How to show that an ideal of $F[x]$ containing an irreducible polynomial of degree $n$ and a nonzero polynomial of degree $Let $F$ be a field and suppose that $I$ is an ideal of $F[x]$ which contains an irreducible polynomial of degree $n$ and a nonzero polynomial of degree less than $n$. Show that $I=F[x]$.
I can't determine any counterexample so I've been moving forward with trying to prove that this is true. I have that $I$ is an ideal of $F[x]$ gives us $\{a_0+a_{1}x+...+a_{2}x^n|a_0=0\}$. From, here, I don't see how it matters what polynomial the nonzero one is with degree less than $n$ because it will always be in $F[x]$, no?
 A: Let $f(x) \in F[x]$ be the irreducible polynomial of degree $n$, $g(x)$ be a nonzero polynomial of degree less than $n$. Since $F$ is a field, $F[x]$ is a PID, in fact a Euclidean domain. Then $J = \langle f(x) \rangle$ is a maximal ideal, as it is generated by an irreducible element. We know that $J \subset I$ since $I$ contains $f(x)$, so it follows that either $I = J$ or $I=F[x]$. But $g(x) \notin \langle f(x) \rangle$, since every nonzero element of $\langle f(x) \rangle$ has degree at least $n$. Hence $I \neq J$. 
A: You can reason elementarily as follows (avoiding explicit mention of a PID, though in fact that's is what is going on). Let $D$ be a nonzero polynomial in $I$ of minimal degree; this minimal degree must be${}<n$ due to the second given element of$~I$. Then for every element of$~I$, its remainder after Euclidean division by $D$ lies in$~I$ and is of lower degree than$~D$, so equal to$~0$: every element of $I$ is divisible by$~D$. In particular this holds for your irreducible polynomial of degree$~n$. But its only divisors are of degree$~n$ (scalar multiples of itself) or of degree$~0$ (invertible elements). Then $D$ must be an invertible element, whence $I$ is the whole ring $F[x]$.
