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Let $D$ be an integral domain and let $a^m=b^m$ and $a^n=b^n$ where $m$ and $n$ are relatively prime integers, $a,b \in D$.

How do I show $a=b$?

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Hint: If $m$ and $n$ are relatively prime, then there exist integers $x$ and $y$ such that $mx + ny = 1$. –  ShreevatsaR Nov 16 '11 at 13:31
    
Can you show this in the special case where $m = n+1$? –  Hans Engler Nov 16 '11 at 13:31
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Subhint: let $D$ the set of $k$ such that $a^k=b^k$. You know that $n$ and $m$ are in $D$ and you want to show that $1$ is in $D$. What other integers do you know as being in $D$? –  Did Nov 16 '11 at 13:44
    
Sorry: not your $D$... –  Did Nov 16 '11 at 14:49
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maybe you could use this identity : (math.stackexchange.com/q/7473/15660) –  pedja Nov 16 '11 at 14:49
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4 Answers 4

up vote 1 down vote accepted

No generality is lost by supposing $m < n$. So $a^n=b^n$ implies $a^{m+(n-m)}=b^{m+(n-m)}$, or $a^ma^{n-m}=b^mb^{n-m}$. In integral domains, there's a cancellation property, so $a^{n-m}=b^{n-m}$.

The pair $(m,n)$ has now been replaced by the pair $(m,n-m)$. If you iterate that process, replacing the pair you've got with the pair consisting of the larger of the two and the difference---the larger minus the smaller, that's Euclid's algorithm. It ends when you reach the gcd.

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Here the integral domain doesn't have a unit. Will the cancellation property still hold? –  Mohan Nov 16 '11 at 14:25
    
@user Suppose $xy = xz$ with $x \ne 0$. Then $x(y-z)=0$; by the definition of an integral domain, since $x \ne 0$, it must be the case that $y - z = 0$. –  Srivatsan Nov 16 '11 at 14:34
    
@user774025 It's been a while since I've thought these issues through, but I've just look in Herstein's undergraduate text, Topics in Algebra, and he gives a definition of integral domain that doesn't assume that those have a unit, and in that context proves that every integral domain is embedded in a field of quotients. If it's embedded in a field of quotients, then it must have the cancellation property. –  Michael Hardy Nov 17 '11 at 16:04
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Since $m$ and $n$ are coprime then $x m - y n=1$ for some $x,y \in \mathbb Z$. The equality $a^m=b^m$ implies that $a^{xm}=b^{xm}$ so $a^{1+yn}=b^{1+yn}$ which implies that : $$(*)\;\;\; \;a a^{yn}=b b^{yn}$$ Since $a^{n}= b^{n}$ then $a^{yn}=b^{yn}\not = 0$ so we can cancel $a^{yn}$ and $b^{yn}$ from both sides of $(*)$ since we are working in an integral domain, and then we get $a=b$.

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What if $y \lt 0$? then the expression $ a^{yn}$ will be meaningless? –  Mohan Nov 16 '11 at 15:48
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It has meaning in the quotient field of the domain $D$. –  Andrea Nov 19 '11 at 13:29
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Hint $\ $ The set $\,S\,$ of naturals $\,k>0\,$ such that $\,a^k = b^k\,$ is nonempty and closed under positive subtraction, i.e. $\, j>k \in S\,\Rightarrow\,j-k\in S\,$ (by cancelling $\,a^k = b^k\,$ from $a^j = b^j).$ Therefore, by a fundamental lemma, the least element $\,\ell\in S\,$ divides every element of $\,S.\,$ In particular, $\,\ell\,$ is a common divisor of the coprimes $\,m,n\in S,\,$ thus $\,\ell = 1.\,$ Hence $\,1\in S\,\Rightarrow\, a^1 = b^1.\ \ $ QED

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What's the necessity of $D$ to be a domain? As $(m,n) = 1$ we have two integers $x$ and $y$ such that $mx+ny = 1$

so, $a^{mx} = b^{mx}$ and $a^{ny} = b^{ny}$ which implies, $a^{mx+ny} = b^{mx+ny} = a = b$

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Your proof only works if $x,y$ are both positive. –  Martin Brandenburg Mar 2 at 15:47
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