# In modules over finite rings $Rv = Rw \iff R^\times v = R^\times w$

Let $R$ be a finite ring (with unit, possibly non-commutative), and $M$ a left module over $R$. Let $v,w\in M$. Then $$Rv = Rw \iff R^\times v = R^\times w.$$

The proof uses the Artin-Wedderburn classification theorem. My question is if there is a simpler proof. (The hard part is the direction $\Rightarrow$, of course.)

As a remark, the statement is not true any more for infinite rings. Counterexamples with $M={}_R R$ can be found on page 466 in Irving Kaplansky. Elementary divisors and modules. Transactions of the American Mathematical Society 66 (1949), 464–491. Counterexamle (b) is over a commutative Noetherian ring $R$.

• What portions of the theory of semisimple rings (or modules) would you permit in a solution? – Manny Reyes Jul 16 '15 at 16:40
• @MannyReyes: The less, the better. Ideally, none. The statement does not involve the notion semisimple. So I'd hope for a proof wich does not rely on the theory of semisimple rings/modules at all. – azimut Jul 16 '15 at 18:28

Let $r,s\in R$ with $rv=w$ and $sw=v$. Since $R$ is finite, there exists $n>0$ such that $f=(rs)^n$ and $e=(sr)^n$ are idempotent. Note that $ev=v$ and $fw=w$. Let $r'=fre$ and $s'=esf$. Notice that $r'v=w$. Also note that right multiplication by $r'$ gives an $R$-module homomorphism $Rf\to Re$. In fact, it is surjective (as is easily verified using the definitions of $e,f,r'$). Similarly right multiplication by $s'$ gives a surjective $R$-module homomorphism $Re\to Rf$. By finiteness of $R$ we conclude both these homomorphisms are isomorphisms. It follows that $R(1-f)\cong R(1-e)$ by the Krull-Schmidt theorem (I don't know if this counts as more or less elementary than Wedderburn-Artin). Such an isomorphism is given by right multiplication by an element $x\in (1-f)R(1-e)$.
Consider $u=r'+x$. Then $u$ is a unit since right multiplication by $u$ gives an isomorphism from $R=Rf\oplus R(1-f)$ to $R=Re\oplus R(1-e)$ (as it is the direct sum of the two isos $Rf\to Re$ and $R(1-f)\to R(1-e)$) and in a finite ring an element with a one-sided inverse is invertible. Also $uv=(r'+x)v=(r'+x)ev=r'ev=r'v=w$ because $x\in (1-f)R(1-e)$ implies $xe=0$. Thus $w\in R^\times v$.
• Why is $R/Re \cong R/Rf$ ? – darij grinberg Jul 17 '15 at 1:32
• @darijgrinberg, the answer to your question is Krull-Schmidt. Since $R=Re\oplus R(1-e)=Rf\oplus R(1-f)$, if $Re\cong Rf$, then Krull-Schmidt guarantees $R(1-1)\cong R(1-f)$. – Benjamin Steinberg Jun 14 '17 at 14:12