## 33.12 Geometrically regular schemes

A geometrically regular scheme over a field $k$ is a locally Noetherian scheme over $k$ which remains regular upon suitable changes of base field. A finite type scheme over $k$ is geometrically regular if and only if it is smooth over $k$ (see Lemma 33.12.6). The notion of geometric regularity is most interesting in situations where smoothness cannot be used such as formal fibres (insert future reference here).

In the following definition we restrict ourselves to locally Noetherian schemes, since the property of being a regular local ring is only defined for Noetherian local rings. By Lemma 33.11.1 above, if we restrict ourselves to finitely generated field extensions then this property is preserved under change of base field. This comment will be used without further reference in this section. In particular the following definition makes sense.

Definition 33.12.1. Let $k$ be a field. Let $X$ be a locally Noetherian scheme over $k$.

Let $x \in X$. We say $X$ is *geometrically regular at $x$* over $k$ if for every finitely generated field extension $k \subset k'$ and any $x' \in X_{k'}$ lying over $x$ the local ring $\mathcal{O}_{X_{k'}, x'}$ is regular.

We say $X$ is *geometrically regular over $k$* if $X$ is geometrically regular at all of its points.

A similar definition works to define geometrically Cohen-Macaulay, $(R_ k)$, and $(S_ k)$ schemes over a field. We will add a section for these separately as needed.

Lemma 33.12.2. Let $k$ be a field. Let $X$ be a locally Noetherian scheme over $k$. Let $x \in X$. The following are equivalent

$X$ is geometrically regular at $x$,

for every finite purely inseparable field extension $k'$ of $k$ and $x' \in X_{k'}$ lying over $x$ the local ring $\mathcal{O}_{X_{k'}, x'}$ is regular, and

the ring $\mathcal{O}_{X, x}$ is geometrically regular over $k$ (see Algebra, Definition 10.166.2).

**Proof.**
It is clear that (1) implies (2). Assume (2). This in particular implies that $\mathcal{O}_{X, x}$ is a regular local ring. Let $k \subset k'$ 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 regular ring. Hence we deduce (3) from the definition of a geometrically regular ring.

Assume (3). Let $k \subset k'$ 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 regular by assumption and (1) is proved.
$\square$

Lemma 33.12.3. Let $k$ be a field. Let $X$ be a locally Noetherian scheme over $k$. The following are equivalent

$X$ is geometrically regular,

$X_{k'}$ is a regular scheme for every finitely generated field extension $k \subset k'$,

$X_{k'}$ is a regular scheme for every finite purely inseparable field extension $k \subset k'$,

for every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is geometrically regular (see Algebra, Definition 10.166.2), and

there exists an affine open covering $X = \bigcup U_ i$ such that each $\mathcal{O}_ X(U_ i)$ is geometrically regular over $k$.

**Proof.**
Assume (1). Then for every finitely generated field extension $k \subset k'$ and every point $x' \in X_{k'}$ the local ring of $X_{k'}$ at $x'$ is regular. By Properties, Lemma 28.9.2 this means that $X_{k'}$ is regular. Hence (2).

It is clear that (2) implies (3).

Assume (3) and let $U \subset X$ be an affine open subscheme. Then $U_{k'}$ is a regular scheme for any finite purely inseparable extension $k \subset k'$ (including $k = k'$). This means that $k' \otimes _ k \mathcal{O}(U)$ is a regular ring for all finite purely inseparable extensions $k \subset k'$. Hence $\mathcal{O}(U)$ is a geometrically regular $k$-algebra and we see that (4) holds.

It is clear that (4) implies (5). Let $X = \bigcup U_ i$ be an affine open covering as in (5). For any field extension $k \subset k'$ the base change $X_{k'}$ is gotten by gluing the spectra of the rings $\mathcal{O}_ X(U_ i) \otimes _ k k'$ (see Schemes, Section 26.17). Hence $X_{k'}$ is regular. So (1) holds.
$\square$

Lemma 33.12.4. Let $k$ be a field. Let $X$ be a scheme over $k$. Let $k'/k$ be a finitely generated 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

$X$ is geometrically regular at $x$,

$X_{k'}$ is geometrically regular at $x'$.

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

**Proof.**
It is clear that (1) implies (2). Assume (2). Let $k \subset k''$ 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, finitely generated, field extension $k \subset k'''$ (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 of Noetherian local rings and hence faithfully flat. By (2) we see that the local ring on the right is regular. Thus by Algebra, Lemma 10.110.9 we conclude that $\mathcal{O}_{X_{k''}, x''}$ is regular. By Lemma 33.12.2 we see that $X$ is geometrically regular at $x$.
$\square$

The following lemma is a geometric variant of Algebra, Lemma 10.166.3.

