I'm doing an introductory course in abstract algebra at university and I keep seeing variations of the following exercise:
Let $p(x) \in \mathbb{Z}_2[x]$, $p(x) = x^2 + x + 1$ and $I = <p(x)>$.
Is $\mathbb{F} = \mathbb{Z}_2[x]/I$ a field?
I have a few theorems that I use to show that $p(x)$ is irreducible/reducible and because of that $\mathbb{F}$ is/isn't a field, so I can solve these exercises, but I don't understand them.
Specifically, I don't understand what $\mathbb{Z}_2[x]/I$ is and I'm also not sure about $I$. I assume $I$ is simply $<p(x)> = \{p(x)^0, p(x)^1, p(x)^2, ...\}$, but I shomehow get the feeling it should be finite.
Regarding $\mathbb{Z}_2[x]/I$, Wolfram MathWorld tells me this:
A field $K$ is said to be an extension field, denoted $K/F$, of a field $F$, if $F$ is a subfield of K. For example, the complex numbers are an extension field of the real numbers, and the real numbers are an extension field of the rational numbers.
What I understand from this is that $\mathbb{Z}_2[x]/I = \mathbb{Z}_2[x]$. This can't be correct because another exercise in my course tells me that "the elements of $\mathbb{Z}_2[x]/I$ are of the form $a_0 + I, a_0 \in \mathbb{Z}_2$". I suspect they may have something to do with the remainders of dividing the polynomials in $\mathbb{Z}_2[x]$ with the polynomials in (what I assume is) $<p(x)>$, which would indeed by of this form, but I am just really confused by all this.
So my questions are:
Is my understanding of $<p(x)>$ correct?
What actually is $\mathbb{Z}_2[x]/I$? How should I imagine / think about it?
$(p(x))=\{r(x)p(x)\mid r(x)\in\mathbb{Z}_2\}$. Notice that $(p(x))$ does not contain $p(x)^0=1$, otherwise the ideal $(p(x))$ would generate the whole ring. You're right that $p(x),p(x)^2,\dots$ all lie inside $(p(x))$, but it also contain any multiple of $p(x)$.
$\mathbb{Z}_2[x]/(x^2+x+1)$ consists of polynomials with coefficients from $\mathbb{Z}_2$ with the relation $x^2+x+1=0$. This means that $x^2=-x-1=x+1$ (since $-1\equiv_2 1$), so whenever you have a polynomial of degree $2$ or greater, you can reduce it to a linear polynomial.
Thus all the polynomials are on the form $a+bx$ where $a,b\in\mathbb{Z}_2$. Now, there are $2$ choices for $a$ and $2$ choices for $b$, so this is a field of $2^2=4$ elements. It's a field beacuse $x^2+x+1$ is irreducible. $$\mathbb{Z}_2[x]/(x^2+x+1)=\{a+bx\mid a,b\in\mathbb{Z}_2\}$$