5
$\begingroup$

The question is to find the minimal polynomial of $\sqrt{2}+\sqrt{7}$ over $\mathbb{Q}(\sqrt{5})$.

First, I found its minimal polynomial over $\mathbb{Q}$ which is equal to $X^4 - 18X^2 + 25$. I suppose this could already be a candidate for a minimal polynomial over $\mathbb{Q}(\sqrt{5})$ so I tried proving that using the tower property, but I don't think that's the right approach.

$\endgroup$
1
  • $\begingroup$ That would be my candidate as well. Rather than "using the tower property", it is probably necessary to show the quartic is irreducible over $\mathbb{Q}[\sqrt 5]$ by examination of factors, e.g. over $\mathbb{C}$. $\endgroup$
    – hardmath
    Oct 29, 2016 at 19:51

1 Answer 1

Reset to default
4
$\begingroup$

Hint:

First note that the minimal polynomial must divide $x^4-18x^2+25$ and using the tower law we can say that if this is not the minimal polynomial then it has to be quadratic.

If you know what the other roots of $x^4-18x^2+25$ are then you can check all 6 pairs (in fact you only need 3 of them), to see if they have coefficients in $\mathbb{Q}(\sqrt{5})$ or not.

There is a quicker method using the Galois group, but I'm guessing you haven't met this yet?

$\endgroup$
7
  • $\begingroup$ What is the quicker method using the Galois group? The only method I can think of using Galois theory is to prove that $\mathbb{Q}(\sqrt{2}, \sqrt{7}, \sqrt{5})$ is a degree $4$ Galois extension of $\mathbb{Q}(\sqrt{5})$, write down the obvious automorphisms, then argue that $\sqrt{2}+\sqrt{7}$ has three additional distinct Galois conjugates. But I can't think of a nice or short proof that $[\mathbb{Q}(\sqrt{2}, \sqrt{7}, \sqrt{5}): \mathbb{Q}(\sqrt{5})] =4$; would be interested what you had in mind. $\endgroup$
    – Alex Wertheim
    Oct 29, 2016 at 19:58
  • $\begingroup$ @AlexWertheim: With Galois theory in place you can say that the intermediate fields $M$ between $\Bbb{Q}\subset M\subset K=\Bbb{Q}(\sqrt2,\sqrt7) $ are the fields $\Bbb{Q}(\sqrt n)$ with $n\in 2,7,14$. Those are the fixed fields of the non-trivial automorphisms. Anyway, $\sqrt5$ is not in any of them, so $[K(\sqrt5):K]=2$. Hence $[K(\sqrt5):\Bbb{Q}]=8$ et cetera. This generalizes, but do check out Bill Dubuque's answer to that question also. $\endgroup$
    – Jyrki Lahtonen
    Oct 29, 2016 at 21:30
  • $\begingroup$ @JyrkiLahtonen: very nice argument, thanks! $\endgroup$
    – Alex Wertheim
    Oct 29, 2016 at 22:10
  • $\begingroup$ I was thinking of taking $K$ to be the splitting field over $\mathbb{Q}(\sqrt{5})$ and showing that the nontrivial automorphism of $\mathbb{Q}(\sqrt{5})$ fixes $\sqrt{2}+\sqrt{7}$ with some argument about quadratic fields. But we have at least an order 4 subgroup coming from the extension to $K$ so combine this to get a Galois group of order at least 8. $\endgroup$
    – Matt B
    Oct 29, 2016 at 22:27
  • $\begingroup$ I'm probably implicitly using that this Galois group must be abelian but this follows as the group has exponent 2. $\endgroup$
    – Matt B
    Oct 29, 2016 at 22:28

You must log in to answer this question.

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