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Let $G$ be a connected linear algebraic group over a field $k$ of characteristic 0. A paper I'm reading seems to imply that $\overline{G}:= G \times_k \overline{k}$ will also be connected, but I don't see how this is true. Can anyone help?

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  • $\begingroup$ The Galois group of $k$ permutes the connected components of $\bar G$ but leaves invariant the identity element. $\endgroup$ – Ariyan Javanpeykar Dec 1 '14 at 8:41
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Here is a nice fact, which implies your question:

Theorem: Let $X/k$ be a finite type $k$-scheme. Assume that $X$ is connected and has a $k$-point. Then, $X$ is geometrically connected.

Proof: It suffices to show that for all finite extesnsions $K/k$ that $X_K$ is connected (why?). To do this, note that we have the following, definitional, fiber diagram

$$\begin{matrix}X_K & \to & X\\ \downarrow & & \downarrow\\ \text{Spec}(K) & \to & \text{Spec}(k)\end{matrix}$$

Now, since $\text{Spec}(K)\to\text{Spec}(k)$ is both finite and flat, so is $X_K\to X$. But, flatness (together with finite type) implies that $X_K$ is open, and finiteness implies that $X_K\to X$ is closed.

Now, let $C\subseteq X_K$ be a component of $X_K$. Since $X_K$ is Noetherian, we know that this component is clopen (since there are finitely many suchcomponents). Thus, since $X_K\to X$ is both closed and open, we have that the image is also clopen. Since $X$ was assumed to be connected, this implies that the image of $C$ is everything. Thus, all connected components of $X_K$ surject onto $X$.

But, let $p\in X(k)$ be the $k$-point we assumed $X$ had. Then, the fiber $(X_K)_p)$ of $p$ under the map $X_K\to X$ sits in the following fibered diagram

$$\begin{matrix}(X_K)_p & \to & X_K & \to & \text{Spec}(K)\\ \downarrow & & \downarrow & & \downarrow\\ \text{Spec}(k) & \to & X & \to & \text{Spec}(k) \end{matrix}$$

and so, in particular, $(X_K)_p$ is $\text{Spec}(K\otimes_k k)=\text{Spec}(K)$, and thus $(X_K)_p$ is a point.

Remark: intuitively this just says that since base-changing to $K$ should just break apart points identified when all we can see is $k$, the $k$-point, which can't be broken apart in this manner, stays the same.

But, we said that all connected components of $X_K$ must contain $(X_K)_p$. And, since connected components are disjoint, this implies that $X_K$ has only one connected component, and thus is connected. $\blacksquare$

Now, since every algebraic group over a field has a $k$-point by definition, the above tells us that all connected algebraic groups are geometrically connected. Also note that there was no need to assume characteristic $0$ in the above.

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