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UFD's are integrally closed (Theorem)

Theorem: Every UFD is integrally closed.

Proof: Let $ R$ be a UFD, $ K$ its field of fractions, $ u\in K, u$ integral over $ R$. Then for some $ c_0,\ldots,c_{n-1}\in R$,

$\displaystyle u^n+c_{n-1}u^{n-1}+\ldots+c_0=0$
Write $ u=\frac{a}{b}, a,b\in R$, where $ a,b$ have no non-unit common divisor (which we can assume since $ R$ is a UFD). Multiply the above equation by $ b^n$ to get
$\displaystyle a^n+c_{n-1}ba^{n-1}+\ldots+c_0b^n=0$
Let $ d$ be an irreducible divisor of $ b$. Then $ d$ is prime since $ R$ is a UFD. Now, $ d\lvert a^n$ since it divides all the other terms and thus (since $ d$ is prime) $ d\lvert a$. But $ a, b$ have no non-unit common divisors, so $ d$ is a unit. Thus $ b$ is a unit and hence $ u\in R$.



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Cross-references: unit, terms, divides, prime, irreducible, equation, divisor, integral, field of fractions, integrally closed, UFD

This is version 3 of UFD's are integrally closed, born on 2006-03-30, modified 2008-05-01.
Object id is 7790, canonical name is UFDsAreIntegrallyClosed.
Accessed 933 times total.

Classification:
AMS MSC: 13G05 (Commutative rings and algebras :: Integral domains)
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