The polynomial $x^4 + x +1$ is unsplittable under $\mathbb{Z}_2$ .
Given the following $K$:
$K = \mathbb{Z}_2[x] / \mathbb{Z}_2[x] (x^4 + x +1)= {{a+bx+ cx^2 + dx^3 : a,b,c,d \in \mathbb{Z}_2}} $
I'm requested to find $4$ elements of sub-field of order $4$ of $K$.
The given $K$ is a field of order $16$. If $x^4 + x +1$ is unsplittable then I can't use it, I think (!??).
If so, do I need to find some other polynomial of order 4 and try to work with it ?
Something like $t^4 - t$ maybe ?
Regards
EDIT:
Sorry for the delay , I've some problems regarding the topic of finite fields . First , we work with the field of $Z_{2}$ , and we know that $t^4 = t$ , and also that :
$t^4 - t = t(t-1)(t^2 +t +1)$
As we can see , $t=0$ and $t=1$ are in the field , and now we're missing two more .
Now , assume that $t = a+bx+cx^2+dx^3$ , and place it in $t^2 +t +1$ , then : $(a+bx+cx^2+dx^3 )^2+a+bx+cx^2+dx^3+1 = a^2+b^2 x^2+c^2 x^4+d^2 x^6+(a+bx+cx^2+dx^3)==Break=$
Since $x^4=-x-1=x+1 $ and $x^6=x^2⋅x^4=x^2(1+x)=x^2+x^3 $
$==Continues=a+bx^2+cx^4+dx^6+a+bx+cx^2+dx^3+1=a+bx^2+c(x+1)+d(x^2+x^3 )+a+bx+cx^2+dx^3+1=$
{After a lot of arithmetics} $= x(b+c+d)+x^2 (b+c)+(c+d+1)⋅1$
Finally , if b=c=1 and d=0 , we'd get that :
$b+c+d=1+1+0=0$
$b+c=1+1=0$
$c+d+1=1+0+1=2=0$
Therefore the other two elements of this lovely field are : $1+x+x^2 ;x^2+x$
Since they are both closed for addition and multiplication :
Addition : $ (x^2+x+1)+(x^2+x)=2x^2+2x+1=0+0+1=1$
Multiplication : $(x^2+x+1)(x^2+x)=x^4+x^3+x^3+x^2+x^2+x=x^4+2x^3+2x^2+1=x^4+0+0+x={x^4=x+1}=x+1+x=2x+1=0+1=1$
Now my questions are :
Is this correct ?
The addition and multiplication of the two , doesn't one of them supposed to give "0" zero ?
Why is it that $t^4 = t$ , always ?
Thanks !