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

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.

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

levurdondishav4
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
$$\displaystyle{q}{\left({x}\right)}{\left({x}^{{{3}}}+{x}^{{{2}}}+{1}\right)}\in{\left\langle{x}^{{{3}}}+{x}^{{{2}}}+{1}\right\rangle}$$
###### Not exactly what youâ€™re looking for?
Thomas Lynn
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.