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?

  • 3
    $\begingroup$ Use the Euclidean algorithm to show that the two polynomials are co-prime. $\endgroup$
    – brick
    Nov 25, 2014 at 16:15
  • $\begingroup$ BLUF: $I$ is generated by the greatest common divisor of the two polynomials. The other responses more or less emphasize how to prove that fact, but IMO the emphasis should be on the fact itself. $\endgroup$
    – user14972
    Nov 25, 2014 at 16:57

2 Answers 2


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 $I \neq J$.

  • $\begingroup$ Is there anything special about the use of the $\rangle$ because that is unfamiliar to me. Or, was that just your style of notating that to avoid repeated parenthesis? $\endgroup$
    – GiGi
    Nov 25, 2014 at 16:31
  • $\begingroup$ @GiGi: I use $\langle \rangle$ to denote "the ideal generated by". So $J = \langle f(x) \rangle$ should be read as "$J$ equals the ideal generated by $f(x)$. Does this clarify things? $\endgroup$ Nov 25, 2014 at 16:33
  • $\begingroup$ I think so. My book just used repeated parenthesis so I wanted to be sure that it was as simple as that. $\endgroup$
    – GiGi
    Nov 25, 2014 at 16:35
  • $\begingroup$ Thank you. That helped (assuming I understood it correctly). $\endgroup$
    – GiGi
    Nov 25, 2014 at 16:43
  • $\begingroup$ @GiGi: my pleasure. Feel free to ask questions if you need further clarification! $\endgroup$ Nov 25, 2014 at 16:48

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 $F[x]$.


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