# What is $\mathbb{R}[x]$ quotiented by a polynomial $f(x) \in \mathbb{R}[x]$ isomorphic to?

By the CRT, we have that: $$\mathbb{R}[x]/(x^2-2) \simeq \mathbb{R}[x]/(x-\sqrt{2}) \times \mathbb{R}[x]/(x+ \sqrt{2}) \simeq \mathbb{R} \times \mathbb{R}.$$

By considering the evaluation map $$\mathbb{R}[x] \to \mathbb{C}$$ at $$2i$$, it is also clear that: $$\mathbb{R}[x]/(x^2+2) \simeq \mathbb{C}.$$

Now, I am wondering if it is true that $$\mathbb{R}[x]/(x-2)^2 \simeq \mathbb{R}[x]/(x)^2,$$ and if so how I would prove it. My idea was simply considering the quotient explicitly as a set of the form $$\{a_0 + a_1x + f(x)(x^2)\}$$ and identifying it with $$\{a_0 + a_1x + g(x)(x^2 - 4x + 4)\}$$. But if that works, what prevents me from using this same method to show that this quotient is isomorphic to the quotients above — which is obviously false?

• $\Bbb R[x] = \Bbb R[x - 2].$ From another point of view, you can Taylor expand polynomials about $x = 2$ and write them $p(x) = \sum a_i (x - 2)^i.$ – Stahl Dec 12 '18 at 0:28

## 2 Answers

I suggest familiarizing yourself with two universal principles. Look them up elsewhere, I’ll just mention an example:

(1) The universal property of polynomial rings. Homomorphisms $$\mathbb R[x] \to \mathbb R[u]/(u^2)$$ are determined by where you send $$x$$, and any choice is ok.

(2) The universal property of quotients. A homomorphism $$\Bbb R[x] \to \Bbb R[u]/(u^2)$$ factors through $$\Bbb R[x]/(f(x))$$ iff $$f(x) \mapsto 0$$. In the ring $$\Bbb R[u]/(u^2)$$ being 0 is the same as being divisible by $$u^2$$ as a polynomial.

Thus to check if $$\varphi: x\mapsto x-2$$ is a homomorphism $$\Bbb R[x]/(x^2)\to \Bbb R[x]/(x-2)^2$$ , you just need to make sure $$\varphi(x^2) = 0$$, which it is since $$(x-2)^2 = 0$$.

In the case of $$\Bbb R[x]/(x^2-2)$$ you would send $$x\mapsto p(x)$$ and you would need to check that $$p(x)^2$$ is 0, in other words it is divisible by $$x^2-2$$. This implies that $$(x^2-2)|p(x)$$ hence the homomorphism sends $$x\mapsto 0$$ and is not an isomorphism.

Similar logic works for $$x^2+2$$.

When you are identifying $$\{a_0 + a_1x + f(x)(x^2)\}$$ with $$\{a_0 + a_1x + g(x)(x^2 - 4x + 4)\}$$, you are mapping $$x^2$$ to $$x^2 - 4x + 4$$, which is a ring homomorphism only because you're implicitly mapping $$x$$ to $$x-2$$, which is the isomorphism that shows $$\mathbb{R}[x]/(x)^2 \simeq \mathbb{R}[x]/(x-2)^2$$. There is no corresponding ring homomorphism that maps $$x^2$$ to $$x^2+2$$ or $$x^2$$ to $$x^2-2$$ (think about where $$x$$ would have to go).