Is $F[x]/q(x)=F[x]/\langle q(x)\rangle$? Let, $F[x]$ is a ring of integer and $q(x) \in F[x]$. What is the difference between $F[x]/q(x)$ and $F[x]/\langle q(x)\rangle$?
By definition  $F[x]/q(x)$ is the set (quotient ring, thus additive group or quotient group to be exact) of residues modulo $q(x)$. Let there are $n$ elements in $F[x]/q(x)$, then for a polynomial $f(x) \in F[x]$, by division algorithm we have -
$$f(x) = p(x) q(x)+r_i(x)$$
Here, $r_i(x)$ is the residue modulo $q(x)$ for $1 \leq i \leq n$, and the residue class $\overline{r_i(x)}$ is an element of $F[x]/q(x)$. Consider the set of all multiple of $q(x)$, this set is the set of ideal $\langle q(x)\rangle$, so if we want to represent all polynomial of $F[x]$ that belongs to residue class $\overline{r_i(x)}$, we write, $\langle  q(x) \rangle +r_i(x)$ which is the residue class $\overline{r_i(x)}$. i.e. $\overline{r_i(x)} = r_i(x) + \langle  q(x) \rangle$.
But at the same time, by definition the elements of $F[x]/\langle q(x)\rangle$ have the form of  $ r_i(x) + \langle  q(x) \rangle$.
Thus, both  $F[x]/q(x)$ and $F[x]/\langle q(x)\rangle$ have elements of the form $ r_i(x) + \langle  q(x) \rangle$ for $i=1, 2, \cdots n$, according to above reasoning  $F[x]/q(x)=F[x]/\langle q(x)\rangle$, is it correct?
Is there any technical fault in above argument?
POSTSCRIPT
I have searched in internet but couldn't find a book or lecture pdf note that explains this, so if possible provide related theorem, book, lecture note to the above question.
 A: This is really an extended comment. I think that your confusion can be explained  by looking at a more simple example.
Let's look at the ring $\mathbb{Z}$, and the element $3\in\mathbb{Z}$.
(i) I think it is standard mathematical notation that
$$
\mathbb{Z}/\langle 3\rangle =\left\{ a+\langle 3\rangle \mid a\in\mathbb{Z}\right\}=
\left\{ 0+ \langle 3\rangle, 1+\langle 3\rangle, 2+\langle 3\rangle\right\}.
$$
I think that it is also very standard to write $\bar{a}:=a+\langle 3\rangle$, so that
$$
\mathbb{Z}/\langle 3\rangle =\left\{\bar{0},\bar{1},\bar{2}\right\}.
$$
(ii) As far as I can understand your explanation you seem to suggest that $\mathbb{Z}/3$ should consist of the three elements $\{0,1,2\}$, the minimal possible non-negative remainders under division by $3$; and that the operations on this set should be the ordinary ones followed by taking the remainder modulo $3$.  It is easy but tedious to show that this is indeed a ring.
Your explanation of why the two rings (the quotient ring defined on the set of cosets, and the ring defined on the three natural numbers $0,1,2$) are isomorphic can be cleaned up and is basically OK.
It seems to me, though, that for almost a century the approach (i) is the one  adopted by mathematicians, and I'd wonder whether anyone using (ii) was a serious player.
