Using the First Isomorphism to show $R[x]/((x-a)^2)$ is isomorphic to $R[x]/(x^2)$

Question

How can I use the First Isomorphism to show $R[x]/((x-a)^2)$ is isomorphic to $R[x]/(x^2)$. I tried to find an isomorphism $\phi : R[x] \rightarrow R[x]/(x^2)$ and define this by $\phi (f) = f(a)$ but then I don't know how to show this is surjective as I don't understand what elements in $R[x]/(x^2)$ actually look like.

Answer 1

You should take$$\begin{array}{rccc}\phi\colon&\mathbb R[x]&\longrightarrow&\mathbb R[x]/(x^2)\\&p(x)&\mapsto&p(x+a)+(x^2).\end{array}$$Then, yes, $\ker\phi=\bigl((x-a)^2\bigr)$. And this map $\phi$ is surjective.

Answer 2

Note that $f\mapsto f(a)$ is not a map into $R[x]/((x-a)^2)$. It's a map into $R$. It's also not true that its kernel is $((x-a)^2)$. Its kernel is $(x-a)$.

Instead, define $\phi(f)(x)=f(x-a)$ which is a map $R[x]\to R[x]$ which is clearly an isomorphism (with inverse $f(x)\mapsto f(x+a)$.) Thus, composing with the quotient map yields a surjective ring homomorphism $\overline{\phi}:R[x]\to R[x]/((x-a)^2)$. Hence, $R[x]/((x-a)^2)\cong R[x]/\textrm{ker}(\phi)$.

However, it's clear that $\textrm{ker}(\phi)=(x^2)$, since $(x-a)^2| f(x-a)$ if and only if $x^2|f(x)$.

Answer 3

The map $x\mapsto x+a$ (together with the canonical embedding $R\to R[x]$) induces a homomorphism $R[x]\to R[x]$. As this leads to $x-a\mapsto x$, it is clearly onto (in fact, it is an automorphism with the inverse being induced by $x\mapsto x-a$). Compose with the canonical projection $R[x]\to R[x]/(x^2)$ to obtain a homomorphism $\phi\colon R[x]\to R[x]/(x^2)$. Wie have $f(x)\in\ker\phi$ iff $f(x+a)$ is a multiple of $x^2$. The latter is clearly equivalent to $f(x)$ being a multiple of $(x-a)^2$. Thus $\ker\phi=((x-a)^2)$ and by the isomorphism theorem $\phi$ induces an isomorphism $R[x]/((x-a)^2)\to R[x]/(x^2)$.

$$\begin{matrix} 0\to &((x-a)^2)&\to& R[x]&\longrightarrow &R[x]/((x-a)^2)&\to0\\ &&&\downarrow\rlap{{x\mapsto x+a}}&&\downarrow\rlap{\cong}\\ 0\to &(x^2)&\to& R[x]&\longrightarrow &R[x]/(x^2)&\to0\\ \end{matrix}$$