# A closed subset of a prevariety is a prevariety

My question comes from Gathmann's notes https://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2014/alggeom-2014.pdf on page 42 Exercise 5.13.

Let $$Y$$ be a closed subset of a prevariety $$X$$, considered as a ringed space with the structure sheaf of Construction 5.12 (b). Prove for every affine open subset $$U ⊂ X$$ that the ringed space $$U∩Y$$ (considered as an open subset of the ringed space $$Y$$ as in Definition 4.1 (c)) is isomorphic to the affine variety $$U∩Y$$ (considered as an affine subvariety of the affine variety $$U$$).

In particular, this shows that Construction 5.12 (b) makes $$Y$$ into a prevariety, and that this prevariety is isomorphic to the affine variety $$Y$$ if $$X$$ is itself affine (and thus $$Y$$ an affine subvariety of $$X$$).

The construction 5.12 is to define $$\mathcal{O}_Y (U)$$ to be the $$K$$-algebra of functions $$U → K$$ that are locally restrictions of functions on $$X$$, or formally $$\mathcal{O}_Y (U) := \{ϕ : U → K :$$for all $$a ∈ U$$ there are an open neighborhood $$V$$ of $$a$$ in $$X$$ and $$ψ ∈ \mathcal{O}_X (V)$$ with $$ϕ = ψ$$ on $$U∩V\}$$.

My confusion is showing the affine variety $$U∩Y$$ is isomorphic to the ringed space $$U∩Y$$. My initial idea is to use the identity map, but I am not really sure how to check the sheaves are equivalent

After that, a closed subset of a prevariety is indeed a prevariety. I can refer to the proof from https://www.math.upenn.edu/~siegelch/Notes/ag.pdf on page 9 on Proposition 1.19.

Any help is appreciated.

• The underlying topological space being the same, isn't it enough to show that for every open set $V$, $\mathcal{O}_Y(V) \cong \mathcal{O}_{X|Y}(V)$? Commented Nov 13, 2019 at 13:54
• erratum, I meant $\mathcal{O}_{U\cap Y}(V) \cong \mathcal{O}_{X|U\cap Y}(V)$ Commented Nov 13, 2019 at 14:07
• See [Görtz; Wedhorn, Algebraic Geometry I, Lemma 1.55]. Commented Apr 14 at 13:50

The question calls for us to view $$U \cap Y$$ in two ways. First, we view $$U \cap Y$$ as a closed subset of $$U$$. In this case, the sheaf $$O_{U \cap Y}$$ is by definition: For any $$V$$ open in $$U \cap Y$$, $$O_{U \cap Y} (V) = \{ \varphi : V \to K :$$ for all $$a \in V$$ there exists an open neighborhood $$N$$ of $$U$$ containing $$a$$, and $$\psi \in O_{U}(N)$$ with $$\varphi = \psi$$ on $$V \cap N \}$$. Then, we note that $$O_U(N) = O_X(N) |_U = O_X(N)$$ where the second equality is because $$N \subset U$$.
On the other hand, we can view $$U \cap Y$$ as an open subset of $$Y$$. Then for any $$V$$ open subset of $$U \cap Y$$, we have that the sheaf is just the restriction: $$O_{U \cap Y}(V) = O_{Y} (V)$$. Lets apply the definition of $$O_{Y} (V')$$. By definition, for any $$V'$$ open in $$Y$$, $$O_Y(V') = \{ \varphi : V' \to K :$$ for all $$a \in V'$$ there exists an open neighborhood $$N'$$ of $$X$$ containing $$a$$, and $$\psi$$ in $$O_X(N')$$ with $$\varphi = \psi$$ on $$V' \cap N' \}$$.
Then just need to show that the restriction of the last set to $$U \cap Y$$ is equal to the first set. Since $$V'$$ is going to be an open subset of $$U \cap Y$$, we can replace "$$V'$$ open in $$Y$$" with "$$V'$$ open in $$U \cap Y$$". $$N'$$ is an open subset of $$X$$, and $$N' \subset U \cap Y$$, so $$N'$$ is an open neighborhood of $$U$$ also. So we can replace "open neighborhood $$N'$$ of $$X$$" with "open neighborhood $$N'$$ of $$U$$". Then both sets can be seen to be equal.