Are $\frac {\mathbb R[x]} {(x^2 -1)}$ and $\frac {\mathbb R[x]}{(x^2)}$ Isomorphic?

Question

Are $\frac {\mathbb R[x]} {(x^2 -1)}$ and $\frac {\mathbb R[x]}{(x^2)}$ Isomorphic?

Can anyone please give me a hint?

My attempt:

I can find the elements of two rings. But I have no clue to prove or disprove that statement.

Answer 1

Note that $x+(x^2)$ is a nonzero member of $\mathbb{R}[x]/(x^2)$ whose square is zero.

Instead of directly showing that $\mathbb{R}[x]/(x^2-1)$ has no such element, we find it easier to work with an isomorphic ring. Namely, we have the isomorphisms $$\frac{\mathbb{R}[x]}{(x^2-1)}\cong\frac{\mathbb{R}[x]}{(x-1)}\oplus\frac{\mathbb{R}[x]}{(x+1)}\cong\mathbb{R}\oplus\mathbb{R}.$$ Since $\mathbb{R}$ has no such element, neither does $\mathbb{R}\oplus\mathbb{R}$.

The first isomorphism is given by the Chinese Remainder Theorem. The second is induced by two evaluation homorphisms $\mathbb{R}[x]\to\mathbb{R}$; at $1$ for the first component and at $-1$ for the second.