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I am part of a small group of math majors from the University of Maryland. We did NOT pursue careers in mainstream mathematical areas and now closing in on retirement we have formed a small study group so as to "relearn" some of the good theoretical mathematics which has been on hold for 40 years. We are presently studying Stewart and Tall's Algebraic Number Theory and FLT. We have a problem with an exercise from the test. Perhaps it is poorly presented. We are not sure of that either.

It basically says that if $a\ \&\ b$ are non-unit, relatively prime members of an Integral Domain, $D$, for which $ab=c^n$ for some $c$ in $D$ then there is a unit, $e$, in $D$ for which both $ea$ and $e^{-1}b$ are $n$th powers in $D$. It, at first glance, appears to be a straightforward problem, but we (embarrassingly) keep going in circles. Any discourse, hints etc. are well received.

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According to Amazon, this looks like the problem on p.100 of the 3rd edition. This version calls $D$ a unique factorization domain, which is essential to the problem. It is false in general for arbitrary integral domains (a fact which played a key role in the history of work on FLT). –  Erick Wong Jan 24 '13 at 21:40
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As I pointed out when you posted to mathoverflow.net/questions/119703/… the statement isn't true without further hypotheses. You might have taken that as a hint to check to see whether you were leaving something out. –  Gerry Myerson Jan 24 '13 at 23:18
    
I am very sorry. I made a serious typo. The hypothesis is indeed that D is a UFD. We were all aware of that while we tackled the problem. It is my bad that I typed incorrectly. Now how about the corrected problem? –  DaveUM Jan 24 '13 at 23:45
    
I just saw Math Gems remarks. –  DaveUM Jan 24 '13 at 23:46

1 Answer 1

For UFDs the proof is easy by examining exponents in (unique!) prime factorizations. One can also prove it using only gcds (so for gcd-domains), similarly to my proof here for $\rm\,n = 2,\,$ namely:

Theorem $\rm\ (a,b)=1,\ \color{#C00}{c^n = ab}\:\Rightarrow\: a = (a,c)^n,\ b = (b,c)^n\ \ $ for $\rm\:a,b,c\ne 0\:$ in a gcd domain.

$\begin{eqnarray} \rm{\bf Proof}\quad\ (a,c)^n &=&\rm (a^n, a^{n-1}c,\ldots ac^{n-1}, \color{#C00}{\,c^n})\\ &=&\rm (a^n, a^{n-1}c,\ldots ac^{n-1}, \color{#C00}{ab})\\ &=&\rm (\color{#0A0}{a^{n-1}}, a^{n-2}c,\ldots c^{n-1},\, \color{#0A0}b)\,a\\ &=&\rm\ a,\ \ \ by\ \ \ (b,a) = 1\,\Rightarrow\, (\color{#0A0}{b,a^{n-1}})= 1\ \ by\ Euclid's\ Lemma \end{eqnarray}$

Of course it fails for general domains, e.g. a generic counterexample being $\rm\,\Bbb Q[x,y,z]/(xy-z^n).$

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