Fields of polynomials Let $F$ be a field and $f(x)$ be an irreducible polynomial in $F[x]$. Show that the set of polynomials modulo $f(x)$ form a field.
I know this question is just about verifying the field axioms but I'm not entirely sure how to go about setting this problem out. Also, I can't think how I prove the multiplicative inverse axiom.
Any help would be great. Cheers
 A: With a bit of theory, this is just a ring modulo a maximal ideal, hence a field.
Working from "first principles", everything should be straightforward except, as you note, multiplicative inverse.
Let $g(x)\in F[x]$ be a polynomial.
Using the extended Euclidean algorithm we can write $d(x)=u(x)g(x)+v(x)f(x)$, where $d(x)=\gcd(f(x),g(x))$. Since $d(x)$ is a divisor of $f(x)$ we can only have $d(x)=1$ or $d(x)=f(x)$ (up to a constant factor). - This is where we use irreducibility of $f$. In the first case we have $u(x)g(x)\equiv 1\pmod {f(x)}$; in the second case $f(x)\mid g(x)$. Thus either $g(x)\equiv 0\pmod{f(x)}$ or we can find a multiplicative inverse.
Note that one should never neglect one other important field axiom: $1\ne 0$. This follows from the fact that irreducible polynomials are by definition of degree $\ge 1$.
A: Checking the ring axioms is straightforward.  Showing the existence of multiplicative inverses is the nontrivial part.  So choose some nonzero polynomial $p(x) \in F[x]$.  Because $f(x)$ is irreducible, $\gcd (p(x), f(x)) = 1$.  This implies there exists polynomials $m(x)$ and $n(x)$ such that
$$ m(x)f(x) + n(x)p(x) = 1. $$
Now
$$n(x)p(x) \equiv 1 \mod f(x)$$
which implies the existence of a multiplicative inverse of $p(x)$.
