The polynomial $p(x)=x^4+x+1$ can be shown to be irreducible over $\mathbb{Z}_7$. Show that $\mathbb{Z}_7[x]/\langle p(x)\rangle$ is a field. The polynomial $p(x)=x^4+x+1$ can be shown to be irreducible over $\mathbb{Z}_7$. Show that $\mathbb{Z}_7[x]/\langle p(x)\rangle$ is a field.
Since $p(x)$ is irreducible over $\mathbb{Z}_7$, then $\mathbb{Z}_7[x]/\langle p(x)\rangle$ has $7^4$ elements. I know this guarantees the existence of a finite field over $\mathbb{Z}_7$. But I don't know how to relate what I know to the conclusion that $\mathbb{Z}_7[x]/\langle p(x)\rangle$ is a field.
Any help/hints would be appreciated. ^_^
 A: Hint: As the comments indicate, you want to show that $p(X)$ is irreducible. So suppose not, then $p(X) = f(X)g(X)$ for some $f, g\in \mathbb{Z}_7[X]$, then we may assume without loss of generality that $f(X)$ and $g(X)$ are both monic (why?). Now,


*

*If either $f(X)$ or $g(X)$ has degree 1, then $p(X)$ must have a root on $\mathbb{Z}_7$. Check that this is impossible.

*If $f(X) = x^2 + ax+b$ and $g(X) = x^2+cx + d$, you can compare coefficients to get
$$
c+a = 0; \qquad ac+d+b = 0
$$
$$
ad+bc = 1; \qquad bd = 1
$$
Can you solve these to get some sort of contradiction? (Remember, you are living over a field, so things should get easier).


Added : Once you know $p(X)$ is irreducible, then you want to show that for any $f(X) \in \mathbb{Z}_7[X]$. Try modifying this solution to suit your needs.
A: I think this is easiest to prove in its full generality; the specifics of ${\mathbb Z}_7$ and $p(x) = x^4 + x + 1$ play no role at all and only distract.
Theorem. Let $K$ be a field and let $p(x) \in K[x]$ be irreducible. Then $K[x]/(p(x))$ is a field.
Proof. (assuming that you already know that $K[x]/(p(x))$ is a commutative ring). We have to show that every non-zero element of $K[x]/(p(x))$ has an inverse.
Take $0 \neq a \in K[x]$. Then $a = \overline{q(x)}$ for some $q(x) \in K[x]$ and we can take $q(x)$ to be of degree less than $\deg(p(x))$. Because $p(x)$ is irreducible (and because $p(x) \not\mid q(x)$ because of the degree), $p(x)$ and $q(x)$ are relatively prime. Hence, there are $f(x), g(x) \in K[x]$ such that $$f(x) p(x) + g(x) q(x) = 1.$$ In $K[x]/(p(x))$, this equation becomes $$\overline{g(x)} \; \overline{q(x)} = 1,$$ so $\overline{g(x)}$ is the inverse of $\overline{q(x)} = a$.
