# Generators for Galois Extension

Original Question is Supposing that $F$ is a field of characteristic $\neq2$ and $K/F$ a Galois extension with $Gal(K/F)$ isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$ to prove that $K=F(\sqrt{D_1},\sqrt{D_2})$ for some $D_1,D_2\in F$ with the property that, none of $D_1,D_2$ or $D_1D_2$ is a square in $F$.

It seem to be very obvious, but i am not able to write in detail why is this so...

As Galois group $\mathbb{Z}_2 \times \mathbb{Z}_2$ has an element of order 2, corresponding irreducible polynomial(whose roots are domain element and image of that element) would be of order 2, say $ax^2+bx+c$, as we have assumed it is irreducible, $b^2-4ac$ is not a square. For getting splitting field corresponding to this polynomial we adjoin $\sqrt{b^2-4ac}$ we let $D_1=b^2-4ac$.

For another element of galois group with order 2, for similar reasons as above, i have a polynomial $px^2+qx+r$ and we adjoin $\sqrt{q^2-4pr}$ to get splitting field and we set $D_2=q^2-4pr$.

As third element would just be composition of two other elements(non identity elements), we do not have to bother about that element.

By this I conclude that $K=F(\sqrt{D_1},\sqrt{D_2})$.

I would like to know is my justification sufficient to claim given result.

I would like to look for a slight generalization, if i have a Galois group (of small order for time being) can i expect to list down the generators of corresponding Galois extension.

Any help/suggestion would be appreciated.

Thank You.

P.S: I am very thankful to the Team who started Mathematics Stack exchange and for other users who were ready to help me with my doubts. I can see drastic change in my attitude to solve problems with in less than 20 days after joining in this forum.

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To be explicit: you are associating to elements $g\in G$ the minimal polynomial of some $\alpha\in K$ such that $F(\alpha)=K^{\langle g\rangle}$ (the latter being the fixed field of $g$ via the Galois correspondence). You could be much more explicit about this. You have $F(\sqrt{D_1},\sqrt{D_2})\subseteq K$; equality can be obtained by comparing dimensions over $F$. Finally one can also prove $D_1,D_2,D_1D_2$ are not squares in $F$ by obtaining a contradiction on the dimension over $F$ assuming the contrary. The reason the characteristic cannot be $2$ is so that the quadratic formula applies. –  anon Aug 12 '13 at 10:43
@anon Or because $\sqrt{D_1D_2}$ is the fix field of $(1,1)$, and so non-trivial over $F$. –  Alex Youcis Aug 12 '13 at 10:46
@anon : yes, initially i have not written F is a field of characteristic $\neq$ 2 but, at the time of writing quadratic formula i realized that i have to add the condition of char$\neq$ 2. :) –  Praphulla Koushik Aug 12 '13 at 10:49
I did not get answer for my question yet :( –  Praphulla Koushik Aug 12 '13 at 13:48