1. ## Normal Closure....

Hi guys,

How do you find the normal closure Q(Sqrt(1+Sqrt(5))) over Q?

I have absolutely no idea how to do this! :-s I think it has something to do with chains of subfields & conjugate, but if some could please explain the general method & how to apply it to this question I would be extremely greatful! :-)

2. Originally Posted by AAM
Hi guys,

How do you find the normal closure Q(Sqrt(1+Sqrt(5))) over Q?

I have absolutely no idea how to do this! :-s I think it has something to do with chains of subfields & conjugate, but if some could please explain the general method & how to apply it to this question I would be extremely greatful! :-)

The normal closure of $\mathbb{Q}(\sqrt{1+\sqrt{5}})$ is the splitting field of $\text{min}(\sqrt{1+\sqrt{5}},\mathbb{Q})$ over $\mathbb{Q}$?

Is that enough or you have any other questions on why that is so?

3. Ok, so does that means the normal closure is:

Q(Sqrt1+Sqrt(5)),Sqrt(1-Sqrt(5))) then?

& could you please exaplin why this is the case? The definition for normal closure that I have is that it is the smallest extension of the current extension which is normal.

Many thanks. :-)

4. Originally Posted by AAM
Ok, so does that means the normal closure is:

Q(Sqrt1+Sqrt(5)),Sqrt(1-Sqrt(5))) then?

& could you please exaplin why this is the case? The definition for normal closure that I have is that it is the smallest extension of the current extension which is normal.

Many thanks. :-)
Let $x=\sqrt{1+\sqrt{5}}\implies x^2 - 1 = \sqrt{5} \implies x^4 - 2x^2 - 4 = 0$. This polynomial is also irreducible.

The zeros of this polynomial are: $x = \pm \sqrt{\sqrt{5}+1},\pm i\sqrt{\sqrt{5}-1}$.
Therefore, the normal closure is: $N = \mathbb{Q}\left( \sqrt{\sqrt{5}+1}, i\sqrt{\sqrt{5}-1} \right)$

Notice that $\left( \sqrt{\sqrt{5} + 1} \right) \left( i \sqrt{\sqrt{5}-1} \right) = i\sqrt{4} = 2i$.

Thus, it is also okay to write: $N = \mathbb{Q}\left( \sqrt{\sqrt{5}+1}, i , \sqrt{\sqrt{5}-1}\right)$.
-----

Here is an explanation of why we take the normal closure to be splitting field. First we define what we mean by "normal closure". Let $K/F$ be an algebraic field extension, the normal closure of $K$ over $F$ is the splitting field of $\{ \min(a,F) | a\in K \}$ over $F$. The idea is that the normal closure is the "smallest" normal extension over $F$ containing $K$. By "smallest" we mean that if $E$ is a normal extension over $F$ which satisfies $K\subseteq E\subseteq N$ then $E=N$. To prove this let $a\in K$ and consider $\min (a,K)$, since $E/K$ is normal it immediately means that $\min (a,K)$ splits over $E$ and so $E$ contains all the roots of $\min (a,K)$. But $N$ is the field generated by the roots of this polynomials, hence, $N\subseteq E$ which forces $N=E$. As a consequence we can prove that the normal closure of $K=F(a_1,...,a_n)$ is the splitting field over $\min(a_j,F)$ for each $j=1,2,...,n$. Let $E$ be this splitting field, then obviously $K\subseteq E\subseteq N$ with $E/K$ normal, therefore it forces $E=N$. Your problem is a special case when $K = F(a)$, so the normal closure is the splitting field of $\min (a,F)$ over $K$.

5. Fantastic! :-D

Thank you SO much theperfecthacker! :-)

In an question online I've seen a chain of fields like:

Q SUBFIELD OF Q(Sqrt(3)) SUBFIELD OF Q(Sqrt(1+2Sqrt(3)),Sqrt(1-2Sqrt(3))).

Is this true in general?

ie -

Q SUBFIELD OF Q(Sqrt(A)) SUBFIELD OF Q(Sqrt(a+bSqrt(A)), Sqrt(a-bSqrt(A))) ?

If so, this means that

Q(Sqrt(A))(Sqrt(a+bSqrt(A)),Sqrt(a-bSqrt(A)
= Q(Sqrt(a+bSqrt(A)),Sqrt(a-bSqrt(A)) , yes?

Furthermore, is Q(Sqrt(A)) the ONLY quadratic subfield of Q(Sqrt(a+bSqrt(A)),Sqrt(a-bSqrt(A))) ? If not, what are the others? Indeed, what are all the subfields of Q(Sqrt(a+bSqrt(A)),Sqrt(a-bSqrt(A))) ?

Many thanks. :-)