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
Field trace
(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!
===Application=== A [[quadratic equation]], {{nowrap|1=''ax''{{i sup|2}} + ''bx'' + ''c'' = 0}} with ''a'' β 0, and coefficients in the finite field <math>\operatorname{GF}(q) = \mathbb{F}_q</math> has either 0, 1 or 2 roots in GF(''q'') (and two roots, counted with multiplicity, in the quadratic extension GF(''q''<sup>2</sup>)). If the [[characteristic (algebra)|characteristic]] of GF(''q'') is [[parity (mathematics)|odd]], the [[discriminant]] {{nowrap|1=Ξ = ''b''<sup>2</sup> β 4''ac''}} indicates the number of roots in GF(''q'') and the classical [[quadratic formula]] gives the roots. However, when GF(''q'') has [[parity (mathematics)|even]] characteristic (i.e., {{nowrap|1=''q'' = 2<sup>''h''</sup>}} for some positive [[integer]] ''h''), these formulas are no longer applicable. Consider the quadratic equation {{nowrap|1=''ax''{{i sup|2}} + ''bx'' + c = 0}} with coefficients in the finite field GF(2<sup>''h''</sup>).<ref>{{harvnb|Hirschfeld|1979|loc=pp. 3-4}}</ref> If ''b'' = 0 then this equation has the unique solution <math>x = \sqrt{\frac{c}{a}}</math> in GF(''q''). If {{nowrap|''b'' β 0}} then the substitution {{nowrap|1=''y'' = ''ax''/''b''}} converts the quadratic equation to the form: :<math>y^2 + y + \delta = 0, \text { where } \delta = \frac{ac}{b^2}.</math> This equation has two solutions in GF(''q'') [[if and only if]] the absolute trace <math>\operatorname{Tr}_{GF(q)/GF(2)}(\delta) = 0.</math> In this case, if ''y'' = ''s'' is one of the solutions, then ''y'' = ''s'' + 1 is the other. Let ''k'' be any element of GF(''q'') with <math>\operatorname{Tr}_{GF(q)/GF(2)}(k) = 1.</math> Then a solution to the equation is given by: :<math> y = s = k \delta^2 + (k + k^2)\delta^4 + \ldots + (k + k^2 + \ldots + k^{2^{h-2}})\delta^{2^{h-1}}.</math> When ''h'' = 2''m''' + 1, a solution is given by the simpler expression: :<math> y = s = \delta + \delta^{2^2} + \delta^{2^4} + \ldots + \delta^{2^{2m}}.</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)