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.
 A: In fact, $(f(x))$ is a maximal ideal, and this is equivalent to $\mathbb{Z}_2[x] / (f(x))$ being a field. Anyway, in this quotient $\bar{X}^2 + \bar{X} + \bar{1} = \bar{0}$, so, $\bar{X}^2 = \bar{X} + \bar{1}$, and by induction any power of $\bar{X}$ can be written as a $\mathbb{Z}_2$-linear combination of $\bar{1}$ and $\bar{X}$. (Neither of these is a multiple of the other, otherwise, by the below comment, $\bar{0}$ or $\bar{1}$ would be a root of $f$ in $\mathbb{Z}_2$, and hence $f$ would not be irreducible, a contradiction.) Thus, the field $\mathbb{Z}_2[x] / (f(x))$ is a vector space over $\mathbb{Z}_2$ with basis $\bar{1}, \bar{X}$, and its elements are precisely $\bar{0}, \bar{1}, \bar{X}, \bar{X} + \bar{1}$.
You can always write $f(x)$ as a product of linear factors over the splitting field $K$ of $f$ (in fact, this is why it's called a splitting field). By construction $f(\bar{X}) = \bar{X}^2 + \bar{X} + \bar{1} = \bar{0}$, so $\bar{X}$ is a root of $f$, and hence $x - \bar{X}$ is a factor of $f$ in $K[x]$. Since $\deg f = 2$, it hence has two linear factors. (We don't need to know the other factor to answer the question as written, but we can show that $f(x) = (x - \bar{X})(x - (\bar{X} + \bar{1}))$ in $K[x]$, which we can find readily with polynomial long division.)
A: 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.
