More specific, we have an integral domain $D$, $a,b\in D$ such that $a^n=b^n$ and $a^m=b^m$ for two positive integers $m,n$ that are relative primes. I need to prove that $a = b$. Also, my teacher insist that this could be proved without the exponent laws, since we haven't proved them in general for rings.
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what are the exponent laws? – Anguepa Sep 16 '18 at 15:19
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4Use Bezout's identity for $n$ and $m$. Only multiplicative monoid is needed. – ajotatxe Sep 16 '18 at 15:19
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sorry, that's how i commonly refer to $x^n*x^m =x^{n+m}$ and $(x^n)^m=x^{nm}$ – Armando Rosas Sep 16 '18 at 15:21
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The integral domain assumption is unnecessary. – Gal Porat Sep 16 '18 at 15:21
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2@GalPorat No, this is not unnecessary. In the ring $M_n(K)$ we can take $A=0$ and $B\neq 0$ a nilpotent matrix with $B^n=B^m=0$. – Dietrich Burde Sep 16 '18 at 16:03
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Another duplicate is this question. – Dietrich Burde Sep 16 '18 at 16:05
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You are right of course, early morning sloppiness on my part :) – Gal Porat Sep 17 '18 at 05:05