Thinking about the cosets of $\mathbb{Z}[X]/(f(x))$ Let's think about
$$\mathbb{Z}[X]/(f(x))$$
First of all, (f(x)) is the ideal generated by $f(x)$ on $\mathbb{Z}[x]$. Let's consider $f(x) = x^d+1$. Then, if we multiply all the elements of $\mathbb{Z}[X]$ by $x^d+1$, we get all polynomials of degree greater than or equal to $d$, plus the polynomial $0$.
So $(f(x)) = \{\mbox{polynomials with degree $\ge d$, and polynomial  0}\}$.
Now I'm trying to think about the cosets, that is, $\mathbb{Z}[X]/(f(x)) = \{p(f(x)), p\in \mathbb{Z}[x]\}$
I should think about all possible cosets and then consider which ones are equivalent so thus discover how this ring looks like. However, I see $p(f(x))$ gives me again all polynomials with degree $\ge d$.
I think something is deeply wrong in my understanding of cosets, as this quotient should be isomorphic to the space of all polynomials of degree up to $d-1$.
 A: The thing is that $(f(x))$ doesn't contain every polynomial in $\mathbb{Z}[X]$, as it has been pointed out in the comments by @Alan. You must think of the ideal as the set of multiples of $f(x)$, as $(2)$ in $\mathbb{Z}$ is not every integer, but rather only the even ones.
Now, to think about cosets is to think about "sending $(f(x))$ to the class of zero", in the sense that two polynomials are in the same coset if they differ by a multiple of $(f(x))$. Assume, for example, $f(x) = x^d+1$. Then, both $g(x) = x^d$ and $h(x) = -1$ are in the same coset, as $x^d = -1 + (x^d+1)$. One neat way to think about this in the particular case of polynomials, is to think about the quotient as an "arithmetic rule to reduce the degree".
Assume we have a polynomial, for example, $x^2-x+2$, and we think about $\mathbb{Z}[X]/(x^2-x+2)$. You must think about this as "sending this polynomial to zero". Then, you abuse notation and write $x^2-x+2=0$ (there is a formal way of doing this, but I'm trying to be ilustrative). You can "leave the $x^2$ term alone", and have $x^2 = x-2$, and use this rule to reduce the degree of polynomials. If you have $x^3-x$ and want to think about its coset, you realize $x^3-x = x(x^2-1)$, and you send $x^2$ to $x-2$, meaning the coset of $x^3-x$ is the same as that of $x(x-3)=x^2-3x$, and applying it again, it is the same as $-2x-2$. Quotients by polynomials of degree $d$ make you think about a space of polynomials of degree less than $d$, and give you a way of reducing the degree in case you go over.
