
## 10.47 Geometrically connected algebras

Lemma 10.47.1. Let $k$ be a separably algebraically closed field. Let $R$, $S$ be $k$-algebras. If $\mathop{\mathrm{Spec}}(R)$, and $\mathop{\mathrm{Spec}}(S)$ are connected, then so is $\mathop{\mathrm{Spec}}(R \otimes _ k S)$.

Proof. Recall that $\mathop{\mathrm{Spec}}(R)$ is connected if and only if $R$ has no nontrivial idempotents, see Lemma 10.20.4. Hence, by Lemma 10.42.4 we may assume $R$ and $S$ are of finite type over $k$. In this case $R$ and $S$ are Noetherian, and have finitely many minimal primes, see Lemma 10.30.6. Thus we may argue by induction on $n + m$ where $n$, resp. $m$ is the number of irreducible components of $\mathop{\mathrm{Spec}}(R)$, resp. $\mathop{\mathrm{Spec}}(S)$. Of course the case where either $n$ or $m$ is zero is trivial. If $n = m = 1$, i.e., $\mathop{\mathrm{Spec}}(R)$ and $\mathop{\mathrm{Spec}}(S)$ both have one irreducible component, then the result holds by Lemma 10.46.2. Suppose that $n > 1$. Let $\mathfrak p \subset R$ be a minimal prime corresponding to the irreducible closed subset $T \subset \mathop{\mathrm{Spec}}(R)$. Let $I \subset R$ be such that $T' = V(I) \subset \mathop{\mathrm{Spec}}(R)$ is the closure of the complement of $T$. Note that this means that $T' = \mathop{\mathrm{Spec}}(R/I)$ (Lemma 10.16.7) has $n - 1$ irreducible components. Then $T \cup T' = \mathop{\mathrm{Spec}}(R)$, and $T \cap T' = V(\mathfrak p + I) = \mathop{\mathrm{Spec}}(R/(\mathfrak p + I))$ is not empty as $\mathop{\mathrm{Spec}}(R)$ is assumed connected. The inverse image of $T$ in $\mathop{\mathrm{Spec}}(R \otimes _ k S)$ is $\mathop{\mathrm{Spec}}(R/\mathfrak p \otimes _ k S)$, and the inverse of $T'$ in $\mathop{\mathrm{Spec}}(R \otimes _ k S)$ is $\mathop{\mathrm{Spec}}(R/I \otimes _ k S)$. By induction these are both connected. The inverse image of $T \cap T'$ is $\mathop{\mathrm{Spec}}(R/(\mathfrak p + I) \otimes _ k S)$ which is nonempty. Hence $\mathop{\mathrm{Spec}}(R \otimes _ k S)$ is connected. $\square$

Lemma 10.47.2. 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 connected, and

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

Proof. For any extension of fields $k \subset k'$ the connectivity of the spectrum of $R \otimes _ k k'$ is equivalent to $R \otimes _ k k'$ having no nontrivial idempotents, see Lemma 10.20.4. 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 $R \otimes _ k \overline{k}$ having no nontrivial idempotents. For any field extension $k \subset k'$, there exists a field extension $\overline{k} \subset \overline{k}'$ with $k' \subset \overline{k}'$. By Lemma 10.47.1 we see that $R \otimes _ k \overline{k}'$ has no nontrivial idempotents. If $R \otimes _ k k'$ has a nontrivial idempotent, then also $R \otimes _ k \overline{k}'$, contradiction. $\square$

Definition 10.47.3. Let $k$ be a field. Let $S$ be a $k$-algebra. We say $S$ is geometrically connected over $k$ if for every field extension $k \subset k'$ the spectrum of $S \otimes _ k k'$ is connected.

By Lemma 10.47.2 it suffices to check this for finite separable field extensions $k \subset k'$.

Lemma 10.47.4. Let $k$ be a field. Let $R$ be a $k$-algebra. If $k$ is separably algebraically closed then $R$ is geometrically connected over $k$ if and only if the spectrum of $R$ is connected.

Proof. Immediate from the remark following Definition 10.47.3. $\square$

Lemma 10.47.5. Let $k$ be a field. Let $S$ be a $k$-algebra.

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

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

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

Proof. This follows from the characterization of connectedness in terms of the nonexistence of nontrivial idempotents. The second and third property follow from the fact that tensor product commutes with colimits. $\square$

The following lemma will be superseded by the more general Varieties, Lemma 32.7.4.

Lemma 10.47.6. Let $k$ be a field. Let $S$ be a geometrically connected $k$-algebra. Let $R$ be any $k$-algebra. The map

$R \longrightarrow R \otimes _ k S$

induces a bijection on idempotents, and the map

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

induces a bijection on connected components.

Proof. The second assertion follows from the first combined with Lemma 10.21.2. By Lemmas 10.47.5 and 10.42.4 we may assume that $R$ and $S$ are of finite type over $k$. Then we see that also $R \otimes _ k S$ is of finite type over $k$. Note that in this case all the rings are Noetherian and hence their spectra have finitely many connected components (since they have finitely many irreducible components, see Lemma 10.30.6). In particular, all connected components in question are open! Hence via Lemma 10.23.3 we see that the first statement of the lemma in this case is equivalent to the second. Let's prove this. As the algebra $S$ is geometrically connected and nonzero we see that all fibres of $X = \mathop{\mathrm{Spec}}(R \otimes _ k S) \to \mathop{\mathrm{Spec}}(R) = Y$ are connected and nonempty. Also, as $R \to R \otimes _ k S$ is flat of finite presentation the map $X \to Y$ is open (Proposition 10.40.8). Topology, Lemma 5.7.5 shows that $X \to Y$ induces bijection on connected components. $\square$

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