# Proving 2 quotient rings are isomorphic

So I want to show $$\mathbb{R}[x]/((x-r)^2)$$ is isomorphic to $$\mathbb{R}[x]/(x^2)$$ where $$r \in \mathbb{R}$$. I thought of constructing a ring homomorphism $$\phi : \mathbb{R}[x] \to \mathbb{R}[x]/(x^2)$$ where $$ker( \phi)$$ is the ideal principal $$((x-r)^2)$$ but then I run into problems because $$(x-r)^2$$ is reducible in $$\mathbb{R}[x]$$ so I can't say $$((x-r)^2) \subseteq ker( \phi) \implies ((x-r)^2) = ker( \phi)$$ (If it was irreducible am I right in saying the opposite inclusion would hold too?) So I gave up on that approach and just tried to explicitly construct an isomorphism and obviously prove that is in fact an isomorphism but I'm struggling with this too, any hints or advice or even solutions would be very very helpful

## 2 Answers

Your idea is right, just use polynomial division to conclude. Let $$F$$ be a field and consider the homomorphism $$\varphi\colon F[x]\to F[x]/\langle x^2\rangle$$ induced by $$\varphi(x)=r+x+\langle x^2\rangle$$. This homomorphism exists because of the properties of polynomial rings. Clearly $$\varphi((x-r)^2)=(x+r-r)^2+\langle x^2\rangle=\langle x^2\rangle$$ so $$(x-r)^2\in\ker\varphi$$. Now let $$f(x)\in\ker\varphi$$; then $$f(x)=g(x)(x-r)^2+ax+b$$ and $$\varphi(f)=g(x+r)(x+r-r)^2+a(x+r)+b+\langle x^2\rangle= ax+ar+b+\langle x^2\rangle$$ and therefore $$ax+ar+b\in\langle x^2\rangle$$, implying $$a=b=0$$.

No properties about $$\mathbb{R}$$ excepts that it is a field are needed, so let's work over a general field $$K$$. Let $$r \in K$$. Using the universal property of polynomial rings, there exists a unique morphism of rings $$\varphi : K[x] \rightarrow K[x]$$ fixing $$K$$ point-wise and satisfying $$\varphi(x) = x+r$$. The composition of $$\varphi$$ with the quotient map $$K[x] \rightarrow K[x]/(x^2)$$ is surjetive and has kernel containing $$((x-r)^2)$$, hence there is an induced morphism $$\bar{\varphi}: K[x]/((x-r)^2) \rightarrow K[x]/(x^2)$$. Similarly, the (extension of) $$x \mapsto x-r$$ induces a morphism $$\bar{\psi}$$ in the other direction, which is inverse to $$\bar{\varphi}$$. (both maps are $$K$$-linear and multiplicative so the identities $$\bar{\psi} \circ \bar{\varphi}= \text{id}$$ and $$\bar{\varphi} \circ \bar{\psi}= \text{id}$$ can be checked on the class of $$x$$, where they hold by construction)

And yes, if $$I \subset K[x]$$ is a proper ideal and containing an irreducible $$f \in K[x]$$, then $$I =(f)$$, because: $$I$$ is principal, so $$I = (g)$$ for some $$g$$ and hence $$f = g h$$ and since $$f$$ is irreducible, and $$g$$ is no unit (because $$I$$ is proper) $$h$$ is a unit, that is $$h \in K^{\times}$$)

• hmm, where you say 'has kernel containing....' and then use the first isomorphism theorem, how are you able to do that? I thought the subring you quotient out had to be the entire kernel not just a subset? – Displayname Jan 18 at 21:47
• No it suffices to have the containment as stated to get an induced morphism. Let $f: A \rightarrow B$ be a morphism of rings and let $I \subset \ker{f}$ be an ideal in the kernel. Then $\bar{f}(a +I) := f(a)$ is well-defined. – m.s Jan 18 at 21:50
• Ah, I saw that I had typed the word isomorphism in my answer in that part o the argument. I now deleted it an edited my answer. – m.s Jan 18 at 22:01
• ahhh good haha, I was checking through lecture notes and thinking nothing makes sense....but yeah I think I get it now, thanks. – Displayname Jan 18 at 22:03