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.
 A: 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}
A: Note that, since $k$ is a field, $k[X]$ is Euclidean, so for any $p\in K[X],$ there exists $q\in K[X]$ and $r\in K[X]$ such that the degree of $r$ is at most $1$ and $p(x)=q(x)(x^2-1)+r(x)$. Accordingly, we see that the quotient is 2-dimensional as a $k$-vector space.
Now, we can construct a $k$-linear map $\varphi: k^2 \to k[X]/(X^2-1)$ by $(1,1)\mapsto 1$ and $(1,-1)\mapsto x$. Note that this is clearly a $k$-linear isomorphism by the above dimension argument (and the two polynomials $1$ and $x$ are $k$-linearly independent independent in the quotient). Now, we should check that it extends to a ring-homomorphism, but this follows, since $x^2=1$ and $(1,-1)^2=(1,1)$.
This proves the isomorphism.
A: If $x^2=1$ in $k[x]$ then the equivalence class of any element has a representative of the form $a+bx$ for $a,b\in k$.
