What is the Simplest Form of a Field Extension of the Reals?

[Kiwi1]

I am happy with my solutions of questions 1-4 below, but need some help on question 5.

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5. The squares in the reals are simply the positive reals and the non-squares are the negative reals so the quotient of two no squares is the quotient of two negative reals that is a positive real of in fact a square in the reals. - QED.

I solved part 4 by arguing that any extension is of the form \(F(\sqrt{a})\) by question 1 and any two such extensions are isomorphic by question 3.

Now to the main part of question 5.
If I can argue that any simple extension of the reals has the form \(\mathbb{R}(a+ib)\) where \(a\in \mathbb{R}, b\in \mathbb{R}^\times\) then I can easily argue that any two such extensions are isomorphic.

Certainly, extensions like \(\mathbb{R}(a+ib)\) exist because the reals are a sub-field of the complex numbers but how do I exclude the possibility of some obscure element being appended in some field bigger than the complex numbers? This seems a bit like saying the only way to create a simple field extension is to use \(F(z) \cong F[x]/p(x)\) that certainly is one way, but I can't be sure it is the only way.

I am also worried that I have not used the argument about a/b being a square and it seems that the question want's me to use that information?

If I am going to use an argument around squares it seems like I would be better showing that \(\mathbb{R}(a+ib)\) contains a-ib and calculating (a+ib)(a-ib) rather than a square? Perhaps I create two cosets by setting \(H=\{a+ib:b\in \mathbb{R}^+\}\), this doesn't work because H is not a group.

Can someone please give me some guidance? In particular will I need to follow through all of the steps laid our by questions 1-4 or do I simply show that two extensions are isomorphic?

Note: \(\mathbb{C} \cong \mathbb{R}(i)\)

 

[Deveno]

It seems something is missing, here-are you allowed to use the Fundamental Theorem of Algebra, or its real corollary: any real polynomial factors into quadratic or linear factors?

In that case, I believe you are supposed to show that any simple extension of $\Bbb R$ is of the form $\Bbb R(\sqrt{r})$, where $r < 0$. Refer to part 2.

 

[Kiwi1]

Thanks Deveno.

The Fundamental theorem of algebra and its corollary were both introduced in this chapter of my book. So yes I can use them.

I haven't worked through the detail yet, but I expect it will be straightforward now that I know:

\(\mathbb{R}(z) \cong \mathbb{R}/p(x)\) where p(x) is (quadratic with characteristic not equal to 2) if \(z \notin \mathbb{R}\).