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

2. 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.

3. 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

4. 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