Show $R[x]/(x)\cong R$

Question

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 $f$.

Answer 1

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 $\ker\varphi \subseteq (x)\tag{1}$

Conversely, if $f \in (x)$ then $f=xg(x)$ where $g(x)$ is some polynomial.

Clearly $\varphi(f)=0$.

Therefore, $(x) \subseteq \ker\varphi \tag{2}$

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 $

Answer 2

Find $f$ such that $Ker(f)$ is $(x)$ and $Im(f)=R$.

Answer 3

Use the first isomorphism theorem for rings along with the map $f(x)\mapsto f(0)$. This evaluation map is a very usefull thing when you are working in polynomial rings and want to look at quotients.

Answer 4

If you think about it too abstractly, you will miss how obvious this is. In the quotient ring $R[x]/(x)$, every power of $x$ is zero. So if you take a typical polynomial $f(x) = a + bx + cx^2 + dx^3 + \operatorname{etc.}$, its image in $R[x]/(x)$ is the same as the image of its constant term:

$$f(x) + (x) = a + (x).$$

In this way, you see that the only elements of $R[x]/(x)$ are constants. Thus the map which sends an element $a \in R$ to the term $a + (x) \in R[x]/(x)$ is the isomorphism of rings you are looking for.