Consider a polynomial x^3-x-1 and let $\alpha$, $\beta$, $\gamma$ be three zeros of the polynomial where $\alpha \in \mathbb{R}$. Since $\beta \not \in \mathbb{Q}[\alpha]$ and the degree of the splitting field of the polynomial over $\mathbb{Q}$ is $6$, the minimal polynomial of $\beta$ over $\mathbb{Q}[\alpha]$ should be of degree $2$. To find the polynomial (by the way, I know it should be $x^2+\alpha x+ \frac{1}{\alpha}$.) I have coded the following

sol = x /. Solve[x^3 == x + 1, x];
al = sol[[1]];
be = sol[[2]];
ga = sol[[3]];
MinimalPolynomial[be, x, Extension -> al]

However, Mathematica returns an unexpected message;

Unexpected result

What is wrong with the code? What would be a right way to get correct answer?

  • $\begingroup$ Look at the roots of: x2+αx+1α. β is not among them. $\endgroup$ Oct 7, 2021 at 9:55
  • $\begingroup$ @DanielHuber If you try "FullSimplify[be^2 + al be + 1/al]", you would get zero. $\endgroup$
    – seoneo
    Oct 7, 2021 at 10:29
  • $\begingroup$ You are right. I only tried Simplify, but it needs FullSimplify. $\endgroup$ Oct 7, 2021 at 12:14
  • 1
    $\begingroup$ In the vein of user64494's comment, FactorList[MinimalPolynomial[be][x], Extension -> al] seems to contain your polynomial (with $1/\alpha$ represented as $\alpha^2-1$). I wonder if there's a good way to identify it as what you want. $\endgroup$
    – thorimur
    Oct 8, 2021 at 3:13
  • 1
    $\begingroup$ @thorimur Thanks for your careful comment. By the help of the paragraphs I got a better understanding. $\endgroup$
    – seoneo
    Oct 8, 2021 at 12:09

1 Answer 1


MinimalPolynomial[s, x, Extension->a] finds the characteristic polynomial of the element $s \in \mathbb{Q}[a]$ over the field $\mathbb{Q}[a]$.


As has been already suggested, you can get the minimal polynomial over a field extension using Factor(List).

In[7]:= minpoly[a_, e_, x_] :=                                                  
   Select[First/@FactorList[MinimalPolynomial[a, x], Extension->e],             
      PossibleZeroQ[#/.x->a, Method->"ExactAlgebraics"]&][[1]]                  

In[8]:= minpoly[be, al, x]                                                      

              2                      3                           3       2
Out[8]= -1 + x  + x Root[-1 - #1 + #1  & , 1] + Root[-1 - #1 + #1  & , 1]
  • $\begingroup$ Many thanks for your valuable solution! $\endgroup$
    – seoneo
    Oct 12, 2021 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.