1

I want to determine the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$.

I calculated the polynomial $x^4-10x^2+1$.

Now I used Eisensteins theorem to test if it is irreducible. It is not applicable in this form so I substituted $x$ with $x+1$.

$(x+1)^4 -10(x+1)^2 +1$

$x^4+4x^3 -4x^2 - 16x -8$

Here the prime $2$ divides all coefficient but the one of $x^4$ and by Eisensteins Theorem the polynomial is therefore irreducible over $\mathbb{Q}$.

But is this sufficient, I guess that it might still be reducible having coefficients of $\sqrt{2}$ or $\sqrt{3}$?

How can I test it for this numbers or am I on the wrong path and need to test it in another way?

Jyrki Lahtonen
  • 133,153
Dino Rossegger
  • 220
  • Hi! Isn't more common notation for field adjoint $\mathbb{Q}(\sqrt2,\sqrt3)$? And btw, you mess up two things clearly. Your polynomial is the minimal polynomial of $\sqrt2+\sqrt3$ over $\mathbb Q$ -- do you search for a minimal polynomial over the other field? – yo' Jan 13 '14 at 14:29
  • 3
    $x^4 + 4x^2 + 4$ also has the property that the prime $2$ divides all coefficients but the one of $x^4$. But this polynomial is $(x^2+2)^2$ and therefore not irreducible.

    You are not applying Eisenstein's Theorem correctly.

     – Konstantin Ardakov Jan 13 '14 at 14:36
  • 2
    I've corrected your formulas. However, the result (still) does not match the condition for the Schönemann-Eisenstein theorem, because $2^2$ divides the final coefficient $-8$. Nonetheless, if you correctly computed the minimal polynomial it will automatically be irreducible over$~\Bbb Q$. – Marc van Leeuwen Jan 13 '14 at 14:39
  • Yes I mixed things up here. I am looking for the minimal polynomial over $\mathbb{Q}$. – Dino Rossegger Jan 13 '14 at 14:56
  • 1
    The zeros of your polynomial are $x_1=\sqrt2+\sqrt3$, $x_2=\sqrt2-\sqrt3$, $x_3=-\sqrt2+\sqrt3$ and $x_4=-\sqrt2-\sqrt3$. To prove that it is irreducible over $\Bbb{Q}$ it suffices to check that A) none of the zeros is rational (i.e. the polynomial has no linear factors), B) none of the polynomials $(x-x_i)(x-x_j), i\neq j$, has rational coefficients (i.e. no quadratic factors). These are all relatively straightforward I think. – Jyrki Lahtonen Jan 13 '14 at 14:59

1 Answers1

1

A standard approach consists in going through the following steps. Set $E = \mathbb{Q}(\sqrt{2}, \sqrt{3})$.

  1. Prove that $\sqrt{2} \in E$, so $K = \mathbb{Q}(\sqrt{2}) \subseteq E$.
  2. Prove that $\sqrt{3} \in E$.
  3. Prove that $\lvert K : \mathbb{Q} \rvert = 2$.
  4. Prove that $\lvert E : K \rvert = \lvert K(\sqrt{3}) : K \rvert = 2$.
  5. Prove that $\lvert E : \mathbb{Q} \rvert = 4$.
  6. Conclude that the minimal polynomial of $\sqrt{2} + \sqrt{3}$ over $\mathbb{Q}$ has degree $4$.
Andreas Caranti
  • 68,873