## 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 10.18.3). 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 10.44.4.

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 10.44.4 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 10.44.4. $\square$

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.

3. If $X$ and $Y$ are geometrically reduced over $k$, then $X \times _ k Y$ is geometrically reduced.

4. If $k$ is perfect and $X$ and $Y$ are reduced, then $X \times _ k Y$ is reduced.

5. Add more here.

Proof. To prove (1) combine Lemma 33.6.2 with Algebra, Lemma 10.43.5. To prove (2) combine Lemma 33.6.4 with Algebra, Lemma 10.43.5. To prove (3) note that $(X \times _ k Y)_{\overline{k}} = X_{\overline{k}} \times _{\overline{k}} Y_{\overline{k}}$ and use (2) as well as Lemma 33.6.4. To prove (4) use (3) combined with Lemma 33.6.3. $\square$

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 10.44.2. At this point Lemma 33.6.8 part (3) implies that $(X_{k'})_{red}$ is geometrically reduced. $\square$

Comment #8773 by Jay Pottharstq on

It seems common enough to say "the fiber product of two reduced schemes over a perfect field is reduced". This follows from, say, 020I + 035Z, but maybe it can get its own tag?

Comment #8843 by Yassin Mousa on

It might by helpfull to clearify that $\bar{k}$ denotes a separable algebraic closure of $k$ in 33.6.4. At leat this is the notation that is used in 33.8.8. Of course this is equivalent to the fact that the base change to an algebraic closure is reduced. Maybe one could use the common notation $k_{s}$ instead.

Comment #9242 by on

@#8773: I have added this to Lemma 33.6.7. See these changes.

@#8843: Hmm, we don't use Lemma 33.8.8 on this page (it is about geometrically irreducible things). So $\overline{k}$ denotes the algebraic closure -- I have added that to the statement.

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