Can I make the spectrum of ring, Spec(R), the topological ring or topological group?

Question

Sorry for a novice question.

My intuition is that for every ring $R$ in algebraic geometry, $\text{Spec}(R)$ is a concept like a manifold.

In manifold theory, we can consider the notion of the Lie group.

Finally, I find 'The algebraic group theory' like the Lie group in manifold theory. (is it correct intuition?)

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My question: Is there an easier way to make $\text{Spec}(R)$ into a topological group or ring? Or is there a theory for that?

Answer 1

Yes! There are several important examples of groups that are also affine schemes (the spectrum of a ring). These are called affine group-schemes, and I bet that you already know many examples of these.

One particularly nice example is the matrix group $\mathsf{GL}_n$. To define it, consider the polynomial ring $\mathbf{Z}[x_{11}, \dots, x_{nn}, t]$ with $n^2 + 1$ variables. Call $\det$ the polynomial obtained by taking the determinant of the matrix whose entries are the variables $x_{ij}$ and form the quotient ring $R = \mathbf{Z}[x_{11}, \dots, x_{nn}, t]/(t\cdot \det -1)$. Then, one can show that $\mathsf{GL}_n = \mathrm{Spec}(R)$ is an affine-group scheme. (Think first about the case $n = 1$).

A great (free) reference to start learning about these is Milne's notes: https://www.jmilne.org/math/CourseNotes/iAG200.pdf

More generally, you might be interested in learning about the functorial perspective of a group object in a category; of which Lie groups, affine group-schemes, topological groups, groups, algebraic groups, and more are all examples.

https://en.wikipedia.org/wiki/Group_object#:~:text=In%20category%20theory%2C%20a%20branch,the%20group%20operations%20are%20continuous.

Answer 2

All the comments and answers pointing towards the theory of group schemes are what you should be looking at if you want to generalize the theory of Lie groups to the context of schemes.

But to answer your question more litteraly, it is almost never possible to give a topological group structure to the topological space $\operatorname{Spec}(R)$, because a topological group is homogeneous: all its points have the same topological properties (specifically, the homoeomorphism group of the underlying topological space acts transitively). But an affine scheme $\operatorname{Spec}(R)$ is never homogeneous unless it is discrete, because any non-trivial irreducible component will have a generic point and some non-generic points, which are not topologically equivalent.