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

  • Thread starter Thread starter Kiwi1
  • Start date Start date
  • Tags Tags
    Extension
Kiwi1
Messages
106
Reaction score
0
I am happy with my solutions of questions 1-4 below, but need some help on question 5.

View attachment 5165

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

Attachments

  • chap27_F.jpg
    chap27_F.jpg
    87.4 KB · Views: 113
Physics news on Phys.org
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.
 
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 \(z \notin \mathbb{R}\).
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 5 ·
Replies
5
Views
802
Replies
21
Views
1K
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 24 ·
Replies
24
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K