Is $\mathbb{Q}[X]/(X^2+1) \cong \mathbb{Q}[X]/(X^2-1)$

Question

Is

$$\mathbb{Q}[X]/(X^2+1) \cong \mathbb{Q}[X]/(X^2-1)$$

I know that $\mathbb{Q}[X]/(X^2+1) \cong \mathbb{Q}(i)$ but I can't say that $\mathbb{Q}[X]/(X^2-1) \cong \mathbb{Q}(1) = \mathbb{Q}$ since $X^2-1$ is not irreducible.

Answer 1

Hint: Show that the coset of $x-1$ is a zero divisor in the latter ring. Hence you can conclude that it is not a field.

Answer 2

$\mathbb{Q}[x]/(x^2-1)$ is not a field. Its a ring with additive group $\mathbb{Q}\oplus\mathbb{Q}$.