The morphism of rings inducing the diagonal morphism.

Question

How to see that the diagonal morphism for affine schemes $\Delta: Spec(R) \rightarrow Spec(R \otimes R)$ is induced by the morphism of rings $R \otimes R \rightarrow R$, $r \otimes r' \rightarrow r \cdot r'$?

I know this is elementary, but I'm stuck and even don't see this for the induced map on the topological spaces.

Using the universal property of the fibered product $Spec(R) \times Spec(R) = Spec (R \otimes R)$, it is the map induced by the identity maps $Spec (R) \rightarrow Spec (R)$. — karl_christ, Oct 10 '13 at 10:50
Well, the ring morphism $R \otimes R \to R$ you describe is defined by the dual universal property. — Zhen Lin, Oct 10 '13 at 10:53
@Zhen Lin: Could you explain that? I'm not familiar enough with category theory to see that. — karl_christ, Oct 12 '13 at 17:01

score 5 · Accepted Answer · answered Oct 11 '13 at 01:48

To be begin with, let's recall how, if $R$ is a ring of functions on a space $X$, then $R\otimes R$ becomes a ring of functions on $X \times X$. Namely, if $f(x)$ and $g(x)$ are two functions on $X$, then $f\otimes g$ gets mapped to the function $(x,y) \mapsto f(x)g(y)$.

Okay, now lets restrict this function to the diagonal. We then get the function $$x \mapsto (x,x) \mapsto f(x)g(x) = (fg)(x).$$

Composing these two, we see that if we map $f\otimes g$ to a function on $X\times X$, and then restrict to the diagonal, we obtain the product function $fg$. This answers your question.

The morphism of rings inducing the diagonal morphism.

1 Answers1