Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Quadratic reciprocity
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
===Gaussian integers=== In his second monograph on [[quartic reciprocity]]<ref>Gauss, BQ Β§ 60</ref> Gauss stated quadratic reciprocity for the ring <math>\Z[i]</math> of [[Gaussian integer]]s, saying that it is a corollary of the [[quartic reciprocity|biquadratic law]] in <math>\Z[i],</math> but did not provide a proof of either theorem. [[Peter Gustav Lejeune Dirichlet|Dirichlet]]<ref>Dirichlet's proof is in Lemmermeyer, Prop. 5.1 p.154, and Ireland & Rosen, ex. 26 p. 64</ref> showed that the law in <math>\Z[i]</math> can be deduced from the law for <math>\Z</math> without using quartic reciprocity. For an odd Gaussian prime <math>\pi</math> and a Gaussian integer <math>\alpha</math> relatively prime to <math>\pi,</math> define the quadratic character for <math>\Z[i]</math> by: :<math>\left[\frac{\alpha}{\pi}\right]_2 \equiv \alpha^\frac{\mathrm{N} \pi - 1}{2}\bmod{\pi} = \begin{cases} 1 & \exists \eta \in \Z[i]: \alpha \equiv \eta^2 \bmod{\pi} \\ -1 & \text{otherwise} \end{cases} </math> Let <math>\lambda = a + b i, \mu = c + d i</math> be distinct Gaussian primes where ''a'' and ''c'' are odd and ''b'' and ''d'' are even. Then<ref>Lemmermeyer, Prop. 5.1, p. 154</ref> :<math> \left [\frac{\lambda}{\mu}\right ]_2 = \left [\frac{\mu}{\lambda}\right ]_2, \qquad \left [\frac{i}{\lambda}\right ]_2 =(-1)^\frac{b}{2}, \qquad \left [\frac{1+i}{\lambda}\right ]_2 =\left(\frac{2}{a+b}\right).</math>
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)