Polynomial division over not a field

Question

I am trying to understand the following example that shows that the polynomial division theorem doesn't hold when the coefficients aren't from a field $F$.

So the statement is that it $F$ isn't a field (let's say it's a ring as it is in the example), then if $f,g\in F[x]$, the representation $f=gq+r, \deg(r)<\deg(g)$ isn't unique. And the example is $$f=\overline3x^3\in\mathbb{Z}_6,g=\overline3x^2-\overline1\in\mathbb{Z}_6$$ $$f=\overline3x^3=(\overline3x^2-\overline1)(x+\overline2)+\overline3x+\overline3=g(x+\overline2)+\overline3x+\overline3$$ $$f=(\overline3x^2-\overline1)(x+\overline4)+x+\overline4=g(x+\overline4)+x+\overline4$$

They are trying to show that $q$ and $r$ are not uniquely defined, so we can find many $q$ and $r$. I don't understand the aritmetic in $\mathbb{Z}_6$, though. Isn't $$(\overline3x^2-\overline1)(x+\overline2)+\overline3x+\overline3=\overline3x^3+\overline6x^2-\overline1x-\overline2+\overline3x+\overline3=\\=\overline3x^3+\overline2x+\overline1\ne f$$ How is this equal to $f$?

Answer 1

I think you are looking at some typos. I will omit the overbar, and just take all constants to lie in $\mathbb{Z}_6$. I think it should be $$f(x) = 3x^3 = (3x^2-1)(x+2) + ({\color{red}{x+2}}).$$ (That is, the remainder should not be $3x+3$; it should be $x+2$).

Indeed, we have $$\begin{align*} (3x^2-1)(x+2) +(x+2) &= 3x^3 + 6x^2 -x-2 + (x+2)\\ &= 3x^3 -x-2+x+2 \\ &= 3x^3 = f(x), \end{align*}$$ and the other one is $$f(x) = (3x^2-1)(x+4) + ({x+4}),$$ because $$\begin{align*} (3x^2-1)(x+4) + (x+4) &= 3x^3 - x + 12x^2 - 4 + (x+4)\\ &= 3x^3 - x - 4 + x + 4\\ &= 3x^3. \end{align*}$$ So you have two different quotients and remainders when dividing $3x^3$ by $3x^2-1$: you can have both quotient and remainder be $x+2$, or both quotient and remainder be $x+4$.