Definition 10.42.1. Let $k$ be a field. Let $S$ be a $k$-algebra. We say $S$ is *geometrically reduced over $k$* if for every field extension $k \subset K$ the $K$-algebra $K \otimes _ k S$ is reduced.

## 10.42 Geometrically reduced algebras

The main result on geometrically reduced algebras is Lemma 10.43.3. We suggest the reader skip to the lemma after reading the definition.

Let $k$ be a field and let $S$ be a reduced $k$ algebra. To check that $S$ is geometrically reduced it will suffice to check that $\overline{k} \otimes _ k S$ is reduced (where $\overline{k}$ denotes the algebraic closure of $k$). In fact it is enough to check this for finite purely inseparable field extensions $k \subset k'$. See Lemma 10.43.3.

Lemma 10.42.2. Elementary properties of geometrically reduced algebras. Let $k$ be a field. Let $S$ be a $k$-algebra.

If $S$ is geometrically reduced over $k$ so is every $k$-subalgebra.

If all finitely generated $k$-subalgebras of $S$ are geometrically reduced, then $S$ is geometrically reduced.

A directed colimit of geometrically reduced $k$-algebras is geometrically reduced.

If $S$ is geometrically reduced over $k$, then any localization of $S$ is geometrically reduced over $k$.

**Proof.**
Omitted. The second and third property follow from the fact that tensor product commutes with colimits.
$\square$

Lemma 10.42.3. Let $k$ be a field. If $R$ is geometrically reduced over $k$, and $S \subset R$ is a multiplicative subset, then the localization $S^{-1}R$ is geometrically reduced over $k$. If $R$ is geometrically reduced over $k$, then $R[x]$ is geometrically reduced over $k$.

**Proof.**
Omitted. Hints: A localization of a reduced ring is reduced, and localization commutes with tensor products.
$\square$

In the proofs of the following lemmas we will repeatedly use the following observation: Suppose that $R' \subset R$ and $S' \subset S$ are inclusions of $k$-algebras. Then the map $R' \otimes _ k S' \to R \otimes _ k S$ is injective.

Lemma 10.42.4. Let $k$ be a field. Let $R$, $S$ be $k$-algebras.

If $R \otimes _ k S$ is nonreduced, then there exist finitely generated subalgebras $R' \subset R$, $S' \subset S$ such that $R' \otimes _ k S'$ is not reduced.

If $R \otimes _ k S$ contains a nonzero zerodivisor, then there exist finitely generated subalgebras $R' \subset R$, $S' \subset S$ such that $R' \otimes _ k S'$ contains a nonzero zerodivisor.

If $R \otimes _ k S$ contains a nontrivial idempotent, then there exist finitely generated subalgebras $R' \subset R$, $S' \subset S$ such that $R' \otimes _ k S'$ contains a nontrivial idempotent.

**Proof.**
Suppose $z \in R \otimes _ k S$ is nilpotent. We may write $z = \sum _{i = 1, \ldots , n} x_ i \otimes y_ i$. Thus we may take $R'$ the $k$-subalgebra generated by the $x_ i$ and $S'$ the $k$-subalgebra generated by the $y_ i$. The second and third statements are proved in the same way.
$\square$

Lemma 10.42.5. Let $k$ be a field. Let $S$ be a geometrically reduced $k$-algebra. Let $R$ be any reduced $k$-algebra. Then $R \otimes _ k S$ is reduced.

**Proof.**
By Lemma 10.42.4 we may assume that $R$ is of finite type over $k$. Then $R$, as a reduced Noetherian ring, embeds into a finite product of fields (see Lemmas 10.24.4, 10.30.6, and 10.24.1). Hence we may assume $R$ is a finite product of fields. In this case it follows from Definition 10.42.1 that $R \otimes _ k S$ is reduced.
$\square$

Lemma 10.42.6. Let $k$ be a field. Let $S$ be a reduced $k$-algebra. Let $k \subset K$ be either a separable field extension, or a separably generated field extension. Then $K \otimes _ k S$ is reduced.

