In the context of field theory, let's say we are finding the minimal polynomial of the $\sqrt[3] 2$ over $\mathbb{Q}$.

Clearly a candidate will be $x^3-2$, which is irreducible by Eisenstein's criterion. Can we then immediately conclude that it is the minimal polynomial?

What I am worried is, is there such thing as an irreducible polynomial that is somehow not the minimal polynomial? (I know minimal polynomial implies irreducibility, but not sure about the converse: does irreducible polynomial implies minimality?)

Thanks for any help.


3 Answers 3


Assume we're looking for the minimal polynomial of $\alpha$, $\mu$ say, and suppose we have another polynomial $f$ satisfying $f(\alpha)=0$.

We can write $f=g\mu+r$, where $g$ and $r$ are polynomials and $\mathrm{deg}(r)<\mathrm{deg}(\mu)$ (Euclidean division).

Clearly, $r(\alpha)=0$, but then $r=0$ since $\mu$ was chosen such that it is the non-zero polynomial of minimal degree with this property. Hence, $\mu$ divides $f$.

Therefore we know that any "candidate" for the minimal polynomial is a multiple of the minimal polynomial and if our candidate is irreducible, then it is the minimal polynomial and we're done.

Note that this works over any field, not just $\mathbb{Q}$.

  • $\begingroup$ This has cleared my doubts. Thanks! $\endgroup$
    – yoyostein
    Nov 7, 2015 at 7:48
  • $\begingroup$ @Oliver Braun You're proof makes sense, but I'm wondering one thing. Consider your second last sentence. If we rephrase this as saying "the minimal polynomial divides any "candidate", why is it not possible to have $f(x)=\mu(x)g(x)$ where $f(x), \mu(x) \in \mathbb{Q}[x]$ but $g(x)$ is not? Then couldn't $f$ still be irreducible but have $\mu$ as a strictly smaller degree minimal polynomial? $\endgroup$
    – A.B
    Feb 2, 2021 at 17:57

Suppose there is a polynomial $p(x)$ of degree $\lt 3$, with rational coefficients, which is not a constant times our polynomial $q(x)$ of degree $3$, such that $p(\sqrt[3]{2})=0$. If the gcd $d(x)$ of $p(x)$ and $q(x)$ has degree $\ge 1$, we have contradicted irreducibility.

If the gcd is $1$, there are polynomials $A(x)$ and $B(x)$, with rational coefficients, such that $A(x)p(x)+B(x)q(x)=1$. This is impossible, for it would imply that $\sqrt[3]{2}$ is a root of the constant polynomial $1$.

The same argument works in general.


For completeness, the only (obvious) difference is that, by definition, a minimal polynomial is a monic irreducible polynomial, whereas an irreducible polynomial need not to be monic.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .