If Pricipal Ideal generated by $a\in R$ then Subring Let $a\in R$. Prove that principal ideal $aR\equiv \\{ ar:r\in R \\}$ is a subring * * * _Closed add._ Let $ar_1,\in aR$ and $ ar_2 \in aR$ so $$\begin{aligned} ar_1+ar_2=a(r_1+r_2) && \text{by left distribution prop of rings } \end{aligned} $$ _Closed mult_ $$\begin{aligned}ar_1*ar_2=a(r_1 a r_2)&& \text{} \end{aligned} $$ Since $r_1,a,r_2 \in R$ and mult closed under $R$: $r_1ar_2 \in R$. So $ar_1*ar_2 \in aR$ $\exists 0_r \in a R$ since $0_r*a=0_r$ _Additive inverse_ since $\forall r\in R$,$\exists -r \in R:r+(-r)=0_R $ so, $a(-r) \in R$. So, $$\forall ar \in R, \exists a (-r) \in aR:ar+a(-r)=a(r-r)=a*0_R=0_R$$ * * * Did I do something wrong?? also, I welcome any other ways of showing it.

What you did is correct. However, an ideal (principal or not) is a subring only if you don't require your rings to have a multiplicative identity. If you do, then a proper ideal is never a subring.