I'm trying to understand an exercise from Hartshorne's book, Algebraic Geometry, about normalization of an integral scheme. So, I'm reduced to prove the affine case and then glue. In dealing with this case, I did not know how to express the dual property for rings. Being specific, Let $A$ be an integral domain, and $\tilde A$ its integral closure in its total ring of quotients. I guess that the universal property enjoyed by $\tilde A$ is this:
Given a ring $B$ integrally closed, and a ring homomorphism $f:A\longrightarrow B$, there is a unique ring homomorphism $\tilde f:\tilde A\longrightarrow B$, such that $\tilde f\iota=f$, where $\iota$ is the inclusion of $A$ in $\tilde A$.
As I mentioned, I don't know if this property is the right one, or if I have to put something more on the morphism $f$. I tried without success to show that in fact, $\tilde A$ has the property that I wrote. I guess I'm missing something.
I've looked in Matsumura's book and Eisenbud's book, but I did not find this formulation. I would be very grateful if someone can tell me what is the correct property, and in case I was correct, how to prove it.
Best Answer
As the comment by @Krish suggests, we need $f$ to be injective.
First, let me give you a counterexample, if $f$ is not injective:
Consider the normalization $\mathbb Z[\sqrt 5] \subset \mathbb Z[\frac{1+ \sqrt{5}}{2}]$ and the map $f:\mathbb Z \to \mathbb Z/p \mathbb Z$, where $p$ is a prime number, that is still prime in $\mathbb Z[\frac{1+ \sqrt 5}{2}]$. In that case, there is no ring map from $\mathbb Z[\frac{1+ \sqrt 5}{2}]$ to $\mathbb Z/p\mathbb Z$ at all, in particular you cannot extend $f$.
If $f: A \to B$ is injective, you get a map of quotient fields and then it is trivial that the image of the restriction of this map to $\overline A$ is contained in $\overline B=B$. This gives you the desired map $\overline A \to B$.