## 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

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

$R \to S$ is flat,

$R \to S$ is of finite presentation,

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

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

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 irreducible^{1}.

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.

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

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

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.
$\square$

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