In a finite ring $R$ with identity show that $ab =1$ implies $ba = 1$, where $a,b \in R$.
I am having difficulty in doing this since there is no condition that there is no zero divisors and how will I use the finiteness of the ring!!
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If the ring is finite, Let's call n the order of ba.
$$ba = b(ab)a = b(ab(ab))a = b(ab(ab(ab)))a = \cdots = (ba)^n = 1$$
Tryss
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You seem to preassume that $b$ and $a$ are units. This however is yet to be shown. Why not $b^n=0$ for some $n$? – drhab Mar 03 '15 at 15:04
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Wait, what did I write? -_-' – Tryss Mar 03 '15 at 15:05
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@drhab: We can't have $b^n=0$ because $a^nb^n=1$. – hmakholm left over Monica Mar 03 '15 at 15:05
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@HenningMakholm I realize that in this context $b^n=0$ is not possible but still I am wondering: why $a^nb^n=1$? That $(ab)^n=1$ is immediately clear, but the ring is not necessarily commutative. – drhab Mar 03 '15 at 15:14
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@drhab: $a^nb^n=a^{n-1}abb^{n-1}=a^{n-1}b^{n-1}$, and then proceed by induction. – hmakholm left over Monica Mar 03 '15 at 15:26
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@HenningMakholm Ah, of course. I am starting to get blind I think. Thank you. – drhab Mar 03 '15 at 15:29
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@Tryss: How do we know that there's any $n>0$ such that $(ba)^n=1$? That's not the case for an arbitrary element in a finite ring, for example in $\mathbb Z_6$ the sequence $2, 2^2, 2^3, \ldots$ never reaches $1$ (nor for that matter $0$). – hmakholm left over Monica Mar 03 '15 at 16:10