I am trying to understand the proof of the fact that UFDs are integrally closed. In addition to the lecture notes I have, there are at least two solutions on MSE:
One is here: How to prove that UFD implies normal?
and the other is here: UFDs are integrally closed
In all instances, I am failing to understand where and how exactly we are using the "unique factorization" property of the domain, which makes me think maybe I have never properly understood the definition of a UFD. Let's stick to the notation used in the answer of the first link:
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I am thinking one place where we used "unique factorization" is at the end where we are concluding $b$ is a unit. Since $R$ is a UFD, we can write $b=u b_{1} b_{2} … b_{n}$ where $u$ is a unit and $b_{i}$'s are irreducibles. Since $b$ does not have any irreducibles in its factorization, we get that $b=u,$ i.e., b is a unit. However, I don't think I used "uniqueness" here. I am not sure if this is an actual thing, but if $R$ was just a "factorization domain" (meaning, any nonzero element could be factored into product of irreducibles, but not necessarily in a unique way), I think we would have the same conclusion.
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The other place where we use unique factorization might be at the beginning of the proof, where we take $a$ and $b$ as having no common irreducible factors, i.e. they are relatively prime. I am thinking that if an element of $R$ can have more than one factorization into irreducibles, then the following can happen: Suppose $\dfrac{s}{t}=\dfrac{abcd}{abef}$ where $s, t\in R$ and $abcd$ is one of the possible factorizations of $s$ into irreducibles, and similarly $abef$ for $t$. Then $\dfrac{s}{t}=\dfrac{cd}{ef}$ ($c, d, e, f$ all distinct). Even though $c, d, e, f$ are distinct, the elements $cd, ef \in R$ could have different factorizations into irreducibles, say $cd=klm$ and $ef =knp$. But then $\dfrac{s}{t}=\dfrac{klm}{knp}=\dfrac{lm}{np},$ and we are back to the similar situation we started with, and there is no guarantee that this process will end. i.e., we don't know if $\dfrac{s}{t}$ can be written as $\dfrac{a}{b}$ with $a, b$ relatively prime.
Is there anything I am failing to see?
Best Answer
In a unique factorization domain, if $b$ divides $a^n$, and $r$ is an irreducible that divides $b$, then $r$ also divides $a$.
To see this, you factor $a$ into irreducibles, $a=ur_1\cdots r_k$. Then the factorization of $a^n$ into irreducibles must be $$a^n = u^n r_1^n\cdots r_k^n$$ because that is a factorization, and hence the factorization. If $b$ divides $a^n$ and $r$ divides $b$, then $r$ divides $a^n$, and hence there exists $c$ such that $rc=a^n$. Expressing $c$ as a product of irreducibles, and using unique factorization, we conclude that $r=r_i$ for some $i$, so $r$ actually divides $a$, and not merely $a^n$.
This chain of reasoning fails without unique factorization, even if the domain is atomic (every elements can be written as a product of irreducibles): for example, $\mathbb{Z}[\sqrt{-5}]$ is an atomic domain that is not a UFD. Let $b=2$, and let $a=1+\sqrt{-5}$. Using the norm map $N\colon \mathbb{Z}[\sqrt{-5}]\to\mathbb{Z}$, defined by $$N(x+y\sqrt{-5}) = x^2+5y^2$$ it is easy to verify that both $a$ and $b$ are irreducible. It is also plain that $b$ does not divide $a$. But $b$ does divide $a^2 = -4+2\sqrt{-5}$. Thus, we have a situation in which an irreducible divides $b$ (namely $2$), and $b$ divides a power of $a$ (namely, $a^2$), but the irreducible does not divide $a$.
Thus, in the argument you quote, the claim that because $a$ and $b$ have no common irreducible factor, and $b$ divides $a^n$, it follows that $b$ is a unit does not follow without unique factorization. Your thinking in point 1 that you could deduce that is incorrect, because we cannot deduce that $b$ is not divisible by any irreducibles; that deduction comes from concluding that any irreducible factor of $b$ must also divide $a$, but as the example above shows, without unique factorization into irreducibles this assertion would not follow.
A different way to see the initial claim is that in a UFD irreducibles are prime: if $r$ is an irreducible and divides a product $xy$, then $r$ must divide either $x$ or $y$. If your ring is atomic but not a UFD, this does not hold. In $\mathbb{Z}[\sqrt{-5}]$, all of $2$, $3$, $1+\sqrt{-5}$ and $1-\sqrt{-5}$ are irreducibles, but none are prime since $$2\times 3 = (1+\sqrt{-5})(1-\sqrt{-5}).$$ That's what sinks you.