Let $R$ be a finite ring, and assume $\exists x,y\in R$ such that $ xy=1$. How can I show it implies $yx=1$?
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
Hint: $xy=1$ implies that left multiplication by $y$ is one-to-one. Can you draw a conclusion whether or not there is a $z$ such that $yz=1$?
If so, you can complete the argument by showing that $z=x$.
Let $f_y\colon:R\rightarrow R,\ z\mapsto yz$ then: $$f_y(z)=f_y(t)\iff yz=yt\Rightarrow x(yz)=x(yt)\Rightarrow (xy)z=(xy)t\Rightarrow z=t$$ hence $f_y$ is one to one. Now since $R$ is finite then the map $f_y$ is bijective hence there's a unique $z\in R$ s.t. $f_y(z)=yz=1$ so $x(yz)=(xy)z=z=x$ and conclude.
Hint $\ $ As often occurs, this result on numbers is a special case of a result on functions. namely, consider $\rm\:x,y\:$ as left-multiplication maps $\rm\:f(r) = xr,\ g(r) = yr,\:$ then apply the following
Lemma $\rm\ fg = 1\ \Rightarrow\ gf = 1\ $ for maps $\rm\:f,g\:$ on a finite set $\rm\:R.$
$\rm(1)\ \ \ fg = 1\ \Rightarrow\ g\ is\ 1\!-\!1\:$ by $\rm\:f\:$ of $\rm\:g(a) = g(b)\ \Rightarrow\ a = b $
$\rm(2)\ \ \ g\ is\ 1\!-\!1\ \Rightarrow\ g\:$ is onto, since $\rm\:R\:$ is finite
$\rm(3)\ \ \ g\ is\ onto\ \Rightarrow\ gf = 1\:$ by $\rm\ a = g(b) = g(fg(b)) = gf(a)$
Remark $\ $ In fact we may view the ring as the set of such maps (left-regular representation), where the elements of $\rm\:R\:$ are essentially viewed as $1$-dimensional matrices. Then the above is analogous to a well-known result about matrices, e.g. see my post here where I prove $\rm\ AB = I\:\Rightarrow\; BA = 1,\:$ or, equivalently, $\rm\:B\:$ injective $\rm \Rightarrow$ $\rm\: B\:$ surjective, by exploiting the pigeonhole principle. See also other posts in that thread which clarify the fundamental role played by the pigeonhole principle. See also this question on Dedekind-finite rings, i.e. rings where $\rm\:xy = 1\:\Rightarrow\: yx = 1.$