Identifying $\mathbb{R}[x]/(x^2-2)$ I'm revising for an exam and I've come across this question. We're trying to identify the quotient ring $R =  \mathbb{R}[x]/(x^2-2)$. This is the same as $\mathbb{R}[\alpha]$ where $\alpha^2 -2 = 0$, and I can show that $\{1,\alpha\}$ forms a $\mathbb{R}$-linear basis for the ring.
I have also shown that the multiplication law is given by
$$
(a + b \alpha) (c + d \alpha) = (ac + 2bd) + (ad + bc) \alpha.
$$
Now I believe that $R \cong \mathbb{R} \times \mathbb{R}$. I think this follows from the Chinese Remainder Theorem, but unfortunately we have not covered this theorem in my course. Is there an simpler way of showing this isomorphism?
Thanks for any help!
 A: Consider the ring-homomorphism $c:\mathbb{R}[x]\to\mathbb{R}\times\mathbb{R}$ given by $c(f)=(f(\sqrt{2}),f(-\sqrt{2}))$.  Since $c(x^2-2)=0$, this induces a homomorphism $\bar{c}:R\to \mathbb{R}\times\mathbb{R}$.  Now note that $R$ and $\mathbb{R}\times\mathbb{R}$ are both 2-dimensional $\mathbb{R}$-vector spaces in obvious ways, and $\bar{c}$ is a linear map between them.  So to show that $\bar{c}$ is a bijection (and hence a ring-isomorphism), it suffices to show that it is injective.  Equivalently, it suffices to show that $\ker(c)=(x^2-2)$, since $\ker(\bar{c})=\ker(c)/(x^2-2)\subset R$.
So suppose $f\in\mathbb{R}[x]$ and $c(f)=0$.  This means that $f(\sqrt{2})=f(-\sqrt{2})=0$, so $f(x)$ is divisible by $x-\sqrt{2}$ and $x+\sqrt{2}$.  Since polynomials have unique factorization and $x-\sqrt{2}$ and $x+\sqrt{2}$ are relatively prime, this means $f(x)$ is divisible by $(x-\sqrt{2})(x+\sqrt{2})=x^2-2$.
More generally, this argument shows that if $F$ is a field and $g(x)\in F[x]$ is a degree $n$ polynomial with $n$ distinct roots in $F$, then $F[x]/(g(x))\cong F^n$ (with the isomorphism being given by evaluating polynomials at each of the roots).
