/sci/ - Science & Math

[Anonymous No.3458834]

If <span class="math">K[/spoiler] is an extension field of rational numbers <span class="math">Q[/spoiler] such that <span class="math">[K:Q] = 2[/spoiler], prove K = <span class="math">Q/ ( \sqrt{d})[/spoiler] for some square-free integer d.

Now why would that be? 8 is not square-free, still the polynomial <span class="math">p(x) = x^2 - 8[/spoiler] could be used to create a vector space over Q of dimension two, with basis <span class="math">\{1_Q , \sqrt{8} \}[/spoiler]. Where am I wrong?

[Anonymous No.3458839]

>>3458834
I almost know enough to keep up. Sorry, but I won't be much help.

[Anonymous No.3458929]

halp /sci/!
Pointing out my logical flaw with a single post of a few words will earn me hours of thinking :)

[Anonymous No.3458937]

>>3458834
Same anon as:>>3458839

Ok, work with me. An extension field of a field (F, +, *) is a new field (E, +, *) so that E contains F. Right?

What does
[K:Q]=2
mean?

And remind me what
Q / sqrt(d)
means.

A "square-free" integer is an integer X so that there is no integer Y so that Y*Y = X?

[Anonymous No.3458963]

>Ok, work with me. An extension field of a field (F, +, *) is a new field (E, +, *) so that E contains F. Right?
Yes

>What does
[K:Q]=2
mean?
It means that the vector space is two-dimensional ie we can describe all elements of it by <span class="math">a1_Q + bu[/spoiler] for <span class="math">a,b \in Q[/spoiler] and 1 and u being the basis.

>And remind me what
Q / sqrt(d)
means.
Smallest possible field containing Q and element <span class="math">\sqrt{d}[/spoiler]

>A "square-free" integer is an integer X so that there is no integer Y so that Y*Y = X?
It means you can not divide X by any p^2, p prime.

[Anonymous No.3458970]

>>3458963
Yeah, I'm going to be no help. I'd need half a day or a day to review this shit. Sorry.

[Anonymous No.3459012]

>>3458834
>8 is not square-free
over Z_4 you mean?

[Anonymous No.3459024]

Ok I understood it I think.

Basis {1, sqrt(8)} = {1 , 2*sqrt(2)}. Since K is an extension of rational numbers, we can obviously express any element of K in base {1, sqrt(2)} as well, and 2 is obviously square-free. Then any [K:Q] is indeed isomorphic to some Q/(sqrt(d)) with d square-free.

[Anonymous No.3459039]

>>3459024
Damnit. This is the stupid anon. I see what you're doing now. That makes sense.

[Anonymous No.3459046]

I'm doing physics, but here is an idea:

Maby it means that if you consider the field K = Q/(\sqrt{D}) where coefficients m,n \in Q, and D is NOT square free, like p=8, say

D = p1^2 · p2^2 · d, where d is square free

then any element is

A = m · 1 + n · sqrt(D) = m · 1 + n·p1·p2 · sqrt(d)

where you can now, without restrictions, choose m=M, n=N/(p1·p2), such that

A = N·1 + M · sqrt(d),

i.e. the field is K = Q/(\sqrt{d}), with d square free.

[Anonymous No.3459098]

>>3459024
I think this guy has it right. Any d^0.5 for d that isn't square free can be reduced to the product of some numbers and the roots of some square free numbers.