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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 :

  1. Is this correct ?

  2. The addition and multiplication of the two , doesn't one of them supposed to give "0" zero ?

  3. Why is it that $t^4 = t$ , always ?

Thanks !

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    You are asked to figure out the set of four combinations of $(a,b,c,d)$ such that the resulting set of elements of K is a subfield of $K$. Any subfield will contain the prime field, in this case $\mathbf{Z}_2$, so $0=0+0x+0x^2+0x^3$ and $1=1+0x+0x^2+0x^3$ are given. What about the other two? You need a little bit of theory. Have you constructed the field of 4 elements as an example in class?2012-03-16
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    No ,the lecturer just gave the rules and some theory regarding finite fields , but did not give any examples . I know how to construct fields , let's say 'build a field of order 9 , or of order 27' .2012-03-16
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    The four roots of $t^4 - t$ in $K$ will indeed be a field of 4 elements, (this is a fact you *know*, right?) but you still need to find the roots.... Based on the way you asked the question, I feel like you haven't really wrapped your head around the question: it would be worth thinking over some more what you're given and what you're being asked to do. e.g. what did you have in mind by "using" or "working with" a polynomial? Try digesting Jyrki's comment and how it relates to the problem)2012-03-16
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    @ron, ok. So what is your construction of the field of 4 elements? You adjoin an element satisfying a certain polynomial equation to the field $\mathbf{Z}_2$, right? Can you find such an element within this field? It is a lot of work to try them out in sequence, I know. What else do you know? Did you learn that the multiplicative group of a finite field is cyclic? It is possible to take advantage of that here!2012-03-16
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    I did. So if I take $t=a + bx + cx^2 + dx^3$ and place that in $t^4-t$ , I'd get a polynom that should be cyclic ?2012-03-16
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    Could it be that you're confusing the degree of a polynomial, the order of a group or field, and the order of an element? You write "some other polynomial of order $4$" and then give $t^4-t$ as an example. That's a polynomial of *degree* $4$. The (multiplicative) *order* of an element $a$ is the least integer $n$ such that $a^n=1$, and the order of a group or field is the number of its elements.2012-03-16

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