Since $k$ is a field, the ring of polynomials $k[x]$ will be a PID, and hence we can describe what the prime ideals of $k[x,y]\cong k[x][y]$ will be as is explained in the linked post. Now, given any commutative ring $R$ and some ideal $I$ of it, the quotient $R/I$ will be a flat $R$-module if $I^2=I$ (to see this, tensor the exact sequence $0\to I\to R$). If moreover we assume that $I$ is finitely generated, we get that $I=(e)$ where $e$ is an idempotent of $R$. Consider $k[x,y]/(x-y)$ as an $k[x,y]-$module.

It is clear that $x-y$ is irreducible as a polynomial over $k[x]$ in $y$, hence it is prime.

How would we go about showing that $V((x-y))$ is clopen (or not) in $\text{Spec} \ k[x,y]$?

The algebraic variety is the diagonal of $k^2$, but I do not know if this tells us anything about the structure in the spectrum.

I apologize in advance if the question has been asked or if I'm not understanding the definitions properly, I've started learning about these things recently and am still a little confused about everything.

I appreciate any answer!

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    Why do you think that $V((x-y))$ is open? (the set of all primes ideals containing $(x-y)$) – reuns Apr 09 '22 at 13:27
  • I don't think it's true that $I^2=I$ is enough for $R/I$ to be flat, and I don't understand what relation this has with $V((x-y))$. – Sergey Guminov Apr 09 '22 at 13:47
  • @reuns I don't know if $V((x-y))$ is open. I thought that looking at that set would tell me something about the ideal $(x-y)$ (so that, if it turns out that the set isn't clopen, the ideal isn't idempotent.) It would probably be much easier to check if $(x-y)^2=(x-y)$ directly, but I wanted some intuition. – kefirofil Apr 09 '22 at 14:56
  • @SergeyGuminov It is necessary, right? (from the exact sequence). And if it is idempotent, then $V(I)$ should be clopen in $\text{Spec} R$. (the second link). – kefirofil Apr 09 '22 at 14:57
  • @kefirofil. It is necessary, but I still do not see how this is relevant, since $(x-y)^2\neq (x-y)$. – Sergey Guminov Apr 09 '22 at 15:08
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    If $\operatorname{Spec} A$ has any nontrivial clopen sets, then $A$ has a nontrivial idempotent as shown in the linked duplicate. But $k[x,y]$ is a domain and therefore the only idempotents are $0$ and $1$. – KReiser Apr 09 '22 at 17:46

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