0

Let $R$ be a ring, possibly noncommutative, in which $xy=0$ implies $x=0$ or $y=0$. If $a, b \in R$ and $a^n=b^n$ and $a^m=b^m$ for two relatively prime positive integers $m$ and $n$, prove that $a=b$

I guess I know the commutative case, but no idea for the noncmmutative case.

My attempt:

$gcd(m,n) = 1$, so there exist integers $x,y$ s.t. $mx+ny=1$

so $$ \begin{aligned} a^{mx+ny}=a \\ (a^m)^x \cdot(b^n)^y=a \end{aligned} $$ similarly, $$ \begin{aligned} b^{mx+ny}=b \\ (a^m)^x \cdot(b^n)^y=b \end{aligned} $$

then I get $a=b$

I know this is certainly wrong as I didn't use any property of ring or assumptions in the questions. Any hints?

mathshungry
  • 837

1 Answers1

1

Hint One of $x,y$ is negative and you cannot use $a^{mx}=(a^m)^x$ when $x$ is negative. Use instead canceling in this case.

It is probably easier to write $$mx+ny=1 \Rightarrow mx=1-ny$$ in the case $y<0$.

N. S.
  • 132,525