Find a fourth degree polynomial in Z2[x] whose roots are the four elements of the field Z2[x]/(x^2+x+1) (given to be [0] [1] [x] [x+1])

Question

anonymous · Answer

note that by Z2[x] I mean |dw:1396236471848:dw|

anonymous · Answer

i think this might be easier than it looks

anonymous · Answer

Well I think this has to do with something called an extension field, which the textbook did not go over before this question appeared (its in the next section). Also I do not seem to understand it cause I keep getting what appears to be an incorrect answer

anonymous · Answer

if i am reading it correctly, you only have 3 elements of $\mathbb{Z}_2[x]/x^2+x+1$ namely $\{0,a,1,a+1\}$ the equivalence classes  or in the square bracket notation 
$$\{[0],[1],[x],[x+1]\}$$

anonymous · Answer

make that 3 non zero elements!

anonymous · Answer

these elements form a group under multiplication, and since the order of the group is 3, then each has order 3
that means in my notation $a^3=1$ for all or $a^3-1=0$  so your polynomial is $x^3-1$
in order to add the zero element, the polynomial will be $x^4-x$

anonymous · Answer

so can you take me through the process of getting 0 by plugging in [x] then? I keep getting that it is [x+1] which is not 0

anonymous · Answer

the equivalence class that i wrote as $a$ and you have written as $[x]$ is the root of that polynomial $x^2+x+1$

anonymous · Answer

the other root is $[x+1]$ and that is an easy check, because 
$$f(a+1))=(a+1)^2+a+1+1=a^2+2a+1+a+1+1=a^2+a+1=0$$

anonymous · Answer

ooops that kind of ran over 
$$f(a+1)=a^2+2a+1+a+1+1$$
$$=a^2+a+1=0$$ because $a$ is a root of $x^2+x+1$ and $2, 2a=0$

anonymous · Answer

yes but when I plug it into x^4 + x in I just get [x^4 + x} which (if I did my division right...is equivalent to [x+1]

anonymous · Answer

I must be doing my division wrong trying to get the equivalent class....

anonymous · Answer

maybe not exactly sure

anonymous · Answer

did you do an example in class where you take 
$$\mathbb{R}/x^2+1$$?

anonymous · Answer

you are introducing other zeros in to $\mathbb{R}$ namely $i$ and $-i$
this is the same kind of idea i think

anonymous · Answer

Sounds familar, alright thank you

anonymous · Answer

yw