## 10.47 Geometrically irreducible algebras

An algebra $S$ over a field $k$ is geometrically irreducible if the algebra $S \otimes _ k k'$ has a unique minimal prime for every field extension $k'/k$. In this section we develop a bit of theory relevant to this notion.

Lemma 10.47.1. Let $R \to S$ be a ring map. Assume

1. $\mathop{\mathrm{Spec}}(R)$ is irreducible,

2. $R \to S$ is flat,

3. $R \to S$ is of finite presentation,

4. the fibre rings $S \otimes _ R \kappa (\mathfrak p)$ have irreducible spectra for a dense collection of primes $\mathfrak p$ of $R$.

Then $\mathop{\mathrm{Spec}}(S)$ is irreducible. This is true more generally with (b) $+$ (c) replaced by “the map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ is open”.

Proof. The assumptions (b) and (c) imply that the map on spectra is open, see Proposition 10.41.8. Hence the lemma follows from Topology, Lemma 5.8.14. $\square$

Lemma 10.47.2. Let $k$ be a separably closed field. Let $R$, $S$ be $k$-algebras. If $R$, $S$ have a unique minimal prime, so does $R \otimes _ k S$.

Proof. Let $k \subset \overline{k}$ be a perfect closure, see Definition 10.45.5. By assumption $\overline{k}$ is algebraically closed. The ring maps $R \to R \otimes _ k \overline{k}$ and $S \to S \otimes _ k \overline{k}$ and $R \otimes _ k S \to (R \otimes _ k S) \otimes _ k \overline{k} = (R \otimes _ k \overline{k}) \otimes _{\overline{k}} (S \otimes _ k \overline{k})$ satisfy the assumptions of Lemma 10.46.7. Hence we may assume $k$ is algebraically closed.

We may replace $R$ and $S$ by their reductions. Hence we may assume that $R$ and $S$ are domains. By Lemma 10.45.6 we see that $R \otimes _ k S$ is reduced. Hence its spectrum is reducible if and only if it contains a nonzero zerodivisor. By Lemma 10.43.4 we reduce to the case where $R$ and $S$ are domains of finite type over $k$ algebraically closed.

Note that the ring map $R \to R \otimes _ k S$ is of finite presentation and flat. Moreover, for every maximal ideal $\mathfrak m$ of $R$ we have $(R \otimes _ k S) \otimes _ R R/\mathfrak m \cong S$ because $k \cong R/\mathfrak m$ by the Hilbert Nullstellensatz Theorem 10.34.1. Moreover, the set of maximal ideals is dense in the spectrum of $R$ since $\mathop{\mathrm{Spec}}(R)$ is Jacobson, see Lemma 10.35.2. Hence we see that Lemma 10.47.1 applies to the ring map $R \to R \otimes _ k S$ and we conclude that the spectrum of $R \otimes _ k S$ is irreducible as desired. $\square$

Lemma 10.47.3. Let $k$ be a field. Let $R$ be a $k$-algebra. The following are equivalent

1. for every field extension $k'/k$ the spectrum of $R \otimes _ k k'$ is irreducible,

2. for every finite separable field extension $k'/k$ the spectrum of $R \otimes _ k k'$ is irreducible,

3. the spectrum of $R \otimes _ k \overline{k}$ is irreducible where $\overline{k}$ is the separable algebraic closure of $k$, and

4. the spectrum of $R \otimes _ k \overline{k}$ is irreducible where $\overline{k}$ is the algebraic closure of $k$.

Proof. It is clear that (1) implies (2).

Assume (2) and let $\overline{k}$ is the separable algebraic closure of $k$. Suppose $\mathfrak q_ i \subset R \otimes _ k \overline{k}$, $i = 1, 2$ are two minimal prime ideals. For every finite subextension $\overline{k}/k'/k$ the extension $k'/k$ is separable and the ring map $R \otimes _ k k' \to R \otimes _ k \overline{k}$ is flat. Hence $\mathfrak p_ i = (R \otimes _ k k') \cap \mathfrak q_ i$ are minimal prime ideals (as we have going down for flat ring maps by Lemma 10.39.19). Thus we see that $\mathfrak p_1 = \mathfrak p_2$ by assumption (2). Since $\overline{k} = \bigcup k'$ we conclude $\mathfrak q_1 = \mathfrak q_2$. Hence $\mathop{\mathrm{Spec}}(R \otimes _ k \overline{k})$ is irreducible.

