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This is my homework question:

Let $F$ be a field and $f(x), g(x) \in F[x]$ be polynomials. Show that $N = \{r(x)f(x) + s(x)g(x) \, | \, r(x), s(x) \in F[x] \}$ is an ideal in $F[x]$. Show that if $f(x), g(x)$ have different degrees and $N \ne F[x]$, then $f(x), g(x)$ cannot both be irreducible over $F$.

I have successfully shown that $N$ is an ideal, but I believe I've found a counterexample to the second claim. What if $f(x) = 0$ and $g(x) = x$? Then (1) $f$ and $g$ have different degrees, (2) $f$ and $g$ are irreducible, and (3) any element $n(x) \in N$ must satisfy $n(0) = 0$ (so $N \ne F[x]$).

What's wrong with my reasoning?

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    What is your definition of irreducible? Usually a non-zero non-unit in a domain is irreducible if... Notice the non-zero bit. – John Myers Jan 25 '13 at 01:28
  • Do you have a definition of degree for the zero polynomial? What is $\deg (0(x))$? – Sigur Jan 25 '13 at 01:42
  • Do you already know that $F[x]$ is a PID? (since it has a Euclidean algorithm: the polynomial division algorithm). – Math Gems Jan 25 '13 at 03:21
  • For a first course in algebra, it might be instructive to use the Euclidean algorithm directly as it is a very nice algorithm easy to program for a computer and directly computes a common factor of both polynomials exactly as you can do for integers. To avoid division by $0$, the zero of a Euclidean domain is often given a negative Euclidean degree such as -infinity and for polynomials in one variable prime monic polynomials are called irreducible. – Barbara Osofsky Jan 25 '13 at 04:11
  • The general denotation of primes and irreducibles mentioned by @Barbara is a special case of concepts applied to nonunits in general integral domains, where, unlike UFDs, irreducibles (or atoms), i.e. $,p=ab\Rightarrow p|a\ {\rm or}\ p|b,,$ need not be prime, i.e. $,p|ab\Rightarrow p|a\ {\rm or}\ p|b,,$ e.g. see this answer. – Math Gems Jan 25 '13 at 16:44

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By definition, irreducible elements are different from $0$, so your conterexample is wrong.

Coming back to the problem: $F[x]$ is a PID, so $(f,g)$ is a principal ideal generated by the greatest common divisor (GCD) of $f$ and $g$. Assume that $f$ and $g$ are both irreducible. Since $\deg f\neq\deg g$ and $f$, $g$ are irreducible, we get that their GCD is $1$. Thus $(f,g)=(1)$, that is, $(f,g)=F[x]$, a contradiction.