There are 8 choices for the norm that make:

irreducible over

why is this so??

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

please help me with this

thank you very much

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- Mar 26th 2009, 07:53 AMclassic_phoheplease please please help me~~ irreducible polynomial
There are 8 choices for the norm that make:

irreducible over

why is this so??

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

please help me with this

thank you very much - Mar 26th 2009, 08:33 AMNonCommAlg
well, has degree 2. so it's irreducible iff it has no root in thus the polynomial is irreducible for all odd values of in because is always even. so the only thing

you need to prove is that the polynomial has a root for 8 remaining even values of this is clear because in

__A Much Better Question__: given a prime number and find the number of elements of in for which the polynomial is irreducible. - Mar 26th 2009, 09:27 PMclassic_phohe
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 is from of irreducible polynomial then how should i determine what is the 8 values of ?

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

sorry for my inappropriate question statement - Mar 27th 2009, 12:26 AMclassic_phohe
also, from the text that it says the 8 possible are

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

i understand the your explanation that to make the polynomial irreducible, has to be odd value of but i dont get it how the odd values are represented as 1000 to 1111