# Finding a minimal polynomial of an algebraic element using Galois theory

There is a canonical (but difficult) way of determining the minimal polynomial of an algebraic element $\alpha$ in a field $F$, namely by considering the $F$-linear transformation defined by left multiplication by $\alpha$ and computing the minimal polynomial of the matrix of the linear transformation.

I am interested in a Galois-theoretic way of doing this. My motivation is the following: consider the problem of finding the minimal polynomial of $\sqrt2 + \sqrt3$ over $\mathbb Q$. One can do this by "conjugate bashing" and checking that $$(x - (\sqrt2 + \sqrt3))(x - (\sqrt3-\sqrt2))(x - (\sqrt2 - \sqrt3))(x + (\sqrt2 + \sqrt3))$$ is irreducible over $\mathbb Q$. The conjugate bashing is the same thing as applying all the Galois elements. This naive approach fails with $\sqrt[3]{2}$, which has minimal polynomial $x^3 - 2$ and Galois group of order 6. Not all hope is lost, however, since this is a "characteristic polynomial" of the element and contains as a factor the minimal polynomial.

Is there any way to fix this deficiency?

• Yes. Commented May 8, 2016 at 3:55
• What is the number of which you are finding the minimal polynomial? Commented May 8, 2016 at 3:58
• Not sure what that means Commented May 8, 2016 at 4:02
• What is the number you are finding? Commented May 8, 2016 at 4:05
• The algebraic element? Not anything specific. I'm looking for a methodical way of doing this for a general element; if such a method doesn't exist, under what conditions on the Galois group and element can we do this? Commented May 8, 2016 at 4:12

There is no "deficiency". $X^3 - 2$ having a Galois group of order $6$ is not a problem. When you apply all the automorphisms from the Galois group things reappear. Ignore the reappearance and you are done.

But you don't really need the galois group of the minimal polynomial. What you are really doing is, you are looking at all the automorphisms of the algebraic closure and making a list of places your element is sent to. Call that set $S$, then your minimal polynomial is $\prod_{t \in S} (X - t)$.

In your first example, $\sqrt 2 \not \in \mathbb Q (\sqrt 3)$ and vice versa. So you can have an automorphism of $\overline {\mathbb Q}$ that sends $\sqrt 2 \to \pm \sqrt 2$ and $\sqrt 3 \to \pm \sqrt 3$. Therefore your set S consists of $\left \{\pm \sqrt 2 \pm \sqrt 3 \right \}$.

Similarly deal with $\sqrt[3] 2$.

• is it "easy" in general to find all such automorphism of $\overline{\mathbb{Q}}$ ? say with $a= \sqrt[13]{\sqrt{2}+\sqrt{\sqrt {3}+7}}+\sqrt{5}$ ? Commented May 8, 2016 at 4:38
• Well, for practical purposes, I am just guessing the conjugates and extending by isomorphism extension to get an automorphism of $\overline {\mathbb Q}$. The terminology is there just to formalize the approach. I guess it is not easy to guess the conjugates always. But neither is finding the minimal! Commented May 8, 2016 at 4:45
• I think that the method I outlined in my answer would work perfectly well, @user1952009, to get a polynomial for $\sqrt2+\sqrt{\sqrt3+7}$, for instance, and the succeeding steps would be equally easy. You’d still have to show irreducibility. Commented May 8, 2016 at 23:45
• @Lubin : yes, it involves computing $Aut(K/F)$ for some $K$ containing $a$ ? Commented May 8, 2016 at 23:53
• Sorry, @user1952009, I was answering the much easier question of finding the minimal polynomial of your irrationality. Getting the associated Galois group is certainly much harder. Commented May 9, 2016 at 12:51

If $K/F$ is Galois and $\alpha \in K$, then the minimal polynomial of $\alpha$ has as its roots the elements in the orbit of $\alpha$ under the action of $G(K/F)$. That is to say, if $\{a_k\}$ is the set of elements in the orbit of $\alpha,$ then $\displaystyle \min_{\alpha}(x) = \prod_{k} (x-a_k)$.

• Ha! It appears you answered my question several months ago Commented Nov 20, 2016 at 6:09
• LOL! Small world stack exchange is :) Commented Nov 20, 2016 at 6:12

Here’s a method I use that doesn’t involve so much conjugate bashing, but does require a bit of arithmetic, namely to check that $3$ is still square-free in the PID $\Bbb Z[\sqrt2\,]$.

Suppose, for variety, that you want the minimal polynomial of $\sqrt[3]3+\sqrt2$, over $\Bbb Q$. You do know the minimal polynomial of $\sqrt[3]3$ over $\Bbb Q(\sqrt2\,)$, namely $f(X)=X^3-3$, and thus you know the minimal polynomial of $\sqrt[3]3+\sqrt2$, over $\Bbb Q(\sqrt2\,)$, namely $g(X)=f(X-\sqrt2\,)$. Just multiply $g$ by its conjugate (replacing $\sqrt2$ by $-\sqrt2$) to get the desired polynomial over $\Bbb Q$.