Prove that $\frac{\mathbb{F}_2[x]}{(x^2+1)} \cong \frac{\mathbb{F}_2[x]}{(x^2)}$

Question

I need to show that $\frac{\mathbb{F}_2[x]}{(x^2+1)}$ is isomorphic to $\frac{\mathbb{F}_2[x]}{(x^2)}$. However, I am becoming increasingly convinced that they are not.

I believe the members of both rings are $\{0, 1, x, x+1\}$. In the latter, $x^2=0$ because of the ideal.

In the former, though, $x^2+1=0$, so $x^2 = 1$ because $1 + 1 = 0$ in $\mathbb{F}_2$. In fact, no multiplication by any two non-zero elements can result in $0$, since $(x+1)^2=1$, $x(x+1)=x$, and anything times $1$ is itself. This tells me they aren't isomorphic.

Where is my mistake?

Answer 1

Consider $f:F_2[X]\rightarrow F_2[X]$ defined by $f(X)=X+1, f(X^2)=(X+1)^2=X^2+2X+1=X^2+1$, we deduce that $f$ induces a morphism $F_2[X]/(X^2)\rightarrow F_2[X]/(X^2+1)$. Since $f$ is an isomorphism, the morphism induced is an isomorphism.

Answer 2

Look at the morphism $\mathbb{F}_2 \to \mathbb{F}_2[x]/(x^2)$ determined by $x \to x-1$.