# Let $f:A\rightarrow B$ be a homomorphism of integral domains. Is it possible to extend $f$ to the integral closures of $A$ and $B$?

I am working on question II.3.8 in Hartshorne's Algebraic Geometry. Our professor mentioned a very useful algebraic result for the problem but I cannot find a reference anywhere.

Let $f:A\rightarrow B$ be a ring morphism of integral domains, and let $\tilde{A}$ and $\tilde{B}$ be the respective integral closures. Then there exists a morphism $\tilde{f}:\tilde{A}\rightarrow \tilde{B}$.

I think the result is supposed to be "there exists a morphism $\tilde{f}:\tilde{A}\rightarrow \tilde{B}$ that extends $f$", but I'm not sure.

Any insight on why this is true or not or a reference for the result would be much appreciated.

My idea to construct such a morphism is as follows. Let $r$ be an element in $\tilde{A}$. Then $r$ is the root of a polynomial $x^n+\cdots+a_1x+a_0$, with $a_i\in A$. Then let $\tilde{f}(r)$ be the root of the polynomial $x^n+\cdots+f(a_1)x+f(a_0)$ in $\tilde{B}$. I'm not sure if this is indeed a homomorphism however.

• The highlighted result mentioned by your professor is false, as demonstrated by user26857's counterexample. Dec 13, 2015 at 7:30

The result is false in general, as demonstrated by user26857's counterexample.
However here is an elementary proof if $f$ is injective.
This special case where $f$ is injective suffices for solving Hartshorne's Exercise II.3.8, which motivated the OP's question.

Let $K=\operatorname {Frac}(A)$ be the fraction field of $A$ and $L$ be the fraction field $L=\operatorname {Frac}(B)$ of $B$.
The ring morphism $f:A\to B$ has a unique extension to a field morphism $F:K\to L$.
Any element $k\in \tilde{A}$ has an image $F(k)\in L$ which is (by the OP's reasoning) integral over $F(A)=f(A)$ and thus a fortiori integral over $B$ (since $f(A)\subset B$).
Thus $F(k)$ belongs to $\tilde B$ and we have proved that $f$ has a unique extension to $$\tilde f=F\vert\tilde A:\tilde A \to \tilde B .$$

Let $f:A\rightarrow B$ be a homomorphism of integral domains. Is it possible to extend $f$ to the integral closures of $A$ and $B$?

No, it's not!

Let $A=\mathbb Z[\sqrt{-3}]$, $B=\mathbb Z[\sqrt{-3}]/(2,1+\sqrt{-3})$, and $f:A\to B$ be the canonical projection.
Then $B\simeq\mathbb Z/2\mathbb Z$, $\tilde A=\mathbb Z[\frac{1+\sqrt{-3}}{2}]$, and $\tilde B=B$.
Set $x=\frac{1+\sqrt{-3}}{2}$. We have $x^2-x+1=0$.
Suppose $\tilde f:\tilde A\to\tilde B$ is a ring homomorphism. Then $\tilde f(x)\in\mathbb Z/2\mathbb Z$. On the other side, $\tilde f(x^2-x+1)=0$ hence $\tilde f(x)^2-\tilde f(x)+1=0$, a contradiction.

• We should be grateful to user26857 for showing us that the claim in the OP's question is false. Dec 13, 2015 at 7:28
• What is your definition of integral closure of an intergral domain? Mine is: The set of all elements in the algebraic closure of the quotient field of the given domain that are roots of polynomials with coefficients in that domain and with leading coefficient $1$. Dec 13, 2015 at 9:31
• I've never heard about your definition. "Mine" is the same as the one given by Bourbaki in Commutative Algebra, Definition 4, page 308, or Bourbaki, Algebre Commutative. Chapitres 5-7, Definition 4, page 11. Also by Zariski and Samuel, Commutative Algebra, vol. 1, page 256. That is, the set of elements of the field of fractions (or total ring of fractions) which are integral over the ring. Dec 13, 2015 at 9:53
• Moreover, in the exercise invoked by the OP, Hartshorne says: "let $\tilde A$ be the integral closure of $A$ in its quotient field". Dec 13, 2015 at 10:00

Consider the set of ring homomorphisms $\hat f\colon \hat A\to \tilde B$ where $A\subseteq \hat A\subseteq \tilde A$ and $\hat f|_A=f$. The conditions of Zorn's lemma are verified, hence there exists a maximal such homomorphisms $\hat f\colon \hat A\to \tilde B$. Suppose $a\notin \hat A$. Then pick an irreducible monic polynomial $p\in A[X]$ with $p(a)=0$ and pick $b\in\tilde B$ with $f(p)(b)=0$. Then we can extend $\hat f$ to $\hat A[a]$ by declaring $f(\sum c_ia^i)=\sum \hat f(c_i)b^i$, which is well-defined by the choice of $b$ (or: we clearly have an induced homomorphism $\hat A[X]\to \tilde B$ that sends $a\mapsto b$; this factors over $A[X]/(p(X))\cong \hat A[a]$). By maximality of $\hat A$, we conclude $a\in \hat A$, so $\hat A=\tilde A$.

• This proof is incorrect: there is no reason why there should exist $b\in\tilde B$ with $f(p)(b)=0$ . Dec 13, 2015 at 7:14
• @GeorgesElencwajg But $f(p)$ is a non-constant polynomial with coefficients in $B$ and leading coefficient $1$. Hence its roots in the algebraic closure of th equotient field of $B$ are elements of $\tilde B$. Dec 13, 2015 at 9:33
• Sure, but the integral closure of $B$ is included in the quotient field of $B$: you are not allowed to take elements in an algebraic closure of that field. User 26857 has already made the point, which is corroborated by the definitions in books by Atiyah-Macdonald, Samuel, Shafarevich,...I have never seen a book or article adopting your definition of integral closure: have you? (That said I upvoted you: your proof is quite clever, even if it refers to a nonstandard definition of integral closure) Dec 13, 2015 at 10:32
• Darn, now I want to delete my answer, but can't as long as it's accepted ... I've worke too much in algebraically closed fields only, it seems Dec 13, 2015 at 10:46
• Dear Hagen, your gracious acknowledgment of criticism is greatly to your credit: this is much more important in my eyes than any mathematical result. Dec 13, 2015 at 11:07