Assume (3) and let $\overline{k}$ be the algebraic closure of $k$. Let $\overline{k}/\overline{k}'/k$ be the corresponding separable algebraic closure of $k$. Then $\overline{k}/\overline{k}'$ is purely inseparable (in positive characteristic) or trivial. Hence $R \otimes _ k \overline{k}' \to R \otimes _ k \overline{k}$ induces a homeomorphism on spectra, for example by Lemma 10.46.7. Thus we have (4).

Assume (4). Let $k'/k$ be an arbitrary field extension and let $\overline{k}$ be the algebraic closure of $k$. We may choose a field $F$ such that both $k'$ and $\overline{k}$ are isomorphic to subfields of $F$. Then

$R \otimes _ k F = (R \otimes _ k \overline{k}) \otimes _{\overline{k}} F$

and hence we see from Lemma 10.47.2 that $R \otimes _ k F$ has a unique minimal prime. Finally, the ring map $R \otimes _ k k' \to R \otimes _ k F$ is flat and injective and hence any minimal prime of $R \otimes _ k k'$ is the image of a minimal prime of $R \otimes _ k F$ (by Lemma 10.30.5 and going down). We conclude that there is only one such minimal prime and the proof is complete. $\square$

Definition 10.47.4. Let $k$ be a field. Let $S$ be a $k$-algebra. We say $S$ is geometrically irreducible over $k$ if for every field extension $k'/k$ the spectrum of $S \otimes _ k k'$ is irreducible1.

By Lemma 10.47.3 it suffices to check this for finite separable field extensions $k'/k$ or for $k'$ equal to the separable algebraic closure of $k$.

Lemma 10.47.5. Let $k$ be a field. Let $R$ be a $k$-algebra. If $k$ is separably algebraically closed then $R$ is geometrically irreducible over $k$ if and only if the spectrum of $R$ is irreducible.

Proof. Immediate from the remark following Definition 10.47.4. $\square$

Lemma 10.47.6. Let $k$ be a field. Let $S$ be a $k$-algebra.

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

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

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

Proof. Let $S' \subset S$ be a subalgebra. Then for any extension $k'/k$ the ring map $S' \otimes _ k k' \to S \otimes _ k k'$ is injective also. Hence (1) follows from Lemma 10.30.5 (and the fact that the image of an irreducible space under a continuous map is irreducible). The second and third property follow from the fact that tensor product commutes with colimits. $\square$

Lemma 10.47.7. Let $k$ be a field. Let $S$ be a geometrically irreducible $k$-algebra. Let $R$ be any $k$-algebra. The map

$\mathop{\mathrm{Spec}}(R \otimes _ k S) \longrightarrow \mathop{\mathrm{Spec}}(R)$

induces a bijection on irreducible components.

Proof. Recall that irreducible components correspond to minimal primes (Lemma 10.26.1). As $R \to R \otimes _ k S$ is flat we see by going down (Lemma 10.39.19) that any minimal prime of $R \otimes _ k S$ lies over a minimal prime of $R$. Conversely, if $\mathfrak p \subset R$ is a (minimal) prime then

$R \otimes _ k S/\mathfrak p(R \otimes _ k S) = (R/\mathfrak p) \otimes _ k S \subset \kappa (\mathfrak p) \otimes _ k S$

by flatness of $R \to R \otimes _ k S$. The ring $\kappa (\mathfrak p) \otimes _ k S$ has irreducible spectrum by assumption. It follows that $R \otimes _ k S/\mathfrak p(R \otimes _ k S)$ has a single minimal prime (Lemma 10.30.5). In other words, the inverse image of the irreducible set $V(\mathfrak p)$ is irreducible. Hence the lemma follows. $\square$

Let us make some remarks on the notion of geometrically irreducible field extensions.

Lemma 10.47.8. Let $K/k$ be a field extension. If $k$ is algebraically closed in $K$, then $K$ is geometrically irreducible over $k$.

