# 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.