Verification of Proof: Let $R$ be a ring of unity and $a \in R$ satisfy $a^2=1$. $S=\{ara \mid r \in R\}$ is a subring

Question

Here's what I got. The three conditions we have to prove are:

1. $0$ is in $S$: Let $r=0$ and this implies $a0a=0a=0$ which is in $S$

2. $(a-b)$ is in $S$ for all $a,b \in S$: Let $a, b \in S$. this means $ada$ and $brb$ given $r \in R$. $ada-brb=r(a^2-b^2)=r(1-1)=r0=0$ and $0$ is in S as previously shown.

3. $ab$ is in $S$ for all $a,b \in S$: Let $a,b$ be in $S$. Then $ara$ and $brb$ given $r$ is in $R$.
   $arabrb=ar(ab)rb=ar(ba)rb=(arb)(arb)=abrab$ as needed.

Is this correct? if not, what needs to be changed?

Answer 1

You seem to be confused that in the expression $ara$, that both $a$ and $r$ are allowed to vary. However, the definition you have indicates $a$ is fixed.

So, in condition (2) you should be checking that $ara-asa\in S$ for any $r,s\in R$, and in condition (3) you should be checking that $(ara)(asa)\in S$.

Actually, there is something even more important to notice: you can show that $aRa=R$.

So this bit about being a subring is a little bit anticlimactic, considering it's the whole ring.

---

Once you get done showing how this works, try an even more general statement:

If $u$ is any unit in $R$, and $T$ is any subring of $R$, then $uTu^{-1}:=\{utu^{-1}\mid t\in T\}$ is a subring of $R$ isomorphic to $T$.