I know that $\mathbb{Z}[x]$ is a commutative ring with unity and for any two elements $f(x)$ and $g(x)$, the set
$\langle f(x), g(x)\rangle = \{ \alpha(x)f(x) + \beta(x)g(x) : \alpha(x) \land \beta(x) \in \mathbb{Z}[x] \}$
is the ideal generated by $f(x)$ and $g(x)$.
I am trying to visualize what the elements of $\frac{\mathbb{Z}[x]}{ \langle f(x), g(x)\rangle }$ looks like, but I don't know how to do modular reduction in this case.
When we are working with with ideal generated by a single element, for instance, $f(x)$, then I know that two elements $p(x) + \langle f(x)\rangle $ and $q(x) + \langle f(x)\rangle $ are equal in $\frac{\mathbb{Z}[x]}{\langle f(x)\rangle }$ if
$p(x) \equiv q(x) (\mod f(x))$
So, how can I do the same for elements of $\frac{\mathbb{Z}[x]}{ \langle f(x), g(x)\rangle }$ ?
In order to find a polynomial $q(x)$ such that
$p(x) \equiv q(x) (\mod \langle f(x), g(x)\rangle )$
do I have to calculate $p(x) \mod (f(x) + q(x))$ or $(p(x) \mod f(x)) \mod q(x)$ or something else ?
--- EDIT ----
For instance, in the section 2.1 of this paper, the authors define $R_{x-a}$ as $\frac{\mathbb{Z}[x]}{ \langle x^n + 1, x - a\rangle }$, for $a \in \mathbb{Z}$. Then, they present exactly the following sentence:
For $f(x) + \langle x^n + 1, x - a\rangle \in R_{x-a}$, we denote by $[f(x)]_{R_{x-a}} \in \mathbb{Z}$ the unique integer with $[f(x)]_{R_{x-a}} \equiv f(x) (\mod x^n + 1, x - a)$, and $[f(x)]_{R_{x-a}} \in \{0,...,|a^n+1|-1\}$.
So, in this case, any polynomial $f(x)$ is reduced to a zero-degree polynomial $[f(x)]_{R_{x-a}}$ (this is why they say that it belongs to $\mathbb{Z}$).
But how to find this $[f(x)]_{R_{x-a}}$ for any given $f(x)$ ?