Show that a given ring is a field with four elements Let $R = ( \mathbb{Z} / 2 \mathbb{Z} ) [t]$ be the ring of polynomials with coefficients $\mathbb{Z} / 2 \mathbb{Z}$, $f = f(t) = t^2 + t +1$, and $g = t^2 +1$. Show that:
(1) $R/(f)$ is a field with four elements.
(2) $R / (g)$ is not a domain and has four elements.
(3) Neither $R / (f)$ nor $R / (g)$ is isomorphic to the ring $\mathbb{Z} / 4\mathbb{Z}$.

I have no idea how to do this problem. My guess for the first one is first show $R / (f)$ has four elements and then show $(f)$ is maximal, but I don't know how to proceed it.
Any help is appreciated.
 A: An entirely elementary approach is the following. 
First, given any field $F$ and monic polynomial $h$ of degree $n$, $F[x]/(h)$ is the ring of residue classes modulo $h$, that is, the remainders of Euclidean division by $h$ (very much as in the case of $\mathbb{Z}/(n) = \mathbb{Z}/n \mathbb{Z}$.)
So in both cases, as sets, your quotient rings consist of the congruence classes (= cosets) of  the four elements $a + b x$, with $a, b \in F = \mathbb{Z}/(2)$. But the product in the two cases is critically different, as we will see in a second.
Now $f$ is irreducible, so in case (1) you will get a field. This is simply because $x (x + 1) = x^{2} + x \equiv 1 \pmod{f}$, so all elements $\ne 0$ will have an inverse.
Instead, $g = x^2 + 1= (x+1)^{2}$ is reducible, and thus in case (2) there are zero divisors, as $x + 1 \not\equiv 0 \pmod{g}$ but $(x + 1)^{2} \equiv 0 \pmod{g}$.
In both cases, the quotient ring is a vector space over $F$, so the additive period of the non-zero elements is $2$, while in $\mathbb{Z}/4\mathbb{Z}$ there are (two) elements with additive period $4$. You can also tell (1) apart from $\mathbb{Z}/4\mathbb{Z}$ because the latter is not a field, but the multiplicative structures of (2) and $\mathbb{Z}/4\mathbb{Z}$ are the same. 
