Left inverse implies right inverse in a finite ring

Let $$R$$ be a finite ring with identity $$1$$, and assume $$\exists x,y\in R$$ such that $$xy=1$$. How can I show it implies $$yx=1$$?

• $ya=yb\Rightarrow y(a-b)=0\Rightarrow xy(a-b)=0\Rightarrow a-b=0\Rightarrow a=b$ – anon Jul 21 '13 at 10:54
• It is true for any noetherian ring, so it's true in this case! – A_Sh Feb 26 '17 at 21:12

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 $z=x$.

• Easy exercise: modify the proof for the case where $R$ is a finite $k$-algebra, where $k$ is field. Hard exercise: find a proof that works for both at once. (Hint: invent $\mathbb{F}_1$.) – Zhen Lin Apr 29 '12 at 19:37
• @Zhen You may find of interest work by Vasconcelos, Armendariz et. al on analogous questions for finitely generated modules. For references see my 2007/12/12 sci.math post – Bill Dubuque Apr 29 '12 at 20:33
• @rschwieb could you help me with this problem: math.stackexchange.com/questions/2163962/… It's similar. – ALannister Feb 27 '17 at 17:28
• @ALannister I added an answer I always give for that, since I can't seem to find a link to one I wrote previously. – rschwieb Feb 27 '17 at 18:48

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 $\rm\:xy = 1\:\Rightarrow\: yx = 1.$

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.

• Another way to understand this argument is functionally: if left-multiplication by $y$ were non-injective, then post-composing it with left-multiplication by $x$ would necessarily yield a non-injective map, meaning multiplication by $xy$, or $1$, absurd. – anon Jul 21 '13 at 19:41