An Open Subset of A Scheme Is a Scheme

Recently I saw that an open subset of an affine scheme need not be affine. (See here for details).. This led me to the following train of thought:

“Take an open subset U of a scheme X. Every point x of U has an affine neighborhood, taking the intersection of U with these neighborhoods shows that this is a scheme.”

But this is not true because the intersection of U and an affine neighborhood need not be affine, as the above link implies. Throughout the following, the main goal is to reassure myself that an open subset of a scheme is a scheme.


So why is an open subset of a scheme a scheme? First off, we have that if U\subset X is an open subset, then (U,\mathcal{O}|_U) is a locally ringed topological space. This is just because the restriction sheaf has identical stalks to those of its image in X.

Now we will follow the above train of thought. Every point x\in U has an affine neighborhood A_x. Taking the intersection A_x\cap U gives an open subset of A_x. Since this is an open subset of an affine scheme, we may regard it as the union of principal open subsets \bigcup_i D(f_i) where the index is essentially arbitrary, but each f_i\in A_x, so maybe it would be nicer to use the index f_{x,i}.

Anyways, we do have that the set D(f_{x,i}) is affine. (In fact, if R_x is the ring associated to A_x then the affine structure is given by the localization (R_x)_{f_{x,i}}; alas our indices are disadvantageous now. Depending on where you learn this from, this could be from definition as in Liu’s book or a theorem as in Hartshorne).

Finally, we have U=\bigcup_{i,x} D(f_{x.i}) and as these are just restrictions of the structure sheaf of X, we have that U can be covered by affine neighborhoods which proves an open subset of a scheme is a scheme.

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