• Mar 26th 2009, 06:53 AM
classic_phohe
There are 8 choices for the norm $\displaystyle \nu$ that make:

$\displaystyle r(y) = y^2 + y + \nu$ irreducible over $\displaystyle GF(2^4)$

why is this so??

this simple question (but hard for me) has been troubling me for weeks and no one can answer me why.

thank you very much
• Mar 26th 2009, 07:33 AM
NonCommAlg
Quote:

Originally Posted by classic_phohe
There are 8 choices for the norm $\displaystyle \nu$ that make:

$\displaystyle r(y) = y^2 + y + \nu$ irreducible over $\displaystyle GF(2^4)$

why is this so??

this simple question (but hard for me) has been troubling me for weeks and no one can answer me why.

thank you very much

well, $\displaystyle y^2+y+ \nu$ has degree 2. so it's irreducible iff it has no root in $\displaystyle \text{GF}(16).$ thus the polynomial is irreducible for all odd values of $\displaystyle \nu$ in $\displaystyle \text{GF}(16)$ because $\displaystyle y^2+y$ is always even. so the only thing

you need to prove is that the polynomial has a root for 8 remaining even values of $\displaystyle \nu.$ this is clear because in $\displaystyle \text{GF}(16): \ \{y^2+y: \ 0 \leq y \leq 7 \}=\{0,2,4,6,8,10,12,14 \}.$

A Much Better Question: given a prime number $\displaystyle p$ and $\displaystyle n \in \mathbb{N},$ find the number of elements of $\displaystyle \nu$ in $\displaystyle \text{GF}(p^n)$ for which the polynomial $\displaystyle y^2+y+\nu$ is irreducible.
• Mar 26th 2009, 08:27 PM
classic_phohe
Quote:

Originally Posted by NonCommAlg
well, $\displaystyle y^2+y+ \nu$ has degree 2. so it's irreducible iff it has no root in $\displaystyle \text{GF}(16).$ thus the polynomial is irreducible for all odd values of $\displaystyle \nu$ in $\displaystyle \text{GF}(16)$ because $\displaystyle y^2+y$ is always even. so the only thing

you need to prove is that the polynomial has a root for 8 remaining even values of $\displaystyle \nu.$ this is clear because in $\displaystyle \text{GF}(16): \ \{y^2+y: \ 0 \leq y \leq 7 \}=\{0,2,4,6,8,10,12,14 \}.$

A Much Better Question: given a prime number $\displaystyle p$ and $\displaystyle n \in \mathbb{N},$ find the number of elements of $\displaystyle \nu$ in $\displaystyle \text{GF}(p^n)$ for which the polynomial $\displaystyle y^2+y+\nu$ is irreducible.

oh my!! thank you very much for saving me! im very new in this finite field and i dont have enough time and resource to learn the the fundamental of it. this question has been bugging me for weeks and no one could actually help me!

now i have this question:

let say the $\displaystyle GF((2^2)^2)$ is from $\displaystyle GF(2^2)$ of irreducible polynomial $\displaystyle x^2+x+N$ then how should i determine what is the 8 values of $\displaystyle \nu$ ?

and if the entire thing is in normal basis, again how can it done?

sorry for my inappropriate question statement
• Mar 26th 2009, 11:26 PM
classic_phohe
also, from the text that it says the 8 possible $\displaystyle \nu$ are

1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111

i understand the your explanation that to make the polynomial irreducible, $\displaystyle \nu$ has to be odd value of $\displaystyle GF(16)$ but i dont get it how the odd values are represented as 1000 to 1111