Show $R[x]/(x)\cong R$ Let $R$ be a commutative ring. Prove that $R[x]/(x)$ is isomorphic to $R$.  Should I use the first isomorphism theorem for rings?  Can someone give me a hint on how to approach this problem? Also how would I use the first isomorphism theorem to prove this? I know that the theorem says $R /\ker f$ is isomorphic to image of $f$.  

Consider the map: $\varphi$ : $R[x]\rightarrow R$ defined by $\varphi(f) =f(0)$
It is easy to show that $\varphi$ is a ring homomorphism and $\operatorname{Im} \varphi= R$
Claim: $\ker \varphi=(x)$
Indeed, let $f(x)=a_nx^n+a_{n-1}x^{n-1}+....+a_1x+a_0$
So $\varphi(f)=0$ $\Leftrightarrow$ $f(0)=0$ which gives us $a_{0}=0$
Consequently, $f(x)=x(a_nx^{n-1}+....+a1)\subseteq (x)$
Hence $\ker\varphi \subseteq (x)$  ($1$)
Conversely, if $f \in (x)$ then $f=xg(x)$ where $g(x)$ is some polynomial.
Clearly $\varphi(f)=0$.
Therefore, $(x) \subseteq \ker\varphi $ ($2$)
Combining ($1$) and ($2$) we have $\ker\varphi= (x)$
By The first Isomorphism Theorem, we have : $R[x]/Ker\varphi \cong Im\varphi$
So $R[x]/(x) \cong R $
