Extensions Fields and Zeros of Polynomials I've just digested the proof for Kronecker's theorem that states every polynomial $f(x)$ over a field $F$ has a zero in an extension field of $F$, mainly, the factor ring of polynomials $F[x]/<p(x)>$ where $F[x]$ is the ring of polynomials over $F$ and $p(x)$ is an irreducible factor of $f(x)$ and $<p(x)>$ is the principal ideal generated by $p(x)$.
What confuses me is this theorem is saying that a coset of polynomials over $F$ is a zero of a polynomial over $F$. I've never conceived of a coset of polynomials as some singular entity that can be plugged into an equation and then evaluated to 0. 
I understand abstract algebra is abstract and I believe I followed it well up to this point but here my understanding is collapsing a bit. I'm looking for comments that would help me come to a better intuitive grasp of this or else point to further results that clarify this. Please excuse my vagueness. This is the best I can muster after staring at the theorem each morning for several days.  
 A: Are you familiar with rings, ideals, and quotients thereof in general? There, one talks just about elements of rings instead of polynomials. So maybe we can spread out the complexity of Kronecker's theorem a bit if you first familiarize yourself with the construction of the quotient of a ring by an ideal in general, and then apply this to polynomial rings.
The integers $\mathbb Z$ provide a simple example for rings, ideals, and quotients. Here, we have $R = \mathbb Z$, every ideal has the form $I = n\mathbb Z$ for some natural number $n$, and the quotient is then $R/I = \mathbb Z / n \mathbb Z.$
In fact, both the integers and polynomial rings are Euclidean rings, which have very nice properties. So maybe you should read up about Euclidean rings and their ideals and quotients.
A: Suppose you have the construction as in your question.
The coset containing $x$ consists of all the polynomials $x+p(x)q(x)$ for $q(x)$ an arbitrary polynomial.
Now note that $\left(x+p(x)q(x)\right)^n=x^n+p(x)r(x)$ where $r(x)$ is a polynomial whose details don't matter. We can use this to compute $p\left(x+p(x)q(x)\right)=p(x)+p(x)s(x)$ - an element of the ideal generated by $p(x)$. This shows us that computing $p$ in the factor ring gives a result independent of the coset representative we choose, and is therefore well defined, and gives a result in the coset which contains zero.
You might want to think of how it works as rather like arithmetic modulo $p$ where we would conventionally use one representative of the class of integers in each residue class (quite often the least non-negative integer). In $\mathbb F [x]/\langle p(x)\rangle$ we conventionally take the monic polynomial of lowest degree in each class as the representative (and zero as a representative of the ideal itself).
A: (This a very good question to think about.) You have constructed a ring, and found a quotient of it that is field with awkward looking elements.But this field is isomorphic to a subfield of the complex numbers,
and here your usual intuition holds. So what has happened is that your have two ways of viewing this field, rather than one.
It is also possible to identify your field with an algebra of complex matrices.
Finally, I offer Feynman's quote: the nice thing abiut intuition is that you can alwys change it in light of the facts.
