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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 $k$?

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    Isnt finite type also a local property ? Every open affine has a covering by open affines with finitely generated rings ? – 2017-02-11
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    @ReneSchipperus It's certainly true that "locally finite type" is a local property, but it's the quasicompactness that I seem to be missing. – 2017-02-11

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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.

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    I totally forgot that each $A_i$ automatically is Noetherian! This alone actually makes the problem much easier; it is a well-known fact that an open subset of any Noetherian topological is quasicompact, and in particular a scheme of finite type over $k$ must be Noetherian, being a finite union of Noetherian schemes. – 2017-02-11
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    Yeah, that seems to be the key, finite type and noetherian are very close. – 2017-02-11
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    Keep asking those algebraic geometry questions, I am enjoying them. – 2017-02-11
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    I'm glad! Trying to get through Vakil without SE would likely be too big a task for me, so they'll definitely keep coming. – 2017-02-11
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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.

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    This shows the $U$ is locally of finite type, not of finite type. – 2017-02-11
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    If X is of finite type then you have a finite covering . Intersecting it with U , you get a finite covering of U by affines. – 2017-02-11
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    But the problem is that $U\cap U_i$ isn't necessarily affine; you even noted that it can be *covered* by distinguished open subsets of $U_i$, but this cover could potentially be infinite. – 2017-02-11
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    But we know than an affine subscheme is quasi compact , so there should be a finite covering of affine opens for $U \cap U_i$. – 2017-02-11