The Stacks project

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10.46 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 10.46.1. 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 10.40.8. Hence the lemma follows from Topology, Lemma 5.8.12. $\square$

Lemma 10.46.2. 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 10.44.5. 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 10.45.7. 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 10.44.6 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 10.42.4 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 10.33.1. Moreover, the set of maximal ideals is dense in the spectrum of $R$ since $\mathop{\mathrm{Spec}}(R)$ is Jacobson, see Lemma 10.34.2. Hence we see that Lemma 10.46.1 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 10.46.3. Let $k$ be a field. Let $R$ be a $k$-algebra. The following are equivalent

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

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

Proof. Let $k \subset k^{perf}$ be a perfect closure of $k$, see Definition 10.44.5. By Lemma 10.45.7 we may replace $R$ by $(R \otimes _ k k^{perf})_{reduction}$ and $k$ by $k^{perf}$ (some details omitted). Hence we may assume that $R$ is geometrically reduced over $k$.

Assume $R$ is geometrically reduced over $k$. For any extension of fields $k \subset k'$ we see irreducibility of the spectrum of $R \otimes _ k k'$ is equivalent to $R \otimes _ k k'$ being a domain. Assume (2). Let $k \subset \overline{k}$ be a separable algebraic closure of $k$. Using Lemma 10.42.4 we see that (2) is equivalent to $R \otimes _ k \overline{k}$ being a domain. For any field extension $k \subset k'$, there exists a field extension $\overline{k} \subset \overline{k}'$ with $k' \subset \overline{k}'$. By Lemma 10.46.2 we see that $R \otimes _ k \overline{k}'$ is a domain. If $R \otimes _ k k'$ is not a domain, then also $R \otimes _ k \overline{k}'$ is not a domain, contradiction. $\square$

Definition 10.46.4. Let $k$ be a field. Let $S$ be a $k$-algebra. We say $S$ is geometrically irreducible over $k$ if for every field extension $k \subset k'$ the spectrum of $S \otimes _ k k'$ is irreducible1.

By Lemma 10.46.3 it suffices to check this for finite separable field extensions $k \subset k'$.

Lemma 10.46.5. 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 10.46.4. $\square$

Lemma 10.46.6. 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 $k \subset k'$ the ring map $S' \otimes _ k k' \to S \otimes _ k k'$ is injective also. Hence (1) follows from Lemma 10.29.5 (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 10.46.7. 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 10.25.1). As $R \to R \otimes _ k S$ is flat we see by going down (Lemma 10.38.18) 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 10.29.5). 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 10.46.8. Let $k \subset K$ be a field extension. If $k$ is algebraically closed in $K$, then $K$ is geometrically irreducible over $k$.

Proof. Let $k \subset k'$ be a finite separable extension, say 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 $K[T]$. Say $P = P_1P_2$ is a nontrivial factorization of $P$ into monic polynomials. Let $b_1, \ldots , b_ t \in K$ be the coefficients of $P_1$. Then we see that $b_ i$ is algebraic over $k$ by Lemma 10.37.5. Hence the lemma follows. $\square$

Lemma 10.46.9. Let $k \subset K$ be a field extension. Consider the subextension $k \subset k' \subset K$ such that $k \subset k'$ is separable algebraic and $k' \subset K$ maximal with this property. Then $K$ is geometrically irreducible over $k'$. If $K/k$ is a finitely generated field extension, then $[k' : k] < \infty $.

Proof. Let $k'' \subset K$ be the algebraic closure of $k$ in $K$. By Lemma 10.46.8 we see that $K$ is geometrically irreducible over $k''$. Since $k' \subset k''$ is purely inseparable (Fields, Lemma 9.14.6) we see from Lemma 10.45.7 that the extension $k' \subset K$ is also geometrically irreducible. If $k \subset K$ is finitely generated, then $k'$ is finite over $k$ by Fields, Lemma 9.26.10. $\square$

Lemma 10.46.10. Let $k \subset K$ be an extension of fields. Let $k \subset \overline{k}$ be a separable algebraic closure. Then $\text{Gal}(\overline{k}/k)$ acts transitively on the primes of $\overline{k} \otimes _ k K$.

Proof. Let $k \subset k' \subset K$ be the subextension found in Lemma 10.46.9. 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 10.35.20. By Lemma 10.46.7 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, (1) $\overline{k} \otimes _ k K = (\overline{k} \otimes _ k k') \otimes _{k'} K$, 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.

Comments (2)

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 factors, the coefficients of any factors (which lie in ) are separable over , being sums of products of the roots of , hence the coefficients lie in (if is separably closed in ). And the converse holds too. If is geometrically irreducible over and is separable algebraic over with minimal polynomial , then has disconnected spectrum unless .

Comment #412 by on

First of all: yes. Secondly, Lemma 10.46.9 contains the result you mention. So, you can just view Lemma 10.46.8 as part of building theory and not the best result of its kind. I like how the proof of 10.46.8 works now because I think 10.37.5 is wonderful and should be used as often as possible.


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