TITLE: What exactly do elements of $F[x] / (f(x))$ look like?
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.