Unable to understand the relation between Field Extensions, Polynomial Quotient Rings generated by Ideals, & Polynomial Roots I am going through this video 302.S2a: Field Extensions and Polynomial Roots by
Matthew Salomone & there is a lot in this video which confuses me.

*

*I understand the construction by which he generates the field with 8 elements {$0, 1, t, 1 + t, t^2, 1 + t^2, t + t^2, 1 + t + t^2$}. He sets $p(t) = 0$ & constructs a set with those 8 elements. But I am unable to figure out why he is calling this field as $F_2[t]/\langle p(t) \rangle$ - I am assuming this is the same as calling it $F_2[t]/\langle t^3 + t + 1 \rangle$
$F_2[x]/\langle t^3 + t + 1 \rangle$ is the Quotient Ring generated by the ideal $t^3 + t + 1$ - it doesn't contain these 8 elements at all. Each element of the Quotient Ring isn't even a polynomial - it's an equivalence class of polynomials - for e.g. one such element of $F_2[x]/\langle t^3 + t + 1 \rangle$  is this equivalence class $[\bar t]$
$[\bar t]$ = {$...-t^3+1, t, t^3 + 1,  t + 2, ...$}
Another element is this equivalence class below
$[\bar 5]$ = {$..., -t, 1, t^3 +t , 1, ...$}
So now why is he calling that field of 8 elements as $F_2[x]/\langle t^3 + t + 1 \rangle$ - isn't that a totally different ring/field?
I don't understand where the quotient construction using an ideal comes in here at all. We construct this new field by 2 things by just setting $t^3 + t + 1 = 0$ in $F_2[x]$. We don't need to use the ideal or the quotient ring generated by the ideal at all for this.


*A little further into the video, timestamp 7:30 - he says $t$ is a root of the polynomial. Again I don't understand what he means by $t$ is a root of a polynomial $p(t)$ - for a polynomial $p(t)$, we always say $t = something$ is a root - what exactly does saying $t$ is a root mean?

 A: $x^3+x+1=p(x)$ is irreducible in $F_2[x]$ so  $\langle p(x)\rangle$ is a maximal ideal of $F_2[x]$. It follows that $F_2[x]/\langle p(x) \rangle=:S$ is a field. Note that this is where $\langle p(x) \rangle$ being an ideal (maximal ideal in this case) is used.
$S$ actually has $8$ elements in it. The elements of $S$ look like $ax^2+bx+c+\langle p(x) \rangle$, where $a,b$ and $c$ are in $F_2$. The elements of $S$ are often written with the bar notation (in such notation, $\langle p(x) \rangle$ is dropped) as $\overline {ax^2+bx+c}$, i.e., $\overline {ax^2+bx+c}=ax^2+bx+c+\langle p(x) \rangle$.
Now, regard every element in $F_2[x]$ as an element of $S$ (using the bar notation).
With this view point, $\overline x\in S$ is a zero of $p(t)\in S$ as the following shows:
\begin{align}
p(\overline x)=(\overline x)^3+\overline x+1&=(x+\langle p(x)\rangle)^3+(x+\langle p(x)\rangle)+1\\
&=x^3+x+1+\langle p(x)\rangle\\&=p(x)+\langle p(x)\rangle=\langle p(x)\rangle 
\end{align}
Note that RHS is a 'zero' in $S$.
To see that $S$ has $8$ elements, note that if $\overline{q(x)}$, where $q(x)$ is of degree$>2$ lies in $S$ then $q(x)$ can be reduced modulo $p(x)$ to get $q(x)=p(x) a(x)+b(x)$, where $b(x)=0$ or degree b(x)<degree $p(x)=3$. It follows that $\overline {q(x)}= q(x)+\langle p(x) \rangle= b(x)+p(x)a(x)+\langle p(x)\rangle= b(x)+\langle p(x)\rangle=\overline{b(x)}$.
Since $b(x)$ is of degree at most $2$, it follows that $b(x)$ is of the form $dx^2+ex+f$. Note that each of $d,e$ or $f$ has two choices (as they are in $F_2$) so maximum number of elements in $S$ is $2^3=8$.
The only way when the number of elements in $S$ is less than $8$ is when any two elements in $S$ are equal (i.e., there is some sort of 'collapsing'). Now, show that there is no collapsing and you are done.
