Method to adjoin elements to a ring - is the symbol $x$ overloaded? On this presentation a systematic method to adjoin elements to a polynomial ring including the zeros to a polynomial $p(x)$ is presented as the quotient ring $R[x] / p(x).$
In this quotient ring $p(x)$ generates a principal ideal $(p(x))=\{p(x)f(x) \vert f(x) \in R[x]\}.$
I understand how an ideal is the equivalent of a normal subgroup, and how it quotients out the ring $R[x].$
So I think I kind of get the general gist of where he is getting at: $p(x)=0$ multiplied by every element of $R[x]$ is the zero element in the quotient group... More or less...
However, and in a preceding presentation motivating the topic, a crystal clear example with $\mathbb F_2$ was introduced. In it the idea was to adjoin an element $\zeta + 1$ so as to include the solution of the polynomial $x^2+x +1.$ This turned $\mathbb F_2$ into the klein-four group, and $\zeta^2+\zeta+1=0,$ as desired.
Transitioning to the more general formulation above, though, I don't see how the corresponding zero (the solution, $\zeta$) is adjoined, when there is only a variable $x.$

Is this $x$ now both the variable in a polynomial and the solution or root(-s) of $p(x)$? Is it an overloaded symbol? I understand things like adjoining $\sqrt{-1}$ or $\sqrt 2$, but where is the new element here?

 A: I much prefer looking at quotients via a Bishop-style setoid approach. A setoid is a set equipped with an equivalence relation that we think of as the notion of equality for the set. We then require functions to be well defined, i.e. a well-defined function $f:(X,\sim_X)\to(Y,\sim_Y)$ is a function $f:X\to Y$ such that if $x\sim_X x'$ then $f(x)\sim_Y f(x')$. That is, functions take "equal" inputs to "equal" outputs. The nice thing about this approach is that quotienting a setoid just amounts to equipping it with a coarser equivalence relation.
For a (commutative) ring $R$ and an ideal $I$ of $R$ we can define the following equivalence relationship: $r\sim_I r' \iff r-r'\in I$. You can easily prove that an ideal is exactly the equivalence class of $0$ for a congruence on a ring. A congruence is an equivalence relation that respects the operations, e.g. $x\sim x'$ and $y\sim y'$ implies $x+y\sim x'+y'$ etc.
$R[X]/\langle p(X)\rangle$ means "quotient $R[X]$ by the smallest congruence that relates $p(X)$ to $0$". From the setoid perspective, this means $R[X]/\langle p(X)\rangle$ is the setoid where the polynomial $p(X)$ "equals" $0$. This implies, via congruence, that many other things equal $0$ such as $p(X)f(X)$ for all $f(X)\in R[X]$ hence the notion of ideal. So $X$ (and all equivalent polynomials) is indeed the root of $p(X)$ in $R[X]/\langle p(X)\rangle$ since, by definition, $p(X)$ "equals" $0$ in $R[X]/\langle p(X)\rangle$.
For a more traditional approach, $R[X]/\langle p(X)\rangle$ will be the set of equivalence classes, i.e. the cosets, of the congruence induced by the ideal $\langle p(X)\rangle$. If we write $[f(X)]$ for the equivalence class induced by $f(X)\in R[X]$, more traditionally written $f(X)+\langle p(X)\rangle$, then we have $[p(X)]=[0]$ almost by definition. The equivalence relation being a congruence implies that $[f(X)]+[g(X)] = [f(X)+g(X)]$ and $[f(X)][g(X)] = [f(X)g(X)]$ etc. Thus $p([X]) = [p(X)] = [0]$ so $[X]$ is a root of $p$ in $R[X]/\langle p(X)\rangle$.
You can think of it as follows. We have a ring $R$. We want a root for a polynomial with coefficients in $R$. We freely adjoin a formal symbol $X$ to $R$ producing $R[X]$. $R[X]$ is the set of polynomials in the variable $X$ with coefficients in $R$. Another way of looking at this, is $R[X]$ is the set of syntactic terms in the algebraic language of rings, i.e. pieces of syntax. To make this a (commutative) ring, we need to quotient it by the (commutative) ring axioms, e.g. $r+s \sim s+r$. $X$ satisfies no special equations. We want $p(X)=0$ so we "force" that by quotienting as described above. We now have the ring $R[X]/\langle p(X)\rangle$ which will (usually...) contain $R$ as a subring (the constant polynomials) but also contains the formal symbol $X$ which has been forced to be a root of $p$. Freely generating a structure out of syntactic terms and then forcing certain equalities of terms by quotienting is what algebra is all about. (Some structures commonly studied in abstract algebra, such as fields, don't [quite] fall into this pattern.)
