# What is the minimal polynomial of $\sqrt{3} + \sqrt[3] {2}$ over $\mathbb{Q}$? [duplicate]

What is the minimal polynomial of $\sqrt{3} + \sqrt[3]{2}$ over $\mathbb{Q}$?

I know the basic idea of what a minimal polynomial is--it is the lowest degree monic polynomial in $\mathbb{Q}[x]$ that has the above as a root. But how do you go about calculating it? How do you know intuitively if it does or doesn't exist?

## marked as duplicate by Jack, C. Falcon, Marc van Leeuwen, Adam Hughes, Ross MillikanDec 22 '16 at 15:48

• How about choosing a $\Bbb Q$-basis of $\Bbb Q[\sqrt3,\sqrt[3]2]$, writing down the matrix for multiplication by $\sqrt3+\sqrt[3]2$, and then computing its minimal polynomial? – Marc van Leeuwen Dec 22 '16 at 15:27
• The field generated by those two numbers, $\mathbf Q(\sqrt{3},\sqrt[3]{2})$, has degree 6 over $\mathbf Q$. Pick a basis for the field extension and compute the characteristic polynomial for the $\mathbf Q$-linear map "multiplication by $\sqrt{3}+\sqrt[3]{2}$" on that field extension, the same way you'd compute the characteristic polynomial for a linear operator on any finite dimensional vector space if you have a basis for the space. It turns out to be $x^6 - 9x^4 - 4x^3 + 27x^2 - 36x - 23$, which is irreducible mod 7 so it's irreducible over $\mathbf Q$. Thus this is the minimal polynomial. – KCd Dec 22 '16 at 15:28
• @KCd: Is there a reason to go for the characteristic polynomial first rather than directly for the minimal polynomial? – Marc van Leeuwen Dec 22 '16 at 15:31
• For field extensions in characteristic $0$, the degree of the sum of two numbers algebraic over a common field is the product of their individual degrees over that field if the two degrees are relatively prime (like $2$ and $3$). See mathoverflow.net/questions/26832/…. – KCd Dec 22 '16 at 15:32
• @MarcvanLeeuwen it's generally easier to work out the characteristic polynomial from scratch than the minimal polynomial, and my previous comment shows the answer in this case (relatively prime degrees 2 and 3) must be the minimal polynomial after all. – KCd Dec 22 '16 at 15:33

Start with $x=\sqrt{3}+\sqrt[3]{2}$. Then \begin{gather*} x-\sqrt{3} = \sqrt[3]{2} \\ (x-\sqrt{3})^3 = 2 \\ x^3 - 3\sqrt{3}x^2 + 9x - 3\sqrt{3} = 2 \\ x^3 + 9x - 2 = 3\sqrt{3}(x^2+1) \\ (x^3+9x-2)^2 = 27(x^2+1)^2 \\ x^6 + 18x^4 - 4x^3 + 81x^2 - 36x + 4 = 27x^4 + 54x^2 + 27 \\ x^6 - 9x^4 - 4x^3 + 27x^2 - 36x-23 = 0. \end{gather*}
• @lhf that polynomial is irreducible modulo $7$ (see my comments to the question). – KCd Dec 22 '16 at 15:34
$$x-\sqrt 3-\sqrt[3] 2=0\\\implies (x-\sqrt 3)^3=2=x^3-3\sqrt 3x^2 +9x-3\sqrt 3\\\implies (x^3-2+9x)^2=27(x^2+1)^2$$