What quotients of polynomial quotient rings are projective? I am trying to learn about projective modules. I.e. modules from which any homomorphism can be factored through a surjection. 
To do this, I'm considering my favorite ring, which is just messy enough, $R=\mathbb{F}[x]/(p(x))$ where $p(x)$ is a polynomial. So now, I'm trying to figure out necessary and sufficient conditions for a quotient ring, say $P=R/(q(x))$ to be projective (i.e. let's assume it is principal). 
Now it seems to me, I have an extreme, where I know $P$ to be projective. Namely, if I take $q(x)=x$ then $P\cong \mathbb{F}$ which should be projective. However, I'm not sure where to head from here?
I was thinking one could perhaps use the CRT to decompose $R$ and get some further insight comparing it with a decomposition for $q$, but I'm quite frankly stuck. How do I characterize the quotients that are projective?
To be clear, I'm considering $R=\mathbb{F}[x]/(p(x))$-modules.
 A: If $I\lhd F[x]$ such that $F[x]/I$ is a projective $F[x]$ module and $I$ is nontrivial, then the exact sequence 
$$
0\to I\to F[x]\to F[x]/I\to 0
$$ splits.
But that would mean $F[x]\cong I\oplus F[x]/I$, which is not possible, because $I$ intersects all other nontrivial ideals nontrivially. So, the only possibilities for $I$ are $\{0\}$ and $F[x]$.

After you said that you want to use ideals of $R=F[x]/(p(x))$ module.
That depends heavily on $p(x)$. For example if the factorization of $p(x)$ is squarefree, then all modules over $R$ are projective.
The logic from above still stands though: If $I\lhd R$ is an ideal such that $R/I$ is projective, then $I$ is a summand of $R$. Conversely, every summand is the kernel of a projective quotient. 
You can compute all summands of $F[x]/(p(x))$ by factoring $p(x)$ into powers of irreducible elements.
For example, if $R=F[x]/(x^n)$, then no proper ideals are projective, because there are no summands.
On the other hand, $\mathbb Q[x]/(x-1)(x-2)(x-3)$ is isomorphic to $\mathbb Q^3$ and has $6$ nontrivial ideals, all of which are summands.
