Question

Show that $\left\{1,\left[x\right]\right\}$is basic of ${\mathit{Z}}_{\mathbf{2}}\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{/}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}$over ${\mathit{Z}}_{\mathbf{2}}$.

Short answer

Answer:

$\left\{1,\left[\mathrm{x}\right]\right\}$is a basic of $Z_2\left[\mathrm{x}\right]/({\mathrm{x}}^2+\mathrm{x}+1)$over $Z_2$.

Step-by-step solution

Definition of basis.

A set of element in a vector space V is called a basic or a set of basic vectors, if the vectors are linearly independent and every vector in a vector space is a linear combination of this set. In general a basic is a linear independent spanning set.

Showing that {1,x} is a basis of Z2[x]/(x2+x+1).

The set is $\left\{1,\left[x\right]\right\}$. Now as any element of $Z_2\left[x\right]/\left(x^2+x+1\right)$is of the form $f\left(x\right)+\left(x^2+x+1\right)$and $f\left(x\right)\in Z_2$.

For any element of $f\left(x\right)\in Z_2$by division algorithm.

$f\left(x\right)=\left(x^2+x+1\right)q\left(x\right)+r\left(x\right)$. Where $r\left(x\right)$is reminder on dividing $f\left(x\right)$by$\left(x^2+x+1\right)$

Either $\mathit{d}\mathit{e}\mathit{g}\mathbf{}\mathit{r}\left(x\right)\mathbf{=}\mathbf{0}\mathbf{}\mathit{d}\mathit{e}\mathit{g}\mathbf{}\mathit{r}\mathbf{}\left(x\right)\mathbf{=}\mathit{d}\mathit{e}\mathit{g}\left(x^2+x+1\right)$

That is $degr\left(c\right)<2$.

Now

role="math" localid="1656909768648" $$\begin{gathered} \mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}\mathit{q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathit{r}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{[}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{r}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{[}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{]} \end{gathered}$$

Here $r\left(x\right)=a+anda,b\in Z_2$. Therefore $degr\left(x\right)=0or1$. So,

$$\begin{gathered} \frac{Z_2}{\left[x^2+x+1\right]}=\left[a+bx+\left[x^2+x+1\right]:a,b\in Z_2\right] \\ =\left\{I,1+I,x+1\right\},I=\left[x^2+x+1\right] \end{gathered}$$

Therefore,

$$\begin{gathered} \frac{Z_2}{\left[x^2+x+1\right]}=\left\{a+bx+I:a,b\in Z_2\right\} \\ =\left\{a\left(1+I\right)+b\left(x+I\right):a,b\in Z_2\right\} \\ =\left\{a\left(1\right)+b\left(x\right):a,b\in Z_2\right\} \end{gathered}$$

Thus $\left\{1,\left[x\right]\right\}$is a basic for $Z_2\left[\mathrm{x}\right]/({\mathrm{x}}^2+\mathrm{x}+1)$over $Z_2$.