1. ## Galois theory

Thanks
Edgar

Let p1, p2, p3, …be distinct primes.

1) prove by induction on m that if x є Q and √x є Q(√p1…..√pm ) then
x = p1^α1…..pm^αm y^2 for some integers a1....am and some y є Q.

2) Deduce that for any n √p n+1 (n+1 subscript) is not an element of
Q(√p1…..√pn )

3) hence prove by induction on n that [Q(√p1…..√pn ):Q] = 2^n

2. Originally Posted by edgar davids
3) hence prove by induction on n that [Q(√p1…..√pn ):Q] = 2^n
If we can show $\displaystyle p_n\not \in \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n-1}} )$ it means $\displaystyle x^2 - p_n$ is minimal polynomial for $\displaystyle \sqrt{p_n}$. This means $\displaystyle [\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_{n-1}})][\mathbb{Q}(\sqrt{p_n}):\mathbb{Q}] = 2^{n-1}\cdot 2 = 2^n$. So it remains to prove the first step.

Start with the basic case that $\displaystyle \sqrt{x} \in \mathbb{Q}(\sqrt{p_1})$ it means $\displaystyle \sqrt{x} = a + b\sqrt{p_1}$ for some rational numbers $\displaystyle a,b$. Thus, $\displaystyle x = a^2+2ab\sqrt{p_1}+b^2p_1$, for this expression to hold we need that $\displaystyle a=0$ or $\displaystyle b=0$ because RHS is irrational otherwise so $\displaystyle x=a^2$ or $\displaystyle x=b^2p_1$, which means $\displaystyle x=p_1^{a_1} y^2$.

Now suppose that $\displaystyle \sqrt{x} \in \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_k})$ means $\displaystyle x=p_1^{a_1}...p_k^{a_k}y^2$. Say that $\displaystyle \sqrt{x} \in \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_k})(\sqrt{p_{k+1 }})$. It means $\displaystyle \sqrt{x} = p+q\sqrt{p_{k+1}}$ where $\displaystyle p,q\in \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_k})$. Thus, $\displaystyle x = p^2 + 2pq\sqrt{p_{k+1}} + q^2p_{k+1}$. So that means by induction $\displaystyle p$ or $\displaystyle q$ is zero. Which means $\displaystyle x=p_1^{a_1}...p_{k+1}^{a_{k+1}}y^2$.

3. need to find the glaios group of the extension Q(√p1.....√pn):Q and show that G is isomorphic with C2^n

once this has been done then need to use this Galois correspondence to show that
Q(√p1.....√pn) = Q(√p1 + √p2 + √p3 +....+√pn)
It is easier to explain what is going on in a specific case, say, $\displaystyle \mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\sqrt{p_3})/\mathbb{Q}$ rather then in general $\displaystyle n$, but the pattern is the same in general.

The Galois group is isomorphic to $\displaystyle \mathbb{Z}_2^3$ because if $\displaystyle \sigma_i$ are the automorphisms $\displaystyle \sigma_i(\sqrt{p_i}) = - \sqrt{p_i}$ for $\displaystyle i=1,2,3$ and any composition between them is too, which gives us $\displaystyle 8$ altogether, which is the size of the Galois group, and furthermore each automorphism which generates the Galois group has order two, so it is isomorphic to the group $\displaystyle \mathbb{Z}_2^3$. The picture below shows all those fields $\displaystyle E$ such that $\displaystyle \mathbb{Q}\subseteq E\subseteq \mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\sqrt{p_3})$, note there have to have $\displaystyle 12$ fields $\displaystyle E$ just like the subgroups of $\displaystyle \mathbb{Z}_2^3$ (it happens to be impossible to draw a nice subfield diagram in 2 dimension, that has something to do with non-planar graphs).

Now, of course, $\displaystyle \mathbb{Q}(\sqrt{p_1}+\sqrt{p_2}+\sqrt{p_3})$ is a subfield. But it cannot be among any of those proper intermediate subfields. Thus, it can therefore only be the improper subfield, i.e. equal to $\displaystyle \mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\sqrt{p_3})$ itself.

Note, the same pattern happens when you are using $\displaystyle n$ primes. It is just easier to explain the situation in a specific case. To actually prove that the Galois group for $\displaystyle n$ primes is isomorphic to $\displaystyle \mathbb{Z}_2^n$ is to note that the size of the Galois group has to be the degree over the field over $\displaystyle \mathbb{Q}$ which is $\displaystyle 2^n$. Now we need to find its elements. If $\displaystyle \sigma_i (\sqrt{p_i}) = -\sqrt{p_i}$ are automorphism for $\displaystyle i=1,2,...,n$ then $\displaystyle \left< \sigma_1,...,\sigma_n \right>$ generate a group with $\displaystyle 2^n$ elements. This is of course isomorphic to $\displaystyle \mathbb{Z}_2^n$.

4. Originally Posted by edgar davis
Let p1, p2, p3..... be distinct primes

1) Deduce that for any n, √ p (n+1) is not an element of Q (√ p1......√ pn)

(fyi (n+1) is subscript)

2) Show that the extension Q (√ p1......√ pn): Q is normal.
For (1) note that if $\displaystyle \sqrt{p_{n+1}}\in \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})$ then it would mean $\displaystyle \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n},\sqrt{p_{n+1} }) = \mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})$ and so $\displaystyle [\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n},\sqrt{p_{n+1} }:\mathbb{Q}] = 2^n$. But that is not true because we know this degree extension has to be $\displaystyle 2^{n+1}$.

For (2) you have to use a theorem (or maybe some would call it a definition). Given fields $\displaystyle F/K$. We say $\displaystyle F$ is normal (and finite) over $\displaystyle K$ iff for every irreducible polynomial in $\displaystyle K[x]$ that has a zero in $\displaystyle F$ has all its zeros in $\displaystyle F$. It turns out that $\displaystyle F/K$ is a normal (and finite) seperable extension if and only if $\displaystyle F$ is the splitting field over $\displaystyle K$ of some seperable polynomial. Now let $\displaystyle F=\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})$ and $\displaystyle K=\mathbb{Q}$. Note that $\displaystyle F$ is the splitting field over $\displaystyle (x^2-p_1)...(x^2-p_n)$. Furthermore, every (non-constant) polynomial over $\displaystyle K$ is seperable because $\displaystyle K$ is perfect (because $\displaystyle \text{char}(K) = 0$). Thus, $\displaystyle F/K$ is a splitting field extension of a seperable polynomial so it is in fact a normal extension.