# $a$ in ring $R$, adjoin $b$ such that we have relation $b=a$ then will get ring isomorphic to R [duplicate]

I have been told that if we have a ring $R$ and $a$ is an element of $R$ and if we adjoin an element $b$ with the relation $b=a$ then we will get a ring isomorphic to $R$. However, I have not seen a formal proof for this. Is there a textbook or nice proof of this anyone knows of? I can't seem to see exactly how this is true.

Formally, adjoining $b=a$ is forming a quotient: $R[x]/(x-a)$. Then $b$ is the class of $x$.
Now $p(x) \mapsto p(a)$ is a surjective homomorphism $R[x] \to R$ whose kernel is the ideal $(x-a)$ because $x-a$ is monic and we can divide $p(x)$ by $x-a$, obtaining $p(x)=(x-a)q(x)+p(a)$.
Therefore, $R[x]/(x-a) \cong R$.
• Did you mean $R[x]/(x)$? Apr 12 '16 at 1:52
• Thanks one last question what does $b$ of class $x$ mean? Apr 13 '16 at 1:50