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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.

user159517
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  • 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 $\mathbb{C}$. – hardmath Oct 29 '16 at 19:51

1 Answers1

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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?

Matt B
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  • 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. – Alex Wertheim Oct 29 '16 at 19:58
  • @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. – Jyrki Lahtonen Oct 29 '16 at 21:30
  • @JyrkiLahtonen: very nice argument, thanks! – Alex Wertheim Oct 29 '16 at 22:10
  • 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. – Matt B Oct 29 '16 at 22:27
  • I'm probably implicitly using that this Galois group must be abelian but this follows as the group has exponent 2. – Matt B Oct 29 '16 at 22:28
  • I thought of examining the roots, but was hoping that there is a quicker method (we havent covered galois theory yet) as this question was on a test and worth around 1/5 of a point. – user159517 Oct 30 '16 at 13:07
  • @JyrkiLahtonen: now that I know Galois theory, I revisited this question. It seems to me that in order to know the fixed fields of the automorphisms in the Galois group, you'd have to calculate the roots of the polynomial and the whole Galois group, or am I missing something? – user159517 Jan 11 '17 at 18:30