Difficulty understanding the meaning of quotient rings with regards to polynomials. I am having difficulty understanding quotient rings with regards to polynomials. For example, the quotient ring $\mathbb{F_2}[x]/(x^3 + 1)$ maps to the set of all remainders when a polynomial of $x$ (with coefficients in $\mathbb{F}_2$) is divisible by $x^3 + 1$. However, it's hard for me to comprehend the exact elements in this quotient ring.
Obviously if the quotient ring is something like $\mathbb{Z}/5$, I know that it maps to the set $\{0, 1, 2, 3, 4\}$, since that is the set of possible remainders when any number is divided by $5$. However, with polynomials, I feel it's more difficult to understand because for instance, I know that when a polynomial of $x$ is divided by $x^3 + 1$, the remainder obviously has to be less than $x^3 + 1$, but a polynomial such as $x^2 + 1$ is greater than $x^3 + 1$ for the domain $(0, 1)$ but not so when $x > 1$. So I'm wondering if there's an objective algorithm to determine the set of all remainders when a polynomial is divided by another polynomial.

The notion of "size" we use when we quotient polynomial rings is not the absolute value, but the degree. So for instance, something like $100x^2 + 500$ (which is degree $2$) would be considered smaller than $x^5 - 50 x^4$ (which is degree $5$), because we only care about the degree.
As for how we compute residue classes, we use the same division algorithm that we're used to in order to compute a remainder. You can see a series of worked out examples here, say.
Now, if we look at $\mathbb{F}_5 = \mathbb{Z} / 5$, the elements we're left with are those which are "smaller" than $5$. So $\{0,1,2,3,4\}$.
If we look at $\mathbb{F}_3[x] / (x^3 - x^2 + 1)$, then the elements we're left with are those which are "smaller" than $x^3 - x^2 + 1$. That is, those polynomials of degree $< 3$. So the elements are
$$
\begin{align}
0  &&    1 &&    2 \\
x  &&  x+1 &&  x+2 \\
2x && 2x+1 && 2x+2 \\
x^2 && x^2 + 1 && x^2 + 2 \\
x^2 + x  &&  x^2 + x+1 &&  x^2 + x+2 \\
x^2 + 2x && x^2 +  2x+1 && x^2 + 2x+2 \\
2x^2 && 2x^2 + 1 && 2x^2 + 2 \\
2x^2 + x  &&  2x^2 + x+1 &&  2x^2 + x+2 \\
2x^2 + 2x && 2x^2 +  2x+1 && 2x^2 + 2x+2 \\
\end{align}
$$
Notice there are $27$ of these, since a general polynomial looks like $a_2 x^2 + a_1 x + a_0$, where each $a_i \in \mathbb{F}_3$.
In general, if $f$ is a polynomial of degree $n$, then $\mathbb{F}_p [x] / (f)$ will have $p^n$ elements for exactly this reason.
As a quick example of how to do reduction, let's reduce $x^4$ mod $x^3 - x^2 + 1$ in $\mathbb{F}_3[x]$.
The key insight is that since $x^3 - x^2 + 1 = 0$ in the quotient ring, we can rearrange this to see $x^3 = x^2 - 1$.
Now $x^4 = (x) (x^3)$, which, by the above argument is $(x) (x^2 - 1)$, or $x^3 - x$. Now we go again, since we still see an $x^3$, and this becomes $(x^2 - 1) - x$.
So then $x^4 = x^2 - x - 1$, and since we are working in a polynomial ring over $\mathbb{F}_3$, we can rewrite this as $x^2 + 2x + 2$, which would be the final answer.
Obviously it's tedious to have to repeatedly subtract off factors of $x^3$ and replace them, but thankfully we don't need to! In addition to the video I linked earlier, you can see a worked example for the division algorithm here which lets you efficiently compute $p(x)$ mod $f(x)$.

I hope this helps ^_^
