Show that $F[x] / (x^2-1) \cong F \times F$ iff $char(F) \neq 2$

Question

Let $F$ be a field. Show that $F[x] / (x^2-1) \cong F \times F$ iff $char(F) \neq 2$.

Here is my work this far: If $char(F) = 2$, then $x^2-1 = (x-1)^2$, and hence $F[x]/(x^2-1)$ has a non-zero element whose square is $0$. But $F \times F$ is a field, so it doesn't have any non-zero elements that square to zero, and hence they are not isomorphic.

Now, I want to show that they are isomorphic if $char(F) \neq 2$. I have tried to construct an isomorphism between the two, but the obvious map ($a + bx + (x^2 -1) \to (a,b)$) does not seem to be a homomorphism. I can't seem to show that it preserves multiplication, and I was hoping it would be possible to show that it preserves multiplication iff $char(F) \neq 2$. How can I prove the second direction of this?

Answer 1

This is s special case of Chinese remainder theorem. Send a polynomial $f(x)\in F[x]$ to the pair $(f(-1), f(1))$. CRT states it is surjective.

To see surjectivity avoiding CRT you have to solve a linear algebra problem that requires one to find a polynomial having prescribed values at two given points. (That is what Lagrange Interpolation Theorem does in Numercal Mathematics) For it to have a zero at both $+1$ and $-1$ it should be a multiple of $x^2-1$. So the kernel is the ideal generated by $x^2-1$.

Answer 2

This map works: $$\frac{a-b}{2}\cdot x + \frac{a+b}{2}\cdot 1 + (x^2-1)\cdot F[x] \mapsto (a,b)$$

you were close. The map is $\bar f \mapsto ( f(1), f(-1))$