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

Question

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?

Answer 1

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 $x^2+2$.

Answer 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).