$\DeclareMathOperator{\Spec}{Spec}$Doubt about Nike's Lemma about the intersection of $\Spec A$ and $\Spec B$

Question

Nike's lemma. Suppose $\Spec A$ and $\Spec B$ are affine open subschemes of a scheme $X$. Then $\Spec A \cap \Spec B$ is the union of open sets that are simultaneously distinguished open subschemes of $\Spec A$ and $\Spec B$.

It begins with

let $\rho\in \Spec A\cap\Spec B$

By definition...

Definition. An affine scheme is a locally ringed space $(X,O_X)$ which is isomorphic (as a locally ringed space) to the spectrum of some ring. A scheme is a locally ringed space $(X,O_X)$ in which every point has an open neighborhood $U$ such that the topological space $U$, together with the restricted sheaf $O_X|_U$ is an affine scheme.

And that's what I do not understand, to say that $\Spec A\cap \Spec B$ is not empty it means that they are of the same species.

I mean, what I understand by $\Spec A$ and $\Spec B$ being affine subschemes of $X$ is that there are some $U,V\subset X$ open, such that $(U,O_X|_U)$ is isomorphic to $(\Spec A,O_{\Spec A})$ and $(V,O_X|_V)$ is isomorphic to $(\Spec B,O_{\Spec B})$. So $U$ and $V$ are in fact subsets of $X$, but those rings, $A$ and $B$ are different.

I will appreciate some help with that. Thank you.

Answer 1

You have two subsets (it does not matter that they are schemes here) of some bigger set, which is the scheme $X$. They can hence be disjoint, but can also intersect. Take for example $X = \mathbb{P}^1$ and the standard open cover given by two copies of $\mathbb{A}^1$. Then the intersection is given by $\mathbb{A}^1 \setminus \lbrace 0 \rbrace$, which is clearly non-empty.

Answer 2

In algebraic geometry people tend to get quite "sloppy" and "forget" many isomorphisms because it's a bother to them write every time and experienced mathematicians will understand the notation. (However, I also found it confusing when I started learning this material - and since I'm not an experienced mathematician, I often remain confused.)

Let me write out a more precise statement.

Lemma. Let $U,V \subseteq X$ be open subschemes with $U \cong \Spec{A}$ and $V \cong \Spec{B}$. Then, there exists a union $U \cap V = \bigcup_{i \in I} W_i$ of open subschemes such that $W_i \subseteq U$ is a distinguished open in $\Spec{A}$ and $W_i \subseteq V$ is a distinguished open in $\Spec{B}$, i.e. there exists an isomorphism $W_i \cong \Spec{A_f}$ for some $f \in A$ that is compatible with $U \cong \Spec{A}$ and similarly for $B$.

(By compatible I mean that $W_i \hookrightarrow U \to \Spec{A}$ and $W_i \to \Spec{A_f} \to \Spec{A}$ are the same.)