Proof. Assume $k$ is algebraically closed in $K$. By Definition 10.47.4 and Lemma 10.47.3 it suffices to show that the spectrum of $K \otimes _ k k'$ is irreducible for every finite separable extension $k'/k$. Say $k'$ is generated by $\alpha \in k'$ over $k$, see Fields, Lemma 9.19.1. Let $P = T^ d + a_1 T^{d - 1} + \ldots + a_ d \in k[T]$ be the minimal polynomial of $\alpha$. Then $K \otimes _ k k' \cong K[T]/(P)$. The only way the spectrum of $K[T]/(P)$ can be reducible is if $P$ is reducible in $K[T]$. Assume $P = P_1 P_2$ is a nontrivial factorization in $K[T]$ to get a contradiction. By Lemma 10.38.5 we see that the coefficients of $P_1$ and $P_2$ are algebraic over $k$. Our assumption implies the coefficients of $P_1$ and $P_2$ are in $k$ which contradicts the fact that $P$ is irreducible over $k$. $\square$

Lemma 10.47.9. Let $K/k$ be a geometrically irreducible field extension. Let $S$ be a geometrically irreducible $K$-algebra. Then $S$ is geometrically irreducible over $k$.

Proof. By Definition 10.47.4 and Lemma 10.47.3 it suffices to show that the spectrum of $S \otimes _ k k'$ is irreducible for every finite separable extension $k'/k$. Since $K$ is geometrically irreducible over $k$ we see that $K' = K \otimes _ k k'$ is a finite, separable field extension of $K$. Hence the spectrum of $S \otimes _ k k' = S \otimes _ K K'$ is irreducible as $S$ is assumed geometrically irreducible over $K$. $\square$

Lemma 10.47.10. Let $K/k$ be a field extension. The following are equivalent

1. $K$ is geometrically irreducible over $k$, and

2. the induced extension $K(t)/k(t)$ of purely transcendental extensions is geometrically irreducible.

Proof. Assume (1). Denote $\Omega$ an algebraic closure of $k(t)$. By Definition 10.47.4 we find that the spectrum of

$K \otimes _ k \Omega = K \otimes _ k k(t) \otimes _{k(t)} \Omega$

is irreducible. Since $K(t)$ is a localization of $K \otimes _ k k(T)$ we conclude that the spectrum of $K(t) \otimes _{k(t)} \Omega$ is irreducible. Thus by Lemma 10.47.3 we find that $K(t)/k(t)$ is geometrically irreducible.

Assume (2). Let $k'/k$ be a field extension. We have to show that $K \otimes _ k k'$ has a unique minimal prime. We know that the spectrum of

$K(t) \otimes _{k(t)} k'(t)$

is irreducible, i.e., has a unique minimal prime. Since there is an injective map $K \otimes _ k k' \to K(t) \otimes _{k(t)} k'(t)$ (details omitted) we conclude by Lemmas 10.30.5 and 10.30.7. $\square$

Lemma 10.47.11. Let $K/L/M$ be a tower of fields with $L/M$ geometrically irreducible. Let $x \in K$ be transcendental over $L$. Then $L(x)/M(x)$ is geometrically irreducible.

Proof. This follows from Lemma 10.47.10 because the fields $L(x)$ and $M(x)$ are purely transcendental extensions of $L$ and $M$. $\square$

Lemma 10.47.12. Let $K/k$ be a field extension. The following are equivalent

1. $K/k$ is geometrically irreducible, and

2. every element $\alpha \in K$ separably algebraic over $k$ is in $k$.

Proof. Assume (1) and let $\alpha \in K$ be separably algebraic over $k$. Then $k' = k(\alpha )$ is a finite separable extension of $k$ contained in $K$. By Lemma 10.47.6 the extension $k'/k$ is geometrically irreducible. In particular, we see that the spectrum of $k' \otimes _ k \overline{k}$ is irreducible (and hence if it is a product of fields, then there is exactly one factor). By Fields, Lemma 9.13.4 it follows that $\mathop{\mathrm{Hom}}\nolimits _ k(k', \overline{k})$ has one element which in turn implies that $k' = k$ by Fields, Lemma 9.12.11. Thus (2) holds.

