Lemma 10.44.3.

Assume (2). Let $k'/k$ be a field extension. Since $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$ is surjective, also $X_{k'} \to X$ is surjective (Morphisms, Lemma 29.9.4). Let $x' \in X_{k'}$ be any point lying over $x$. The local ring $\mathcal{O}_{X_{k'}, x'}$ is a localization of the ring $\mathcal{O}_{X, x} \otimes _ k k'$. Hence it is reduced by assumption and (1) is proved. $\square$

The notion isn't interesting in characteristic zero.

Lemma 33.6.3. Let $X$ be a scheme over a perfect field $k$ (e.g. $k$ has characteristic zero). Let $x \in X$. If $\mathcal{O}_{X, x}$ is reduced, then $X$ is geometrically reduced at $x$. If $X$ is reduced, then $X$ is geometrically reduced over $k$.

Proof. The first statement follows from Lemma 33.6.2 and Algebra, Lemma 10.43.6 and the definition of a perfect field (Algebra, Definition 10.45.1). The second statement follows from the first. $\square$

Lemma 33.6.4. Let $k$ be a field of characteristic $p > 0$. Let $X$ be a scheme over $k$. The following are equivalent

1. $X$ is geometrically reduced,

2. $X_{k'}$ is reduced for every field extension $k'/k$,

3. $X_{k'}$ is reduced for every finite purely inseparable field extension $k'/k$,

4. $X_{k^{1/p}}$ is reduced,

5. $X_{k^{perf}}$ is reduced,

6. $X_{\bar k}$ is reduced,

7. for every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is geometrically reduced (see Algebra, Definition 10.43.1).

Proof. Assume (1). Then for every field extension $k'/k$ and every point $x' \in X_{k'}$ the local ring of $X_{k'}$ at $x'$ is reduced. In other words $X_{k'}$ is reduced. Hence (2).

Assume (2). Let $U \subset X$ be an affine open. Then for every field extension $k'/k$ the scheme $X_{k'}$ is reduced, hence $U_{k'} = \mathop{\mathrm{Spec}}(\mathcal{O}(U)\otimes _ k k')$ is reduced, hence $\mathcal{O}(U)\otimes _ k k'$ is reduced (see Properties, Section 28.3). In other words $\mathcal{O}(U)$ is geometrically reduced, so (7) holds.

Assume (7). For any field extension $k'/k$ the base change $X_{k'}$ is gotten by gluing the spectra of the rings $\mathcal{O}_ X(U) \otimes _ k k'$ where $U$ is affine open in $X$ (see Schemes, Section 26.17). Hence $X_{k'}$ is reduced. So (1) holds.

This proves that (1), (2), and (7) are equivalent. These are equivalent to (3), (4), (5), and (6) because we can apply Algebra, Lemma 10.44.3 to $\mathcal{O}_ X(U)$ for $U \subset X$ affine open. $\square$

Lemma 33.6.5. Let $k$ be a field of characteristic $p > 0$. Let $X$ be a scheme over $k$. Let $x \in X$. The following are equivalent

1. $X$ is geometrically reduced at $x$,

2. $\mathcal{O}_{X_{k'}, x'}$ is reduced for every finite purely inseparable field extension $k'$ of $k$ and $x' \in X_{k'}$ the unique point lying over $x$,

3. $\mathcal{O}_{X_{k^{1/p}}, x'}$ is reduced for $x' \in X_{k^{1/p}}$ the unique point lying over $x$, and

4. $\mathcal{O}_{X_{k^{perf}}, x'}$ is reduced for $x' \in X_{k^{perf}}$ the unique point lying over $x$.

Proof. Note that if $k'/k$ is purely inseparable, then $X_{k'} \to X$ induces a homeomorphism on underlying topological spaces, see Algebra, Lemma 10.46.7. Whence the uniqueness of $x'$ lying over $x$ mentioned in the statement. Moreover, in this case $\mathcal{O}_{X_{k'}, x'} = \mathcal{O}_{X, x} \otimes _ k k'$. Hence the lemma follows from Lemma 33.6.2 above and Algebra, Lemma 10.44.3. $\square$

Lemma 33.6.6. Let $k$ be a field. Let $X$ be a scheme over $k$. Let $k'/k$ be a field extension. Let $x \in X$ be a point, and let $x' \in X_{k'}$ be a point lying over $x$. The following are equivalent

1. $X$ is geometrically reduced at $x$,

2. $X_{k'}$ is geometrically reduced at $x'$.

In particular, $X$ is geometrically reduced over $k$ if and only if $X_{k'}$ is geometrically reduced over $k'$.

Proof. It is clear that (1) implies (2). Assume (2). Let $k''/k$ be a finite purely inseparable field extension and let $x'' \in X_{k''}$ be a point lying over $x$ (actually it is unique). We can find a common field extension $k'''/k$ (i.e. with both $k' \subset k'''$ and $k'' \subset k'''$) and a point $x''' \in X_{k'''}$ lying over both $x'$ and $x''$. Consider the map of local rings

$\mathcal{O}_{X_{k''}, x''} \longrightarrow \mathcal{O}_{X_{k'''}, x''''}.$

This is a flat local ring homomorphism and hence faithfully flat. By (2) we see that the local ring on the right is reduced. Thus by Algebra, Lemma 10.164.2 we conclude that $\mathcal{O}_{X_{k''}, x''}$ is reduced. Thus by Lemma 33.6.5 we conclude that $X$ is geometrically reduced at $x$. $\square$

Lemma 33.6.7. Let $k$ be a field. Let $X$, $Y$ be schemes over $k$.

