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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.

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It is not true that a dense set is open, and it is true that every non-empty open set is dense only if the scheme (or variety) is irreducible. – Georges Elencwajg Nov 10 '11 at 16:27
Maybe the word "irreducible" should be used. As a guess, what if I take something like the open set $\mathbf{A}^2 - Z(y)$ and add in the origin. That's still dense. Is that open? – Dylan Moreland Nov 10 '11 at 16:27
I do not know much about the Zariski topology, however if it is the co-finite topology then any infinite set is dense, not only open subsets. This is simply because every open set meets every infinite set. – Asaf Karagila Nov 10 '11 at 16:31
@Asaf That reasoning should work for $\mathbf{A}^1$, then. – Dylan Moreland Nov 10 '11 at 16:32
Dear @Dylan and Asaf, what you say is correct in the language of varieties. However in the language of schemes $\mathbb A^1$ does not carry the cofinite topology because the complement of the generic point is not open. More generally, contrary to a widespread misconception, an irreducible scheme never carries the cofinite topology, for the same reason. And yet: in an irreducible scheme every non-empty open set is dense. I acknowledge that this is a little confusing... – Georges Elencwajg Nov 10 '11 at 18:50
up vote 3 down vote accepted

For the sake of having an answer: no. For example, $\mathbb{Z}$ is dense in $\mathbb{A}^1(F)$ for any field $F$ of characteristic zero.

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Thank you very much for answering. Now I know what is wrong with my question. Can I change it a little bit now? I am sorry but in fact, I want to know something more. – ShinyaSakai Nov 11 '11 at 8:41
Oh, I think I should ask another question... Thanks again! – ShinyaSakai Nov 11 '11 at 8:52

Is it improper to answer my own questions? I hope not... I do this just for the convinience of any possible viewers who are interested in the proof of commutativity, i.e.,

Any closed connected algebraic group of dimension $1$ is commutative.

In the third step, I failed because I thought that any dense subset of a closed connected $1$-dimensional algebraic group had finite complement. This is false. A counterexample is given by Yuan in the first answer.

What I neglected is the fact that being a conjugacy class, $Cl_G(x)$ is locally closed. In fact,

Any dense and locally closed subset of a closed connected $1$-dimensional algebraic group have finite complement.

Suppose that $G$ is such a group, and $X$ is a dense and locally closed subset of $G$. Then there exists a closed subset $C$ and an open subset $O$ of $G$, such that $X = C \cap O$. As $X$ is dense in $G$, $C$ is also dense in $G$. But $G-C$ is an open subset of $G$ intersecting $C$ trivially. $G-C$ must be empty, and $C = G$. Then $X = O$ is an open subset, nonempty because of density. If $G - O$ is infinite, then its closure $\overline{G-O}$ must be of dimension at least $1$. This dimension $\leq \mathrm{dim}G =1$. Clearly, $\overline{G-O} = G$. As $O$ is open, $\overline{G-O} = G - O$. From this, $O$ is empty. Contradiction.

I hope I am not mistaken.

The key to the fifth step may be in the identification of $G \cap D(n,K)$ when it is not $\{e \}$. But I still have to think for a while.

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Not at all improper; in fact it’s encouraged to answer your own question if you can. – Brian M. Scott Nov 14 '11 at 6:19

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