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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.

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    $\begingroup$ I think you need $f$ to be injective. If you look at Exercise I, 3.17 of Hartshorne's book, there the universal property of normalization of the affine variety was given. $\endgroup$ – Krish Dec 6 '17 at 3:53
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    $\begingroup$ If $f$ is injective you can extend it to $f : \text{Frac}(A) \to \text{Frac}(B)$ thus to $f:\overline{A} \to \text{Frac}(B)$, and for any $a \in \overline{A}, f(a) \in \overline{B}$ because $a^d+\sum_{n=0}^{d-1} c_n a^n = 0, c_n \in A \implies f(a)^d+\sum_{n=0}^{d-1} f(c_n) f(a)^n = 0,f(c_n) \in B$. If $f$ is not injective, I'm not sure where to start. $\endgroup$ – reuns Dec 6 '17 at 11:26
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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 $\overline A \to B$.

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  • $\begingroup$ Well, he said "$B$ is integrally closed". This suggests that $B$ is a domain. If $B$ is no domain one usually refers to this as normality and defines it via $S_2$ and $R_1$. $\endgroup$ – MooS Dec 6 '17 at 11:29
  • $\begingroup$ This probably a typo, the OP meant $B$ is an integral domain and $\overline{B}$ its integral closure ? Also for $f$ not injective, how would you check if $f$ extends to $\overline{A}$ ? $\endgroup$ – reuns Dec 6 '17 at 11:29
  • $\begingroup$ I meant looking at something in $\ker(f)$ to find if it extends or not. And your counter-example is weird, did you mean $f : \mathbb{Z}[p \sqrt{5}] \to \mathbb{Z}/p \mathbb{Z}$ with $p$ prime in $\mathbb Z[\frac{1+ \sqrt 5}{2}]$ (so the extension is $f : \mathbb Z[\frac{1+ \sqrt 5}{2}] \to \mathbb{Z}/p \mathbb{Z}[\frac{1+ \sqrt 5}{2}] \cong \mathbb{F}_{p^2}$) ? $\endgroup$ – reuns Dec 6 '17 at 11:38

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