We're trying to understand the field \frac{\mathbb{Z}_{2}

hunterofdeath63

hunterofdeath63

Answered question

2021-12-12

We're trying to understand the field Z2[x]x3+x2+1, and in particular why it has eight elements. I know that x3+x2+1 is irreducible in Z2 and so factor is going to be a field.
The issue I'm having is that the notes claim that (x+I) has an inverse in the field, where I=x3+x2+1. They write
(x+I)(x2+x+I)=x3+x2+I=1+I
I don't understand the last line, I know that multiplication of cosets is defined that way, I just don't see why we get 1 at the end. I think it's possible I didn't fully understand cosets and ideals but I can't fully articulate what the issue is. Any help towards filling in the gaps of my understanding would be greatly appreciated.

Answer & Explanation

levurdondishav4

levurdondishav4

Beginner2021-12-13Added 38 answers

Step 1
The might know that
a+I=b+IabI
So
(x3+x2)1=(x3+x2+1)I
So
(x3+x2)+I=1+I
also any element in Z2[x](x3+x2+1) is of the form (ax2+bx+c)+I. There are 2 choices for each a, b, c. So there are 8 distinct elements in Z2[x] by a cubic polynomial, you get a remainder which is a quadratic polynomial of the form ax2+bx+c
So if
p(x)Z2[x]] then p(x)=q(x)(x3+x2+1)+r(x)
where r(x)=ax2+bx+c. For some a, b, cZ2
So
p(x)+I=ax2+bx+c+I
as
q(x)(x3+x2+1)x3+x2+1

Thomas Lynn

Thomas Lynn

Beginner2021-12-14Added 28 answers

This is because
x3+x2+1I
by definition and therefore
(x3+x2+1)+I=I
meaning that,
(x3+x2)+I=(x3+x2)+(x3+x2+1)+I=1+I
because the characteristic is 2.

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