The Stacks project

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.

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.

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.

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

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

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

  4. 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.

  1. 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.

  2. 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.

  3. 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

\[ S \otimes _ k K \to (\prod S_\mathfrak p) \otimes _ k K \to \prod S_\mathfrak p \otimes _ k K \]

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

\[ K \otimes _{k'} A = (K \otimes _ k A) \otimes _{(k' \otimes _ k k')} k' \]

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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