Chinese Remainder Theorem problem in ring theory I'm currently following my 2nd algebra course at uni where we are discussing rings.
The assignment goes like this
$$\text{Show that} \ \mathbb{R}[x]/(x^2-1) \ \text{is isomorphic to} \ \mathbb{R}\times\mathbb{R}$$
$\mathbb{R}[x]$ is the polynomial ring and $\mathbb{R}[x]/(x^2-1)$ is then a quotient ring. A hint to solve it is to use the Chinese Remainder Theorem.
I am stumped. The hint implies to me that $(x^2-1)$ must be the kernel of some ring homomorphism. It also to me implies that there is an ideal in $\mathbb{R}[x]$, let's call it K, so that $\mathbb{R}[x]/K=\mathbb{R}$ and I can't make either of those things to make sense. Is it also right to assume that $\mathbb{R}[x]/(x^2-1)=\{f(x)+x^2-1\vert f(x)\in\mathbb{R}[x]\}=\mathbb{R}[x]$, since all equivalence classes will contain just one element, since for two elements to be in the same equivalence class, the polynomials must be the same? Like if,
$$q(x)+x^2-1=r(x)+x^2-1\Leftrightarrow q(x)=r(x)$$
for $q(x),r(x)\in\mathbb{R}[x]$.

Hint: Consider the map $\phi:\mathbb R[x] \to \mathbb{R}\times\mathbb{R}$ given by $\phi(f)=(f(1),f(-1))$. Prove that $f$ is a surjective ring homomorphism and compute its kernel.
