# Congruence modulo's in a polynomial field being a field?

Define $F[x]/(q)$ for polynomial $q$ by $f=g$ $mod(q) \iff q|(f-g)$. Denote the equivalence classes to be $[f]_q$ and the operations $[f]_q+[g]_q=[f+g]_q$ and $[f]_q*[g]_q=[f*g]_q$ (Sorry, I'm too new to know how much of this canonical notation).

Prove that the set of polynomials $F[x]/(q)$ is a field if and only if $q$ is irreducible.

I frankly have no idea how to proceed. I tried going back to the corresponding one in modular arithmetic, but no luck.

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Do you know some ring theory? For a ring $R$ and an ideal $I$, $R/I$ is a field if and only if $I$ is a maximal ideal. So you could show that the maximal ideals in $F[x]$ are precisely those generated by irreducible polynomials. –  Jeff Jan 17 '13 at 18:30
Are you familiar with ideals and quotient rings? If not, then mimic the proof that $\Bbb Z/q$ is a field $\iff q$ is prime, using the fact that $F[x],\,$ like $\Bbb Z,\,$ enjoys a Euclidean algorithm. –  Math Gems Jan 17 '13 at 18:36
By "tried going back"... does that mean you know how to prove $\mathbb{Z} / (p)$ is a field when $p$ is a prime? What is the most important idea in that proof? –  Hurkyl Jan 17 '13 at 18:38

Here is some scaffolding:

1. The ideals of $R/I$ correspond to the ideals of $R$ between $R$ and $I$.

2. $\Bbb F[x]$ and $\Bbb Z$ are principal ideal rings. (In both cases, this can be proven using a division algorithm.)

3. Combining 1 and 2, the ideals of $\Bbb F[x]/(f)$ correspond to ideals $(g)$ such that $(f)\subseteq(g)\subseteq \Bbb F[x]$.

4. $(f)\subseteq (g)$ iff $g$ divides $f$

5. Combining 3 and 4 and elaborating: ideals between $(f)$ and $\Bbb F[x]$ correspond to divisors of $f$, and those ideals properly containing $(f)$ correspond to factors of "proper" factorizations $f$. (Here I'm using "proper" to mean that the factorization consists of nonunits, and there are at least two nonunits.)

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