1. If $X$ is geometrically reduced at $x$, and $Y$ reduced, then $X \times _ k Y$ is reduced at every point lying over $x$.

2. If $X$ geometrically reduced over $k$ and $Y$ reduced. Then $X \times _ k Y$ is reduced.

Lemma 33.6.8. Let $k$ be a field. Let $X$ be a scheme over $k$.

1. If $x' \leadsto x$ is a specialization and $X$ is geometrically reduced at $x$, then $X$ is geometrically reduced at $x'$.

2. If $x \in X$ such that (a) $\mathcal{O}_{X, x}$ is reduced, and (b) for each specialization $x' \leadsto x$ where $x'$ is a generic point of an irreducible component of $X$ the scheme $X$ is geometrically reduced at $x'$, then $X$ is geometrically reduced at $x$.

3. If $X$ is reduced and geometrically reduced at all generic points of irreducible components of $X$, then $X$ is geometrically reduced.

Proof. Part (1) follows from Lemma 33.6.2 and the fact that if $A$ is a geometrically reduced $k$-algebra, then $S^{-1}A$ is a geometrically reduced $k$-algebra for any multiplicative subset $S$ of $A$, see Algebra, Lemma 10.43.3.

Let $A = \mathcal{O}_{X, x}$. The assumptions (a) and (b) of (2) imply that $A$ is reduced, and that $A_{\mathfrak q}$ is geometrically reduced over $k$ for every minimal prime $\mathfrak q$ of $A$. Hence $A$ is geometrically reduced over $k$, see Algebra, Lemma 10.43.7. Thus $X$ is geometrically reduced at $x$, see Lemma 33.6.2.

Part (3) follows trivially from part (2). $\square$

Lemma 33.6.9. Let $k$ be a field. Let $X$ be a scheme over $k$. Let $x \in X$. Assume $X$ locally Noetherian and geometrically reduced at $x$. Then there exists an open neighbourhood $U \subset X$ of $x$ which is geometrically reduced over $k$.

Proof. Assume $X$ locally Noetherian and geometrically reduced at $x$. By Properties, Lemma 28.29.8 we can find an affine open neighbourhood $U \subset X$ of $x$ such that $R = \mathcal{O}_ X(U) \to \mathcal{O}_{X, x}$ is injective. By Lemma 33.6.2 the assumption means that $\mathcal{O}_{X, x}$ is geometrically reduced over $k$. By Algebra, Lemma 10.43.2 this implies that $R$ is geometrically reduced over $k$, which in turn implies that $U$ is geometrically reduced. $\square$

Example 33.6.10. Let $k = \mathbf{F}_ p(s, t)$, i.e., a purely transcendental extension of the prime field. Consider the variety $X = \mathop{\mathrm{Spec}}(k[x, y]/(1 + sx^ p + ty^ p))$. Let $k'/k$ be any extension such that both $s$ and $t$ have a $p$th root in $k'$. Then the base change $X_{k'}$ is not reduced. Namely, the ring $k'[x, y]/(1 + s x^ p + ty^ p)$ contains the element $1 + s^{1/p}x + t^{1/p}y$ whose $p$th power is zero but which is not zero (since the ideal $(1 + sx^ p + ty^ p)$ certainly does not contain any nonzero element of degree $< p$).

Lemma 33.6.11. Let $k$ be a field. Let $X \to \mathop{\mathrm{Spec}}(k)$ be locally of finite type. Assume $X$ has finitely many irreducible components. Then there exists a finite purely inseparable extension $k'/k$ such that $(X_{k'})_{red}$ is geometrically reduced over $k'$.

Proof. To prove this lemma we may replace $X$ by its reduction $X_{red}$. Hence we may assume that $X$ is reduced and locally of finite type over $k$. Let $x_1, \ldots , x_ n \in X$ be the generic points of the irreducible components of $X$. Note that for every purely inseparable algebraic extension $k'/k$ the morphism $(X_{k'})_{red} \to X$ is a homeomorphism, see Algebra, Lemma 10.46.7. Hence the points $x'_1, \ldots , x'_ n$ lying over $x_1, \ldots , x_ n$ are the generic points of the irreducible components of $(X_{k'})_{red}$. As $X$ is reduced the local rings $K_ i = \mathcal{O}_{X, x_ i}$ are fields, see Algebra, Lemma 10.25.1. As $X$ is locally of finite type over $k$ the field extensions $K_ i/k$ are finitely generated field extensions. Finally, the local rings $\mathcal{O}_{(X_{k'})_{red}, x'_ i}$ are the fields $(K_ i \otimes _ k k')_{red}$. By Algebra, Lemma 10.45.3 we can find a finite purely inseparable extension $k'/k$ such that $(K_ i \otimes _ k k')_{red}$ are separable field extensions of $k'$. In particular each $(K_ i \otimes _ k k')_{red}$ is geometrically reduced over $k'$ by Algebra, Lemma 10.44.1. At this point Lemma 33.6.8 part (3) implies that $(X_{k'})_{red}$ is geometrically reduced. $\square$

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