I have been thinking about the basic notion of separated schemes, and there is something that I can't seem to get right. I have written down a "proof" that would show that any morphism of schemes is separated. Obviously, a mistake must lie somewhere. Could someone please explain me what is wrong in the following ? I thank you very much for your help.
Let $X$ be a scheme over another scheme $S$ via a morphism $f$. Let $\Delta:X\rightarrow X\times _S X$ denote the usual diagonal morphism. We know that $\Delta$ is a closed immersion if and only if its image is closed.
To check whether the image of $\Delta$ is closed, it is enough to check that its intersection with any open subset of an open cover of $X\rightarrow X\times _S X$ is closed inside this open.
Let us consider open affine subsets $U=\operatorname{Spec}(A)\subseteq S$ and $V=\operatorname{Spec}(B)\subseteq f^{-1}(U)$ respectively of $S$ and $X$. The open affine subsets of the form $V\times _U V = \operatorname{Spec}(B\otimes _A B)$ form an open cover of the fibered product $X\rightarrow X\times _S X$.
The intersection of $\Delta (X)$ with $V\times _U V$ is the image of the co-restricted morphism $\Delta: \Delta ^{-1}(V\times _U V) \rightarrow V\times _U V$. In fact, we have $\Delta ^{-1}(V\times _U V)=V$.
This morphism is now the diagonal morphism between two affine schemes. Because the corresponding morphism at the level of rings the map $B\otimes _A B \rightarrow B$ which sends $b\otimes b'$ to $bb'$, it is surjective and hence the diagonal morphism between affine schemes is a closed immersion. It follows that $\Delta (X) \cap V\times _U V$ is closed in $V\times _U V$. As a consequence, $X$ is separated over $S$.
NB: My guess is that
The open affine subsets of the form $V\times _U V = \operatorname{Spec}(B\otimes _A B)$ form an open cover of the fibered product $X\rightarrow X\times _S X$.
would be the wrong part of the proof. Could someone please confirm this?
Let $X \rightarrow \mathrm{Spec}k$ be the double pointed line, i.e. $X= X_1 \cup X_2$ with $X_i \cong \mathbb{A}^1_k, 0_1 \mapsto 0$ and $X_1 \cap X_2 = X_1 \backslash 0_1 = X_2 \backslash 0_2$, together with the obvious scheme structure. Following your argumentation, $X_1 \times X_1 \cup X_2 \times X_2$ would be an open cover of $X \times X$. However, the point $(0_1, 0_2)$ is not contained in $X_1 \times X_1 \cup X_2 \times X_2$.
In particular, observe that the restriction $ \Delta_{|\Delta^{-1}(X_1 \times X_2)} \colon X_1 \cap X_2 \rightarrow X_1 \times X_2$ is not a closed immersion as the corresponding morphism of rings $k[x,y] \rightarrow k[x, x^{-1}], x, y \mapsto x$ is not surjective.