# gcd in F5

• Jun 16th 2011, 07:52 PM
wik_chick88
gcd in F5
calculate $gcd(x^3 + 2x^2 + 3x - 1, 2x^2 - x - 1)$ in $F_{5}$

does $6x^2 \equiv 0$, because we are working in $F_{5}$?
• Jun 16th 2011, 08:28 PM
TheEmptySet
Re: gcd in F5
Quote:

Originally Posted by wik_chick88
calculate $gcd(x^3 + 2x^2 + 3x - 1, 2x^2 - x - 1)$ in $F_{5}$

does $6x^2 \equiv 0$, because we are working in $F_{5}$?

No, if that were the case it would be the additive identity. Just reduce the coefficient mod 5

$6x^2=1x^2=x^2$
• Jun 16th 2011, 08:47 PM
Also sprach Zarathustra
Re: gcd in F5
Quote:

Originally Posted by wik_chick88
calculate $gcd(x^3 + 2x^2 + 3x - 1, 2x^2 - x - 1)$ in $F_{5}$

does $6x^2 \equiv 0$, because we are working in $F_{5}$?

Roots of: g(x)=2x^2 - x - 1=0 are: 1 and -1/2

Say f(x)=x^3 + 2x^2 + 3x - 1,

f(1)!=f(-1/2)!=0

gcd((f(X),g(x))=1