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If we consider G to be the group of non-singular n by n matrices carrying Zariski topology. Then the joint continuity of the multiplication map from G x G to G would make G Hausdorff.(We consider the topology in G x G to be the Cartesian product topology of the Zariski topology of G.) Please explain how ?

(Reference- Kaplansky's book 'An Introduction to Differential Algebra')

Jagdeep Singh
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More generally, the following argument shows that any $T_1$ topological group $G$ is automatically Hausdorff (in fact, a slightly longer argument shows that actually you only have to assume $T_0$). Here "topological group" means that the multiplication map $G\times G\to G$ and the inverse map $G\to G$ are continuous. Given this, consider the map $f:G\times G\to G$ given by $f(g,h)=gh^{-1}$. This map is continuous because the multiplication and inverse maps are. Since $G$ is $T_1$, $\{1\}$ is closed in $G$, and hence $f^{-1}(\{1\})$ is closed in $G\times G$. But $f^{-1}(\{1\})=\{(g,g):g\in G\}$ is the diagonal in $G\times G$, and a space $G$ is Hausdorff iff its diagonal is closed in $G\times G$ (if you are not familiar with this fact, it is a good and simple exercise to prove it).

Eric Wofsey
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  • I think you actually need the result for $T_0$ groups here, because the OP is asking about the Zariski topology, which is usually not $T_1$. – Andreas Blass Aug 13 '15 at 00:33
  • @AndreasBlass: From context it sounds as though they are talking about the Zariski topology on the $k$-points of $GL_n$ for some field $k$, which is always $T_1$. Otherwise, I'm not sure in what sense they even mean that it is a "group" (if you are working with schemes, the underlying set of the scheme product $G\times_{\operatorname{Spec} k} G$ is not just the cartesian product of the underlying set of $G$ with itself). – Eric Wofsey Aug 13 '15 at 01:08
  • Not having the book, I conjectured that the passage cited in the question was intended as a hint why the product topology is not a good idea in the world of schemes. But you're right that he might well be thinking only of closed points, where the product topology is still not a good idea. (In the case of schemes, I'd say the real reason for not using the product topology is that it doesn't have the universal property of a product, but this group argument could be useful for people who don't appreciate category theory.) – Andreas Blass Aug 13 '15 at 01:16