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Question: Suppose that a and b belongs to an integral domain. If $a^{m}=b^{m}$ and $a^{n}=b^{n}$, where m and n are positive integers that are relatively prime, prove that a=b.

Well, m and n are relatively prime so gcd(m,n)=1 holds. In fact, from the fact that the gcd of any positive integers: gcd(m,n)=1=ms+nt for some integers s,t.

Raising the element a to ms+nt: $a^{ms+nt}=a$

A good bit of hint here would be helpful.

Thanks in advance.

Mathematicing
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3 Answers3

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If $ms+nt=1$, then:

$a = a^1 = a^{(ms+nt)} = a^{ms}a^{nt} = (a^m)^s (a^n)^t = (b^m)^s (b^n)^t = b^{ms}b^{nt} = b^{(ms+nt)} = b^1 = b$.

We must interpret $a$ and $b$ as belonging to the field of fractions $\kappa(R)$ here, since the powers $ms,nt$ etc. may possibly be negative. You can still work around this : for example by transferring the negative term to the other side etc.

Sarvesh Ravichandran Iyer
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    Note that generally $,m,$ or $,n,$ is negative so you either need to work in the quotient field, or else you need to take care to keep everything positive, then cancel. If you pass to the quotient field then it is more conceptual to use the order of $,a/b,$ as in my answer. – Bill Dubuque Aug 02 '16 at 04:12
  • @Bill Dubuque If $m$ and $n$ are negative, does that create multiple values somewhere along the chain of equalities? I'm not aware of that, so can you tell me? – Sarvesh Ravichandran Iyer Aug 02 '16 at 04:15
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    @астонвіллаолофмэллбэрг If $m$ (and thus $ms$) is negative, for instance, you're not guaranteed that $a^{ms}$ exists - there's no reason a priori for that to be a meaningful expression in the original integral domain. – Steven Stadnicki Aug 02 '16 at 04:36
  • @StevenStadnicki I see. I forgot we are in an integral domain, and not in the real numbers. Then yes, the quotient approach shown below works. – Sarvesh Ravichandran Iyer Aug 02 '16 at 04:41
  • @астонвіллаолофмэллбэрг Jyrki has added an exposition of the approach I mentioned that avoids fractions by keeping all exponents nonnegative, then cancelling. It is essentially just a rearrangement of the Bezout-based approach in your answer – Bill Dubuque Aug 02 '16 at 13:41
  • Yes, thank you for that. – Sarvesh Ravichandran Iyer Aug 04 '16 at 03:07
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Hint $ $ In the quotient field $\,(a/b)^m = 1 = (a/b)^n\,$ so the order $\,d\,$ of $\, a/b\,$ divides the coprimes $\,m,n\,$ thus $\,d=1,\,$ so $\,a/b = 1,\,$ so $\,a=b.$

Bill Dubuque
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If you haven't covered fields of fractions yet you can use the fact that cancellation law holds in integral domains instead.

Anyway, you know that there exist integers $s$ and $t$ such that $ms+nt=1$. Necessarily then one of them is negative and the other is positive. Moving the negative term to the other side, and interchanging $m,n$ if needed, we can rewrite that to read $$ mu=1+nv $$ for some positive integers $u,v$. Then $$ a\cdot a^{nv}=a^{1+nv}=a^{mu}=(a^m)^u=(b^m)^u=b^{mu}=b^{1+nv}=b\cdot(b^n)^v=b\cdot a^{nv}, $$ and you get the claim by cancelling the factor $a^{nv}$. Leaving the case $a^{nv}=0$ to you :-)

Jyrki Lahtonen
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  • Fully endorse Bill Dubuque's answer. Adding this to cater for the possibility that the exercise comes at a point, where fields of fractions have not been covered yet. Conceivable IMHO. – Jyrki Lahtonen Aug 02 '16 at 04:42