3
$\begingroup$

We know that the extension $\mathbb{Q}(\sqrt{2\sqrt{2}-1}$) over $\mathbb{Q}$ has degree 4 by considering the minimal polynomial mod 3. Now I want to show that $-14$ isn't a square in this field. How do I do this in the easiest way? Please also let me know if you have a method not in this way (e.g. by considering intermediate field extensions in different orders).

Somehow feels like it ought to be true because $-14$ has a factor $7$ in it whereas $\mathbb{Q}(\sqrt{2\sqrt{2}-1}$) involves only $2$'s.

If it helps, I found that $\mathbb{Q}(\sqrt{-14},\sqrt{-2\sqrt{2}-1})$ is the splitting field of $X^4+2X^2-7$ over $\mathbb{Q}$.

$\endgroup$

2 Answers 2

1
$\begingroup$

There are many ways to see this. The easiest is to realise that $K= \mathbb Q(\sqrt{2\sqrt2-1})\subset \mathbb R$, so it cannot possibly contain a root of $-14$.

Alternatively, you could show that $K$ is not a Galois extension, so it cannot be the splitting field of an (irreducible) polynomial.

$\endgroup$
5
  • $\begingroup$ ahhhhh...how could I have been so dumb! But I'm slightly uncomfortable with you saying that K is inside R. I guess you mean that if we took an embedding of K into R (which exists and must fix Q and so -14) then we would get a contradiction, right? $\endgroup$
    – maths
    Apr 14, 2016 at 11:57
  • 1
    $\begingroup$ Usually, when you write $K$ in that form you are picking an embedding. The ambiguity occurs precisely because the extension is not Galois - in a Galois extension, all embeddings have the same image in $\mathbb C$, since an embedding restricts to an automorphism of $K$. Either way, all the embeddings are isomorphic as abstract fields, so if a polynomial has a root in one, it will have a root in all of them. $\endgroup$
    – Mathmo123
    Apr 14, 2016 at 12:05
  • $\begingroup$ btw I tried to edit your answer to be "so it cannot be the splitting field of a polynomial". The "irreducible" isn't important here right? But my edit has been rebuked. $\endgroup$
    – maths
    Apr 14, 2016 at 12:15
  • 1
    $\begingroup$ I think irreducibility matters just so that you know you've picked the right polynomial, but I agree it's not strictly necessary. $\endgroup$
    – Mathmo123
    Apr 14, 2016 at 12:24
  • $\begingroup$ Feel free to chat with me here. $\endgroup$
    – Mathmo123
    Apr 15, 2016 at 12:12
1
$\begingroup$

The splitting field of $X^4+2x^2-7$ over $\mathbb{Q}$ is what you have found. This polynomial is separable, hence the extension is Galois.Furthermore,this polynomial is irreducible. I used corollary 4.5 of this wonderful lecture by Keith Conrad : http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/cubicquartic.pdf

We can see, by what it says, that $-7 \neq 0$ and $(2^2+4*7)(-7) = -224 \neq 0$, hence the Galois group is $D_4$, wherein the extension is $8$.

For proofs of the facts, you can look at the same paper.

$\endgroup$

You must log in to answer this question.

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