Assume (2). Let $k' \subset K$ be the subfield consisting of elements algebraic over $k$. By Lemma 10.47.8 the extension $K/k'$ is geometrically irreducible. By assumption $k'/k$ is a purely inseparable extension. By Lemma 10.46.7 the extension $k'/k$ is geometrically irreducible. Hence by Lemma 10.47.9 we see that $K/k$ is geometrically irreducible. $\square$

Lemma 10.47.13. Let $K/k$ be a field extension. Consider the subextension $K/k'/k$ consisting of elements separably algebraic over $k$. Then $K$ is geometrically irreducible over $k'$. If $K/k$ is a finitely generated field extension, then $[k' : k] < \infty$.

Proof. The first statement is immediate from Lemma 10.47.12 and the fact that elements separably algebraic over $k'$ are in $k'$ by the transitivity of separable algebraic extensions, see Fields, Lemma 9.12.12. If $K/k$ is finitely generated, then $k'$ is finite over $k$ by Fields, Lemma 9.26.11. $\square$

Lemma 10.47.14. Let $K/k$ be an extension of fields. Let $\overline{k}/k$ be a separable algebraic closure. Then $\text{Gal}(\overline{k}/k)$ acts transitively on the primes of $\overline{k} \otimes _ k K$.

Proof. Let $K/k'/k$ be the subextension found in Lemma 10.47.13. Note that as $k \subset \overline{k}$ is integral all the prime ideals of $\overline{k} \otimes _ k K$ and $\overline{k} \otimes _ k k'$ are maximal, see Lemma 10.36.20. By Lemma 10.47.7 the map

$\mathop{\mathrm{Spec}}(\overline{k} \otimes _ k K) \to \mathop{\mathrm{Spec}}(\overline{k} \otimes _ k k')$

is bijective because (1) all primes are minimal primes, (2) $\overline{k} \otimes _ k K = (\overline{k} \otimes _ k k') \otimes _{k'} K$, and (3) $K$ is geometrically irreducible over $k'$. Hence it suffices to prove the lemma for the action of $\text{Gal}(\overline{k}/k)$ on the primes of $\overline{k} \otimes _ k k'$.

As every prime of $\overline{k} \otimes _ k k'$ is maximal, the residue fields are isomorphic to $\overline{k}$. Hence the prime ideals of $\overline{k} \otimes _ k k'$ correspond one to one to elements of $\mathop{\mathrm{Hom}}\nolimits _ k(k', \overline{k})$ with $\sigma \in \mathop{\mathrm{Hom}}\nolimits _ k(k', \overline{k})$ corresponding to the kernel $\mathfrak p_\sigma$ of $1 \otimes \sigma : \overline{k} \otimes _ k k' \to \overline{k}$. In particular $\text{Gal}(\overline{k}/k)$ acts transitively on this set as desired. $\square$

[1] An irreducible space is nonempty.

Comment #410 by Keenan Kidwell on

I'm not sure if this is worth changing, but in the statement of 037P, "algebraically closed" can be replaced with "separably closed," and the same argument works, because if $P$ factors, the coefficients of any factors (which lie in $K$) are separable over $k$, being sums of products of the roots of $P$, hence the coefficients lie in $k$ (if $k$ is separably closed in $K$). And the converse holds too. If $K$ is geometrically irreducible over $k$ and $\alpha\in K$ is separable algebraic over $k$ with minimal polynomial $P$, then $k(\alpha)\otimes_kK=K[T]/(P)$ has disconnected spectrum unless $\deg(P)=1$.

Comment #412 by on

First of all: yes. Secondly, Lemma 10.47.13 contains the result you mention. So, you can just view Lemma 10.47.8 as part of building theory and not the best result of its kind. I like how the proof of 10.47.8 works now because I think 10.38.5 is wonderful and should be used as often as possible.

Comment #5466 by Yoav on

In the proof of lemma 04KP you wrote (1), (1) and (3). It should be (1), (2) and (3).

Comment #6639 by Lukas Kofler on

In the proof of Lemma 10.47.12 you wrote "k'=k(alpha) is a finite separable extension of k contained in k". The last "k" should be "K".

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