The question is:
Let $F$ be a field and $f(x),g(x) \in F[x]$. Verify that $$N=\{r(x)\ f(x)+s(x)\ g(x):r(x),s(x)\in F[x]\}$$ is an ideal of $F[x]$. Then show that if $f(x)$ and $g(x)$ have different degrees and $N\ne F[x]$, then $f(x)$ and $g(x)$ cannot both be irreducible over $F$.
It's easy to verify that $N$ is an ideal of $F[x]$ so I'll leave that bit out. For the second part, I think I can prove the contrapositive, but would just like confirmation that my proof is correct.
The contrapositive of the statement I want to prove is this:
If $f(x)$ and $g(x)$ are both irreducible over $F[x]$, then either $f(x)$ and $g(x)$ have same degree, or $N=F[x]$.
Here's the proof:
Since $F$ is a field, then we know every ideal of $F[x]$ is principal. Since $f(x)$ is irreducible, then $\langle f(x)\rangle$ is a maximal ideal of $F[x]$. But it is clear that $\langle f(x)\rangle \subset N$, so $N=F[x]$.
I can't see anything wrong with what I'm doing, but the reason I have misgivings is because I didn't use all the information given (like the degrees of $f(x)$ and $g(x)$) and I'd like some confirmation.
Thanks!
P.S Can anyone think of a better title for this question? This was a pretty specific question, and I feel like the title's a little awkward.
You proved that if $f$ is irreducible $(f)\subset N$ but this doesn't mean $N=R$. Either $N=R$ or $N=(f)$ is the correct result. Similarly if $g$ is irreducible and $N\neq R$ then $(g)=N$. This means $(f)=(g)$. If $\deg f\neq \deg g$ then one of degrees is bigger, say $\deg f>\deg g$. Then $f=gh$ for some non-constant polynomial $h$. A contradiction.