Why is $k[X]/(X^2 -1)$ isomorphic to $k^2$? For a field $k$, why is the ring $k[X]/(X^2 -1)$ isomorphic to $k^2$ ?
Please also explain how one might notice this is the first place, without being told. Thanks.

That is because we have the Chinese remainder theorem:
\begin{alignat}{2}
k[X]/(X-1)(X+1)&\simeq K[X]/(X-1)\times k[X](X+1)&&\simeq k\times k \\
P\bmod X^2-1&\mapsto\bigl(P\bmod (X-1),P\bmod(X+1)\bigr)&&\mapsto\bigl(P(1),P(-1)\bigr)
\end{alignat}
