Say
$\frac{\mathbb{Z}_{2}\left [ x \right ]}{x^{2}+x+1}=\left \{0,1,\alpha ,1+\alpha \right \}$
is a finite field with its elements listed. I am finding it difficult to understand what it means to have $\alpha$ as residue class of $x$ mod $x^{2}+x+1$ here.
I know how to generate field elements using primitive poly by successive multiplication of ring variable(say x).
If I use $p={{x}^{6}}+{{x}^{4}}-{{x}^{2}}+x+1$, I get $x+1$, so is it in $1+\alpha$ class? Why it can't be just represented in terms of $x$?
Perhaps it is needed a more thorough study in quotient rings in general, and in particular quotients of polynomial rings.
In this particular case we have simply that $\;\alpha:=x+I\;$ , with $\;I:=\langle x^2+x+1\rangle\;$ , so it is just the coset corresponding to $\;x\;$ modulo the ideal generated by that irreducible quadratic in $\;\Bbb F_2[x]\;$ , and it is thus a root of the quadratic in the given quotient (field), since
$$\alpha^2+\alpha+1=(x^2+I)+(x+I)+(x+I)=(x^2+x+1)+I=I=\overline 0=\;\text{The zero in}\;$$
the quotient ring
How to work with it? Very simple:
$$\alpha^2=-\alpha-1=\alpha+1\pmod 2\;,\;\;\text{so for example}$$
$$\alpha\cdot(\alpha+1)=\alpha^2+\alpha=(\alpha+1)+\alpha=2\alpha+1=1$$
so in this case we get that in our field with four element we get $\;\alpha^{-1}=\alpha+1\;$.
Remember! You always work with arithmetic modulo $\;2\;$ (or, in general, modulo the prime you're working with)