What happens if I take a quotient over a reducible polynomial? I know that for adjoinging roots to a field, I need to find irreducible polynomials so that the ideal I am taking the quotion with will be maximal, hence the resut being a field.
Imagine I am working with $F=\mathbb{Z}_7$ and I take a look at the polynomial $g(x):=x^3+6 = (x-1)(x-2)(x-4)$.
What kind of structure is $F[x]/(g)$?
My attempt so far: I thought that there is no further information given by the reducible polynomial (all zeros are in the field $F$ already), so, by the degree-theorem of field-extensions, I thought that $[F(\alpha):F] = 1$, where $\alpha$ denotes a zero of the polynomial $g$.
Is this really a field, not just an integral domain (as a subset of $F[x]$)? 
What happens if we consider a polynomial so that there is an irreducible with degree > 1, but which is still reducible?
I am aware that the minimal polynomial then will not be sent to 0, hence the "factorisation" will not be complete. But it is not clear what the "improper factorisation" looks like.
Thanks!
 A: Let me first remind you the chinese remainder theorem:

Theorem. Let $R$ be a ring and let $I,J$ be ideal of $R$ such that $I+J=R$, then one has the following ring isomorphism: $$\left\{\begin{array}{ccc}R/(I\cap J)&\rightarrow&R/I\times R/J\\x\mod I\cap J&\mapsto&(x\bmod I,x\bmod J)\end{array}\right..$$

Proof. Adapt the proof you already know for $R=\mathbb{Z}$. $\Box$
Now let's see how we can apply this theorem for polynomial quotient rings.
Let $k$ be a field and $f$ be a polynomial with coefficients in $k$. Assume that $f=g_1\cdots g_n$ with $g_1,\dots,g_n$ irreducibles over $k[x]$, then for all $i,j\in\{1,\dots,n\},i\neq j$, using Bezout's theorem, one has: $$(g_i)+(g_j)=k[x].$$
Therefore, using the theorem, one gets: $$k[x]/(f)=\prod_{i=1}^nk[x]/(g_i).$$
In your case, one has: $$\mathbb{Z}_7[x]/(x^3+6)=\mathbb{Z}_7[x]/(x-1)\times\mathbb{Z}_7[x]/(x-2)\times\mathbb{Z}_7[x]/(x-4).$$
A: In general, you will get a vector space over $F$ of dimension $deg \, g$ that has a ring structure (via multiplication $\mod g$) and that has zero divisors. This is a bit like $\mathbb{Z}_m$ where $m$ is composite. For example if $g = g_1g_2g_3$ and the $g_i$ are irreducible, then $g_i \ne 0$ in $F[x]/g$ but $g_1g_2g_3 = g = 0$ in that set. 
Don't confuse this with $F(\alpha)$ where $g(\alpha) = 0$. That is indeed again $F$.
