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Let D be a UFD with quotient field F. If f (x) $\in$ D[x] is monic and b$\in$ F such that f (b) = 0 , then show that b$\in$ D.

All I know is

F is a field and f(b) = 0 therefore $$f(x) = (x-b)q(x)$$ and also f(x) $\in$ D[x] so f(x) can be written as the product of some irreducible elements of D[x].

What should I do to show that b$\in$ D?

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1 Answer

up vote 4 down vote accepted

Hint $\: $ Mimic the proof of the Rational Root Test, which works over any domain where gcds exist, i.e $\rm\:(a,b) = 1,\ 0 = b^n\:\! f(a/b) =\: a^n + b\:(\cdots)\ \Rightarrow\ b\:|\:a^n\ \Rightarrow\ b\:|\:1\:$ by Euclid's Lemma.

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If you're looking for a bigger hammer, the name Eisenstein might help. –  Christopher Creutzig Apr 9 '12 at 20:10
    
@Chris How do you propose to use Eisenstein generally? –  Bill Dubuque Apr 9 '12 at 20:15
    
Scrap that. I thought I had learned Eisenstein for UFDs, but cannot find a reference to back that up right now, so my memory is probably faulty. –  Christopher Creutzig Apr 9 '12 at 20:17
    
@Chris It does work over UFDs, but how do you propose to apply it here? –  Bill Dubuque Apr 9 '12 at 20:20
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Gauss' lemma, which follows from Eisenstein, implies that if a monic polynomial over a UFD $D$ has a zero over the quotient field of $D$, then that zero is an element of $D$, no? (I agree, a pointer to Gauss' lemma would have been more appropriate.) –  Christopher Creutzig Apr 9 '12 at 20:28
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