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

hunterofdeath63 2021-12-12 Answered
We're trying to understand the field \(\displaystyle{\frac{{{\mathbb{{{Z}}}}_{{{2}}}{\left[{x}\right]}}}{{{\left\langle{x}^{{{3}}}+{x}^{{{2}}}+{1}\right\rangle}}}}\), and in particular why it has eight elements. I know that \(\displaystyle{x}^{{{3}}}+{x}^{{{2}}}+{1}\) is irreducible in \(\displaystyle{\mathbb{{{Z}}}}_{{{2}}}\) and so factor is going to be a field.
The issue I'm having is that the notes claim that \(\displaystyle{\left({x}+{I}\right)}\) has an inverse in the field, where \(\displaystyle{I}={\left\langle{x}^{{{3}}}+{x}^{{{2}}}+{1}\right\rangle}\). They write
\(\displaystyle{\left({x}+{I}\right)}{\left({x}^{{{2}}}+{x}+{I}\right)}={x}^{{{3}}}+{x}^{{{2}}}+{I}={1}+{I}\)
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.

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Expert Answer

levurdondishav4
Answered 2021-12-13 Author has 5828 answers
Step 1
The might know that
PSKa+I=b+I\Leftrightarrow a-b\in IZK
So
\(\displaystyle{\left({x}^{{{3}}}+{x}^{{{2}}}\right)}-{1}={\left({x}^{{{3}}}+{x}^{{{2}}}+{1}\right)}\in{I}\)
So
\(\displaystyle{\left({x}^{{{3}}}+{x}^{{{2}}}\right)}+{I}={1}+{I}\)
also any element in \(\displaystyle{\frac{{{\mathbb{{{Z}}}}_{{{2}}}{\left[{x}\right]}}}{{{\left({x}^{{{3}}}+{x}^{{{2}}}+{1}\right)}}}}\) is of the form \(\displaystyle{\left({a}{x}^{{{2}}}+{b}{x}+{c}\right)}+{I}\). There are 2 choices for each a, b, c. So there are 8 distinct elements in \(\displaystyle{\mathbb{{{Z}}}}_{{{2}}}{\left[{x}\right]}\) by a cubic polynomial, you get a remainder which is a quadratic polynomial of the form \(\displaystyle{a}{x}^{{{2}}}+{b}{x}+{c}\)
So if
\(\displaystyle{p}{\left({x}\right)}\in{\mathbb{{{Z}}}}_{{{2}}}{\left[{x}\right]}{]}\) then \(\displaystyle{p}{\left({x}\right)}={q}{\left({x}\right)}{\left({x}^{{{3}}}+{x}^{{{2}}}+{1}\right)}+{r}{\left({x}\right)}\)
where \(\displaystyle{r}{\left({x}\right)}={a}{x}^{{{2}}}+{b}{x}+{c}\). For some \(\displaystyle{a},\ {b},\ {c}\in{\mathbb{{{Z}}}}_{{{2}}}\)
So
\(\displaystyle{p}{\left({x}\right)}+{I}={a}{x}^{{{2}}}+{b}{x}+{c}+{I}\)
as
\(\displaystyle{q}{\left({x}\right)}{\left({x}^{{{3}}}+{x}^{{{2}}}+{1}\right)}\in{\left\langle{x}^{{{3}}}+{x}^{{{2}}}+{1}\right\rangle}\)
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Thomas Lynn
Answered 2021-12-14 Author has 1097 answers
This is because
\(\displaystyle{x}^{{{3}}}+{x}^{{{2}}}+{1}\in{I}\)
by definition and therefore
\(\displaystyle{\left({x}^{{{3}}}+{x}^{{{2}}}+{1}\right)}+{I}={I}\)
meaning that,
\(\displaystyle{\left({x}^{{{3}}}+{x}^{{{2}}}\right)}+{I}={\left({x}^{{{3}}}+{x}^{{{2}}}\right)}+{\left({x}^{{{3}}}+{x}^{{{2}}}+{1}\right)}+{I}={1}+{I}\)
because the characteristic is 2.
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