Field of Quotients of Integral Subdomain in Complex Num

  • Thread starter Thread starter ehrenfest
  • Start date Start date
  • Tags Tags
    Field
ehrenfest
Messages
2,001
Reaction score
1

Homework Statement


Describe the field of quotients of the integral subdomain D = {n+mi|n,m in Z} of the field of complex numbers. "Describe" means give the elements of C that make the field of quotients of D in C.

Homework Equations


The Attempt at a Solution


So any complex number that has the form (nn'+mm'+i(nm'+mn'))/(n'^2+m'^2) will be in the field...but how can I be more descriptive...
 
Last edited:
Physics news on Phys.org
Do you mean a set generators of D?
 
NateTG said:
Do you mean a set generators of D?

I stated the problem the way it is stated in the book, but I guess a set of generators for Quot(D) would work.
 
The "field of quotients" of the sat {m + ni} where m and n are integers (the "Gaussian integers) is, by definition, the set of things of the form (m+ ni)/(a+ bi) where both a and b are also integers. Multiplying numerator and denominator of the fraction by a- bi will make the denominator an integer and give us something of the form (x/p)+ (y/p)i. Looks to me like the field of integers is the set of numbers of the form r+ si where r and s are rational numbers.
 
That seems reasonable but I still need to prove that
\frac{nn'+mm'+i(nm'+mn')}{n'^2+m'^2}
hits every number of the form r+si, where r and s are rational numbers...
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Replies
3
Views
502
Replies
2
Views
7K
Replies
1
Views
2K
Replies
11
Views
2K
Replies
5
Views
618
Back
Top