Lemma 33.12.5. Let $k$ be a field. Let $f : X \to Y$ be a morphism of locally Noetherian schemes over $k$. Let $x \in X$ be a point and set $y = f(x)$. If $X$ is geometrically regular at $x$ and $f$ is flat at $x$ then $Y$ is geometrically regular at $y$. In particular, if $X$ is geometrically regular over $k$ and $f$ is flat and surjective, then $Y$ is geometrically regular over $k$.

**Proof.**
Let $k'$ be finite purely inseparable extension of $k$. Let $f' : X_{k'} \to Y_{k'}$ be the base change of $f$. Let $x' \in X_{k'}$ be the unique point lying over $x$. If we show that $Y_{k'}$ is regular at $y' = f'(x')$, then $Y$ is geometrically regular over $k$ at $y'$, see Lemma 33.12.3. By Morphisms, Lemma 29.25.7 the morphism $X_{k'} \to Y_{k'}$ is flat at $x'$. Hence the ring map

\[ \mathcal{O}_{Y_{k'}, y'} \longrightarrow \mathcal{O}_{X_{k'}, x'} \]

is a flat local homomorphism of local Noetherian rings with right hand side regular by assumption. Hence the left hand side is a regular local ring by Algebra, Lemma 10.110.9.
$\square$

Lemma 33.12.6. Let $k$ be a field. Let $X$ be a scheme locally of finite type over $k$. Let $x \in X$. Then $X$ is geometrically regular at $x$ if and only if $X \to \mathop{\mathrm{Spec}}(k)$ is smooth at $x$ (Morphisms, Definition 29.34.1).

**Proof.**
The question is local around $x$, hence we may assume that $X = \mathop{\mathrm{Spec}}(A)$ for some finite type $k$-algebra. Let $x$ correspond to the prime $\mathfrak p$.

If $A$ is smooth over $k$ at $\mathfrak p$, then we may localize $A$ and assume that $A$ is smooth over $k$. In this case $k' \otimes _ k A$ is smooth over $k'$ for all extension fields $k'/k$, and each of these Noetherian rings is regular by Algebra, Lemma 10.140.3.

Assume $X$ is geometrically regular at $x$. Consider the residue field $K := \kappa (x) = \kappa (\mathfrak p)$ of $x$. It is a finitely generated extension of $k$. By Algebra, Lemma 10.45.3 there exists a finite purely inseparable extension $k \subset k'$ such that the compositum $k'K$ is a separable field extension of $k'$. Let $\mathfrak p' \subset A' = k' \otimes _ k A$ be a prime ideal lying over $\mathfrak p$. It is the unique prime lying over $\mathfrak p$, see Algebra, Lemma 10.46.7. Hence the residue field $K' := \kappa (\mathfrak p')$ is the compositum $k'K$. By assumption the local ring $(A')_{\mathfrak p'}$ is regular. Hence by Algebra, Lemma 10.140.5 we see that $k' \to A'$ is smooth at $\mathfrak p'$. This in turn implies that $k \to A$ is smooth at $\mathfrak p$ by Algebra, Lemma 10.137.19. The lemma is proved.
$\square$

Example 33.12.7. Let $k =\mathbf{F}_ p(t)$. It is quite easy to give an example of a regular variety $V$ over $k$ which is not geometrically reduced. For example we can take $\mathop{\mathrm{Spec}}(k[x]/(x^ p - t))$. In fact, there exists an example of a regular variety $V$ which is geometrically reduced, but not even geometrically normal. Namely, take for $p > 2$ the scheme $V = \mathop{\mathrm{Spec}}(k[x, y]/(y^2 - x^ p + t))$. This is a variety as the polynomial $y^2 - x^ p + t \in k[x, y]$ is irreducible. The morphism $V \to \mathop{\mathrm{Spec}}(k)$ is smooth at all points except at the point $v_0 \in V$ corresponding to the maximal ideal $(y, x^ p - t)$ (because $2y$ is invertible). In particular we see that $V$ is (geometrically) regular at all points, except possibly $v_0$. The local ring

\[ \mathcal{O}_{V, v_0} = \left(k[x, y]/(y^2 - x^ p + t)\right)_{(y, x^ p - t)} \]

is a domain of dimension $1$. Its maximal ideal is generated by $1$ element, namely $y$. Hence it is a discrete valuation ring and regular. Let $k' = k[t^{1/p}]$. Denote $t' = t^{1/p} \in k'$, $V' = V_{k'}$, $v'_0 \in V'$ the unique point lying over $v_0$. Over $k'$ we can write $x^ p - t = (x - t')^ p$, but the polynomial $y^2 - (x - t')^ p$ is still irreducible and $V'$ is still a variety. But the element

\[ \frac{y}{x - t'} \in (\text{fraction field of }\mathcal{O}_{V', v'_0}) \]

is integral over $\mathcal{O}_{V', v'_0}$ (just compute its square) and not contained in it, so $V'$ is not normal at $v'_0$. This concludes the example.

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