**Proof.**
Assume $k \subset K$ is separable. By Lemma 10.42.4 we may assume that $S$ is of finite type over $k$ and $K$ is finitely generated over $k$. Then $S$ embeds into a finite product of fields, namely its total ring of fractions (see Lemmas 10.24.1 and 10.24.4). Hence we may actually assume that $S$ is a domain. We choose $x_1, \ldots , x_{r + 1} \in K$ as in Lemma 10.41.3. Let $P \in k(x_1, \ldots , x_ r)[T]$ be the minimal polynomial of $x_{r + 1}$. It is a separable polynomial. It is easy to see that $k[x_1, \ldots , x_ r] \otimes _ k S = S[x_1, \ldots , x_ r]$ is a domain. This implies $k(x_1, \ldots , x_ r) \otimes _ k S$ is a domain as it is a localization of $S[x_1, \ldots , x_ r]$. The ring extension $k(x_1, \ldots , x_ r) \otimes _ k S \subset K \otimes _ k S$ is generated by a single element $x_{r + 1}$ with a single equation, namely $P$. Hence $K \otimes _ k S$ embeds into $F[T]/(P)$ where $F$ is the fraction field of $k(x_1, \ldots , x_ r) \otimes _ k S$. Since $P$ is separable this is a finite product of fields and we win.

At this point we do not yet know that a separably generated field extension is separable, so we have to prove the lemma in this case also. To do this suppose that $\{ x_ i\} _{i \in I}$ is a separating transcendence basis for $K$ over $k$. For any finite set of elements $\lambda _ j \in K$ there exists a finite subset $T \subset I$ such that $k(\{ x_ i\} _{i\in T}) \subset k(\{ x_ i\} _{i \in T} \cup \{ \lambda _ j\} )$ is finite separable. Hence we see that $K$ is a directed colimit of finitely generated and separably generated extensions of $k$. Thus the argument of the preceding paragraph applies to this case as well. $\square$

Lemma 10.42.7. Let $k$ be a field and let $S$ be a $k$-algebra. Assume that $S$ is reduced and that $S_{\mathfrak p}$ is geometrically reduced for every minimal prime $\mathfrak p$ of $S$. Then $S$ is geometrically reduced.

**Proof.**
Since $S$ is reduced the map $S \to \prod _{\mathfrak p\text{ minimal}} S_{\mathfrak p}$ is injective, see Lemma 10.24.2. If $k \subset K$ is a field extension, then the maps

are injective: the first as $k \to K$ is flat and the second by inspection because $K$ is a free $k$-module. As $S_\mathfrak p$ is geometrically reduced the ring on the right is reduced. Thus we see that $S \otimes _ k K$ is reduced as a subring of a reduced ring. $\square$

Lemma 10.42.8. Let $k'/k$ be a separable algebraic extension. Then there exists a multiplicative subset $S \subset k' \otimes _ k k'$ such that the multiplication map $k' \otimes _ k k' \to k'$ is identified with $k' \otimes _ k k' \to S^{-1}(k' \otimes _ k k')$.

**Proof.**
First assume $k'/k$ is finite separable. Then $k' = k(\alpha )$, see Fields, Lemma 9.19.1. Let $P \in k[x]$ be the minimal polynomial of $\alpha $ over $k$. Then $P$ is an irreducible, separable, monic polynomial, see Fields, Section 9.12. Then $k'[x]/(P) \to k' \otimes _ k k'$, $\sum \alpha _ i x^ i \mapsto \alpha _ i \otimes \alpha ^ i$ is an isomorphism. We can factor $P = (x - \alpha ) Q$ in $k'[x]$ and since $P$ is separable we see that $Q(\alpha ) \not= 0$. Then it is clear that the multiplicative set $S'$ generated by $Q$ in $k'[x]/(P)$ works, i.e., that $k' = (S')^{-1}(k'[x]/(P))$. By transport of structure the image $S$ of $S'$ in $k' \otimes _ k k'$ works.

In the general case we write $k' = \bigcup k_ i$ as the union of its finite subfield extensions over $k$. For each $i$ there is a multiplicative subset $S_ i \subset k_ i \otimes _ k k_ i$ such that $k_ i = S_ i^{-1}(k_ i \otimes _ k k_ i)$. Then $S = \bigcup S_ i \subset k' \otimes _ k k'$ works. $\square$

Lemma 10.42.9. Let $k \subset k'$ be a separable algebraic field extension. Let $A$ be an algebra over $k'$. Then $A$ is geometrically reduced over $k$ if and only if it is geometrically reduced over $k'$.

**Proof.**
Assume $A$ is geometrically reduced over $k'$. Let $K/k$ be a field extension. Then $K \otimes _ k k'$ is a reduced ring by Lemma 10.42.6. Hence by Lemma 10.42.5 we find that $K \otimes _ k A = (K \otimes _ k k') \otimes _{k'} A$ is reduced.

Assume $A$ is geometrically reduced over $k$. Let $K/k'$ be a field extension. Then

Since $k' \otimes _ k k' \to k'$ is a localization by Lemma 10.42.8, we see that $K \otimes _{k'} A$ is a localization of a reduced algebra, hence reduced. $\square$

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