<s>[INST] Polynomial ring addition in $\mathbb{Z_{6}}[x]$ I know this is a very simplest question ever. But, I need help with understanding it. So here it goes... 

Let, $f(x) = \bar{1}+\bar{2}x+\bar{3}x^2$ and $g(x) = \bar{4}+\bar{5}x$ $\in \mathbb{Z_{6}}[x]$. Compute $f+g$.

I got: $f+g = \bar{5}+\bar{7}x+\bar{3}x^2$, or should it be $f+g = \bar{5} + \bar{1}x + \bar{3}x^2$ since $f$ and $g$ both are in $\mathbb{Z_{6}}[x]$. I am just not getting the purpose of $\mathbb{Z_{6}}$ here.
And for $f\cdot g$, I got: $\bar{4} + \bar{13}x + \bar{22}x^2 = \bar{4} + x +\bar{4}x^2$ (mind my arithmetic).
 [/INST] In $\mathbb{Z}_6$, $\overline{1} = \overline{7}$ so $\overline{5} + \overline{7}x + \overline{3}x^2 = \overline{5} + \overline{1}x + \overline{3}x^2$. Usually, one writes the final answer using only $\overline{0}, \dots, \overline{5}$ so the latter may be considered to be a simplified version of the former (in the same way that $\frac{1}{2}$ is a simplified version of $\frac{2}{4}$, even though $\frac{2}{4} = \frac{1}{2}$).
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