An open subscheme of a variety is a variety?

Question

In Vakil's notes, he mentions that an open subscheme $U$ of a $k$-variety $X$ is automatically a $k$-variety. Why is this the case? Here a $k$-variety is defined as a reduced, separated scheme of finite type over $k$.

It is immediately clear that $U$ is separable, because our map $U\to Spec(k)$ is equal to the composition $U\hookrightarrow X\to Spec(k)$, both of which are separable maps (the former because it is monic, and the latter by definition). And it is clear that $U$ is reduced, because stalks of points in $U$ are the same as their stalks in $X$, which are all reduced.

So all I'm unsure about is, how can I see that $U$ is finite type over $k$?

Answer 1

We have a finite sum $$X=\cup Spec A_i$$ where each $A_i$ is a finitely generated $k$-algebra. Note that this implies that $A_i$ is noetherian, as $A$ is the image of $k[x_1, \ldots x_n]$.

Now let $U\subseteq X$ be open, it suffices to show that $U\cap Spec A_i$ is quasi compact for each $i$. Note that this is an open subset of $Spec A_i$.

Thus we have the following problem Let $A$ be noetherian and $U\subseteq Spec A$ open then $U$ is quasi compact. Now there is a not necessarily finite covering

$$U\subseteq \cup D(f_i)$$ for $f_i\in A$.

If we can find a finite sub cover we will be finished. Let $U=Spec A-V(\mathfrak{a})$ Then the above covering statement is equivalent to

$$\sqrt{\mathfrak{a}}\subseteq \sqrt{(f_1,f_2, \ldots)}$$

However the ideal $\sqrt{\mathfrak{a}}$ is finitely generated since $A$ is noetherian and thus for some $k$, $$\sqrt{\mathfrak{a}}\subseteq \sqrt{(f_1,f_2, \ldots, f_k)}$$

Or, maybe better, the chain $\{\sqrt{(f_1,f_2, \ldots, f_k)}\}$ of ideals is finite.

And we have our finite subcover.

Answer 2

Since X has a finitely generated cover$U_i$'s of k- algebras. Then U is the cover of finitely generated k- algebras intersected with U.Each of the open sets $U \cap U_i$ is an open set of $U_i$ which is cover by basic open sets of $U_i$ and hence affine . So thus it is of finite type. I hope it is correct. I am also just learning nowadays so there might be some arguments which are not correct.