Is this an equivalent definition of the notation $f(x) \equiv g(x) \hspace{4mm} ( \text{mod} \hspace{2mm} h(x), \hspace{1mm} n )$?

Question

I understand that the notation $$ f(x) \equiv g(x) \hspace{4mm} ( \text{mod} \hspace{2mm} h(x), \hspace{1mm} n ) $$ means that $f(x)$ is congruent to $g(x)$ in the polynomial quotient ring $\mathbb{Z}_n / h(x)$ (although I am not really sure what exactly this means).

Is this the same as saying $$ \left( f(x) \hspace{4mm} \text{mod} \hspace{2mm} h(x) \right) \hspace{4mm} \text{mod} \hspace{2mm} n \hspace{4mm} = \hspace{4mm} g(x) $$

Answer 1

The left hand side of your equation is an equivalence class in $\mathbb{Z}_n[x]/(h(x))$ whereas the right hand side is a polynomial in $\mathbb{Z}[x]$. Hence it is not correct.

The congruence you define means $\overline{f}-\overline{g}\in (\overline{h})$ where $\overline{f}$ denotes the image of $f$ under the canonical map $\mathbb{Z}[x]\rightarrow\mathbb{Z}_n[x]$.