Is it true that under the Zariski topology, a subset is dense if and only if it is a nonempty open subset?
I know this is true in one direction, i.e., any nonempty open subset is dense, but how about the other?
In fact, I have two problems in proving
Any closed connected algebraic group of dimension $1$ is commutative,
and what I have mentioned before is one of these.
Suppose that $G$ is a closed connected algebraic group of dimension $1$, and $x$ is an element out of $Z(G)$. Denote $Cl_G(x)$ the conjugacy class of $x$ in $G$. I followed the hints:
- The class of $x$ is infinite.
Denote the centralizer of $x$ as $C_G(x)$. This is a closed subgroup $G$ and inequal to $G$, so it is finite. Moreover, $|C_G(x)||Cl_G(x)| = |G|$. $|G|$ is infinite, so is $Cl_G(x)$.
- So $Cl_G(x)$ is dense in $G$.
If $O$ is any open subset of $G$ intersecting $Cl_G(x)$ trivially, then $G-O \subseteq \overline{Cl_G(x)} = G$. So, $O$ is empty, and $Cl_G(x)$ is dense in $G$.
- The complement of $Cl_G(x)$ in $G$ is finite.
I failed in proving this. But if it is true that only nonempty open subsets of $G$ are dense, then obviously, $Cl_G(x)$ would have finite complement.
- Assume $G \subseteq GL(n,K)$, $c_i(y) = i$th coefficient of characteristic polynomial of $y \in G$. Then $\phi(y) = (c_0(y), \cdots, c_n(y))$ defines a morphism $G \rightarrow K^{n+1}$ with finite image.
All the elements in $Cl_G(x)$ have the same characteristic polynomial, so they are mapped to the same sequence $(c_0, \cdots, c_n)$. $Cl_G(x)$ has finite complement, so $\phi(G)$ is finite.
- Conclude that $G$ is unipotent, hence nilpotent.
If $G$ contains no semisimple elements other than the identity, then $G$ must consist of unipotent elements. In this case, $G$ is nilpotent, and $(G,G) \neq G$. But $(G,G)$ is a closed connected subgroup of $G$, and $\mathrm{dim}G =1$. This forces $(G,G)$ to be $1$, thus proving the commutativity of $G$. So, what I have to do is to rule out the possibility of $G$ containing nonidentity semisimple elements. If $K$ is of characteristic $0$ and $1 \neq s \in G$ is semisimple element, then for any $n_1 \neq n_2 \in \mathbb{Z}_+$, $\phi(s^{n_1}) \neq \phi(s^{n_2})$, contradicting the fact that $\phi(G)$ is finite. But when $K$ is of prime characteristic, I don't know what will happen.
Thanks for any help.