Elementary questions about polynomials and field extensions Let $$f(x)=x^2+x+1.$$ This is irreducible in $\mathbb{Z_2}[x]$, and thus $\mathbb{Z_2}[x]/(f(x))$ is a field $K$ where $(f(x))$ is a principle ideal. I don't quite understand how I find that $\overline{0}$,$\overline{1}$, $\overline{X}$, and $\overline{X+1}$ are the elements of this new field.
Also, would I be able to rewrite $f(x)$ as a product of factors of degree $1$ in $K[x]$? If so how would I go about this? 
This is from Dan Saracino's Abstract Algebra: A First Course.

Well, I think it's clear that $\overline{0}$ and $\overline{1}$ are in this field, since these have to be elements of any field.  Also, it's clear that $\overline{x}$ is in this field, since $x \in \mathbb{Z}_{2}[x]$ and $\mathbb{Z}_{2}[x] \subseteq K$ where $K$ is our field.  Also, $\overline{x}$ is clearly different from $\overline{0}$ and $\overline{1}$.
So, we have three distinct elements of $K$ already, and we know this field has $2^{2} = 4$ elements (why? hint: what is the characteristic of this field? how many basis elements are there?), so we just need one more distinct element.  Well, fields are closed under addition, and both $\overline{x}$ and $\overline{1}$ are in $K$, so $\overline{ x + 1}$ should be in $K$.  Now, the question is, is $\overline{x + 1}$ distinct from the other three elements?
Recall that if $R$ is a ring, and $I$ is an ideal, then if $\overline{a}, \overline{b} \in R/I$, we say $\overline{a} = \overline{b}$ if $b - a \in I$.  So, to check if $\overline{x + 1}$ is distinct from $\overline{0}$, $\overline{1}$, and $\overline{x}$, let's see if the above criteria holds or not:


*

*$x + 1 - 0 = x + 1 \not \in \langle x^{2} + x + 1 \rangle$ (why?), so $\overline{x + 1}$ is different from $\overline{0}$.

*$x + 1 - 1 = x \not \in \langle x^{2} + x + 1 \rangle$ (why?), so $\overline{x + 1}$ is different from $\overline{1}$.

*$ x + 1 - x = 1 \not \in \langle x^{2} + x + 1 \rangle$ (why?), so $\overline{x + 1}$ is different from $\overline{x}$.
So $\overline{x + 1}$ is the fourth element of our field, and since our field can only have four elements, we have $\mathbb{Z}_{2}[x]/ \langle f(x) \rangle = \{ \overline{0}, \overline{1}, \overline{x}, \overline{x + 1} \}$.
I'm not completely sure for the second question. Let's wait and see what others have to say.
