Let $F=\mathbb{Z}/2\mathbb{Z}$. Show that $F[x]/(x^2+1)$ is a ring of four elements

Question

Let $F=\mathbb{Z}/2\mathbb{Z}$. Show that $F[x]/(x^2+1)$ is a ring of four elements.

I thought I could use the first isomorphism theorem, but I get stuck. Could anyone help me at this point?

Answer 1

$F[x]/(x^2+1)$ is a ring because it is the quotient of a ring by an ideal, so it is enough to show that $F[x]/(x^2+1)$ contains only four elements.

If $f(x)$ is a polynomial in $F[x]$, then the remainder when $f$ is divided by $x^2+1$ is either the zero polynomial or has degree less than two. Therefore $F[x]/(x^2+1)$ consists of (the equivalence classes of) $0$ and the three non-zero polynomials of degree $\leq 1$, namely $1,x$, and $x+1$.