Let $\overline{\mathbb{Q}}$ be the algebraic closure of the field $\mathbb{Q}$.
I know that $\operatorname{Spec}\mathbb{Z}=\{0\}\cup\{(p): p\text{ is prime}\}$ and the closed points of $\operatorname{Spec}\mathbb{Z}$ is $(p)$. The closure of $\{0\}$ is $\operatorname{Spec}\mathbb{Z}$.
And I have proved that $\overline{\mathbb{Q}}$ is countable. But I am not sure about how does $\operatorname{Spec}\overline{\mathbb{Q}}[x]$ look like. I was thinking if I could find a bijection of closed point and non-closed point of these two sets, then I can say they are homeomorphic in Zariski topology. Is this right?
Best Answer
The prime ideals of $\overline{\Bbb Q}[x]$ are $\{0\}$ and $(x-a)$ for $a\in\overline{\Bbb Q}$. So the spectrum consists of countably infinitely many closed points and one dense point, just as the spectrum of $\Bbb Z$ has. These spectra are therefore homeomorphic.