For the field $F$, integral domain $F[x]$, some element $c$ which is not contained in $F$ and the minimal polynomial of $c$, $f(x)$ I know that we have the isomorphism
$$ F(c) \cong F[x] / (f(x)) $$
where $(f(x))$ is the ideal generated by the polynomial $f(x)$, but I am having a lot of difficulty in comprehending what elements of $F[x] / (f(x))$ actually look like. Can someone provide a few concrete examples, or is there a set-theory type definition which describes all of these elements?
The quotient notation $A/B$ usually represents a set of equivalence classes of $A$ induced by the relation of congruence modulo $B$.
In the current case this means that you partition $F[x] $ into sets of equivalence classes with two polynomials lying in same class if and only if their difference is divisible by $f(x) $. The equivalence classes can be represented by any member of their class and we can make a convention that a typical equivalence class $[p(x)] \in F[x] /(f(x)) $ will be represented as $[r(x)] $ where $r(x)$ is the remainder obtained when $p(x) $ is divided by $f(x) $. When viewed in this way you can consider $F[x] /(f(x)) $ as a set of polynomials with degree less than degree of $f$.
The big deal here is that if $f(x) \in F[x] $ is irreducible over $F$ then the structure $K=F[x] /(f(x)) $ is a field and more importantly this field $K$ has a subfield $F_0$ which is isomorphic to the original field $F$.
In this sense $K$ is a field extension of $F$. It is instructive to consider actual examples. Thus you can take $K=\mathbb {R} [x] /(x^2+1)$ where $\mathbb{R} $ is the field of real numbers.
As mentioned above $K$ is a set of equivalence classes and each class can be represented by a polynomial of degree less than $2$. Let $[a+bx], [c+dx] $ be two such elements of $K$ and define their addition and multiplication as $$[a+bx] +[c+dx]=[(a+c) +(b+d) x], [a+bx] [c+dx] =[(a+bx) (c+dx)] $$ The definition of addition does not present any challenge, but for multiplication the polynomial on right side is a polynomial of degree $2$ and it makes sense to find a corresponding representative polynomial of degree less than $2$.
We have $$(a+bx) (c+dx) =ac-bd+(ad+bc) x+bd(x^2+1)$$ Since the last term is divisible by $x^2+1$ we have $$[(a+bx) (c+dx)] =[(ac-bd) +(ad+bc) x] $$ and the definition of multiplication given above is equivalent to $$[a+bx] [c+dx] =[(ac-bd) +(ad+bc) x] $$ One can check that these operations make $K$ a field.
Further the set of elements of the form $[a] $ behave exactly like real number $a$ and an isomorphism can be establish between the set of these elements and $\mathbb{R} $.
We can now write $$[a+bx] =[a] +[b] [x] $$ Using the isomorphism mentioned above and notation $i=[x] $ we get the familiar form $a+ib$ for the element $[a+bx] $ and we have found out the complex numbers.
Usually the presentation of algebraic numbers (like $\sqrt{2}$) in high school presupposes their existence and the polynomials satisfied by them. So the view of an algebraic element $c$ lying outside $F$ is common. However once we learn a little abstract algebra the more common perspective is to have a field $F$ and consider polynomials over $F$. The irreducible polynomials then lead us to field extensions and algebraic elements like $c$.
I admit this abstract view is a bit difficult to grasp (who wouldn't prefer symbol $\sqrt{2}$ instead of $[x] \in\mathbb {Q} [x] /(x^2-2)$?), but even for a novice (in abstract algebra) like me this is really beautiful and amazing.
Let $d$ be the degree of $f(x)$. Each member of $F[x]/(f(x))$ has a unique model as a polynomial of degree $d-1$, since any coefficient of degree $\geq d$ can be reduced modulo $f$.
So, from how I see it, the best interpretation of $F[x]/(f(x))$ is the polynomials of degree $\leq d$ under the multiplication operation $g\cdot h=(g\cdot h \mod{f})$, by which I mean that first one performs regular polynomial multiplication and then one reduces mod $f$.
When I say that we reduce a polynomial $g$ mod $f$, I mean that we subtract off multiples of $f$ from $g$ until we get something of degree $\leq d-1$. For example, $9x^3+x^2-3x+6$ modulo $x^3+x$ is $x^{2}+6x+6$ since
$$9x^{3}+x^{2}-3x+6\equiv9x^{3}+x^{2}-3x+6-9\left(x^{3}-x\right)\equiv x^{2}+6x+6\mod x^3-x$$
I also struggled with this when I was learning ring theory. So let's start with the ideal
$$(f(x))=\{g(x)f(x):g(x)\in F[x]\}$$
this is just all elements in $F[x]$ which are divisible by $f(x)$.
Now for the quotient, remember in a ring we have two operation $+, \cdot$. For quotient rings, we take the the ideal $(f(x))$ and move it around by addition of other elements in the ring. All elements that are divisible by $(f(x))$ are seen zero since the get absorbed by the ideal (ideals are additive abelian groups).
So what does an element in $F[x]/(f(x))$ look like? Well take an $g(x)\in F[x]$ and consider $$g(x)+(f(x))=\{g(x)+h(x)f(x):h(x)\in F[x]\}$$ and that's it.
Now, let's try an example maybe in the context of what you're talking about. We know of course that $\sqrt{2}\not\in \mathbb{Q}$ but we know that $\sqrt{2}$ is a root of the polynomial $x^2-2\in \mathbb{Q}[x]$ and it is the minimal polynomial of this number. We can show that the field extension $$\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}[x]/(x^2-2)$$ this is done by constructing the isomorphism and is pretty standard.
The point is, a number which we wish to extend a field by to make a larger field, the structure of the field extension sits inside a polynomial quotient ring by the ideal generated by the minimal polynomial.
The isomorphism you have given above gives one of the best ways to think about elements of $F(c)$. Suppose the minimal polynomial $f(x)$ can be written as $$f(x) = x^n + a_{n-1}x^{n-1} +... + a_0,$$ for $a_{n-1},...,a_0\in F$. Then the elements of $F[x]/(f(x))$ are polynomials in $x$ with the relation $$x^n = -a_{n-1}x^{n-1} - ... - a_0.$$ Thus the elements $$1,x,...,x^{n-1},$$ give a basis for $F[x]/(f(x))$. Thus, under the isomorphism $F(c) \cong F[x]/(f(x))$, you can think of elements of $F(c)$ as polynomials in $c$ of degree $\leq n-1$. When you multiply elements of $F(c)$, you can always reduce the resulting product to a polynomial in $c$ of degree $\leq n-1$ using the relation $$c^n = -a_{n-1}c^{n-1} - ... - a_0.$$
For example, elements of $\mathbb{Q}(\sqrt{2})$ look like $$a\sqrt{2} + b,$$ for $a,b\in \mathbb{Q}$.
