Let $L\subset k$ a field extension such that $k$ is algebraically closed. Now consider the algebraic set $Z(\mathfrak a)$ where $\mathfrak a$ is an ideal of $k[T_1,\ldots, T_n]$ but it is generated by polynomials with coefficients in $L$. In other words $\mathfrak a$ is an extended ideal of $\mathfrak b\subset L[T_1,\dots,T_n]$. In symbols we write $\mathfrak a=\mathfrak b^e$.

Now let $$I(Z(\mathfrak a))=\{f\in K[T_1,\ldots,T_n]:f(x)=0\,\forall x\in Z(\mathfrak a) \}$$

Can I conclude that $I(Z(\mathfrak a))$ is an extended ideal of $L[T_1,\ldots,T_n]$?

Note that by Hilbert Nullstellensatz theorem I have that:

$$I(Z(\mathfrak b^e))=\text{rad}(\mathfrak b^e)\supseteq \text{rad} (\mathfrak b)^e$$


1 Answer 1


Remark. An alternative formulation of the question is: let $B$ be a finite type $K$-algebra, let $K \subseteq L$ be an extension, and let $A = B \otimes_K L$. If $B$ is reduced, then so is $A$.

Indeed, if we take $B = K[T_1,\ldots,T_n]/\operatorname{rad}(\mathfrak b)$, then $A = L[T_1,\ldots,T_n]/\operatorname{rad}(\mathfrak b)^e$, and proving that $\operatorname{rad}(\mathfrak b)^e$ is radical is equivalent to proving that $A$ is reduced.

The result is false in positive characteristic:

Example. Let $K = \mathbb F_p (X)$, and consider $K \subseteq \bar K$. Let $n = 1$, and let $\mathfrak b = (T^p - X) \subseteq K[T]$; then $\mathfrak a = (T^p - X) \subseteq \bar K[T]$. Now $$\operatorname{rad}(\mathfrak a) = \left(T-X^{\frac{1}{p}}\right),$$ which is not generated by polynomials over $K$. (In fact, $\mathfrak b$ is radical, so $\operatorname{rad}(\mathfrak b)^e = \mathfrak a$.)

However, the result is true for field extensions $K \subseteq L$ with $K$ perfect. We will break down the proof into a few smaller lemmata.

Remark. It suffices to consider the case where $B$ is a domain. Indeed, the localisation of $B$ at the set of nonzerodivisors is the total ring of fractions $K(B)$. It is the product of the localisations at the minimal primes of $B$. Since $B$ is reduced, each $B_\mathfrak p$ for $\mathfrak p$ minimal is a field; thus $B$ injects into a product of fields. Now if we prove the result for each $B_\mathfrak p$ (with $\mathfrak p$ minimal), then that of $B$ follows since a subring of a product of reduced rings is reduced.

Lemma 1. Let $K \subseteq L$ be finite separable, and let $B$ be a $K$-algebra. If $B$ is a domain, then $A = B \otimes_K L$ is reduced.

Proof. We can write $L = K[x]/f$ for some $f \in K[x]$ separable irreducible, by the primitive element theorem. Then $A = B[x]/f$, which embeds into $(\operatorname{Frac}B)[x]/f$. This is a reduced ring, since $f$ is separable (a quality that is not altered by field extension). $\square$

But: note that $f$ is not necessarily irreducible in $(\operatorname{Frac}B)[x]$.

Corollary 1. Let $K \subseteq L$ be separable algebraic, and let $B$ be a $K$-algebra. If $B$ is a domain, then $A = B \otimes_K L$ is reduced.

Proof. If $x \in A$ satisfies $x^k = 0$ for some $k \in \mathbb Z_{>0}$, then $x$ actually lives in $B \otimes_K M$ for some finite subextension $M \subseteq L$. Thus, $x = 0$ by the lemma. $\square$

Lemma 2. Let $K \subseteq L$ be purely transcendental, and let $B$ be a $K$-algebra. If $B$ is a domain, then so is $A = B \otimes_K L$.

Proof. In this case, $L = K(\{x_i\}_{i \in I})$ is a localisation of $K[\{x_i\}_{i \in I}]$, so $A = B \otimes_K K(\{x_i\}_{i \in I})$ is a localisation of $B[\{x_i\}_{i \in I}]$. $\square$

Note however that $A$ is not equal to $B(\{x_i\}_{i \in I})$. For example, we cannot necessarily divide by $x_i - b$ for all $b \in B$; only for all $b \in K$.

Corollary 2. Let $K \subseteq L$ be a separably generated extension, and let $B$ be a $K$-algebra. If $B$ is a domain, then $A = B\otimes_K L$ is reduced.

Proof. A separably generated extension $K \subseteq L$ can be written as a tower $K \subseteq K(\{x_i\}_{i \in I}) \subseteq L$, with the first extension purely transcendental and the second separable algebraic. For the first piece, the result follows from Lemma 2. For the second piece, the result follows from Corollary 1. $\square$

Corollary 3. Let $K \subseteq L$ be a separable extension, and let $B$ be a $K$-algebra. If $B$ is a domain, then $A = B \otimes_K L$ is reduced.

Proof. This is a limit argument identical to that of Corollary 1. $\square$

Proposition. Let $K$ be perfect, and let $K \subseteq L$ be any field extension. Let $B$ be a $K$-algebra. If $B$ is a domain, then $A \otimes_K L$ is reduced.

Proof. For $K$ perfect, every field extension is separable. Thus, the result follows from Corollary 3. $\square$

Remark. In all the lemmata, corollaries, and propositions above, we can replace the assumption if $B$ is a domain by if $B$ is a finite type reduced $K$-algebra. See the remark above.

The key word in all of the above is geometrically reduced. I just proved that every reduced variety over a perfect field is geometrically reduced, and showed that this is not true over imperfect fields.

Definition. Let $B$ be a finite type $K$-algebra. Then $B$ is geometrically reduced if $B \otimes_K L$ is reduced for every field extension $K \subseteq L$.

The above also shows:

Proposition. Let $B$ be a finite type $K$-algebra. Then the following are equivalent:

  1. $B$ is geometrically reduced;
  2. $B \otimes_K \bar K$ is reduced;
  3. $B \otimes_K K^{\operatorname{perf}}$ is reduced;
  4. $B \otimes_K L$ is reduced for every purely inseparable extension $K \subseteq L$.

Proof. The implications $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4$ are trivial. Suppose $4$ holds, and let $K \subseteq L$ be any field extension. By a limit argument as in Corollary 1, we may assume $L$ is finitely generated. By Tag 04KM, there exists a commutative diagram $$\begin{array}{ccc}K & \to & L \\ \downarrow & & \downarrow \\ K' & \to & L'\end{array}$$ with $K \subseteq K'$ and $L \subseteq L'$ finite and purely inseparable, and $K'\subseteq L'$ separably generated. By assumption, $B \otimes_K K'$ is reduced. By Corollary 2, so is $B \otimes_K L'$. The result follows since $B \otimes_K L \subseteq B \otimes_K L'$. $\square$

All of this can also be found (with a slightly different presentation) in Tag 030V. I got a little carried away trying to understand and restructure all the arguments myself...

  • $\begingroup$ Great answer. I didn't even need to read and understand the question to find what I was looking for! :) $\endgroup$ Dec 11, 2016 at 10:01

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