Let $ R = \mathbb{Z}_8$. Find a nonzero polynomial $f$ in $R[x]$ of degree at most 3 such that each element of $R$ is a root of $f$.

Question

I tried using the method provided by Mark here but I can't seem to come up with a solution for $\mathbb{Z}_8$.

Answer 1

Going in the direction of the link you provided, what about $p(x)=4x(x+1)$?

Answer 2

The linked method of iterating the Factor Theorem to get multiple factors from various roots works fine in fields (and domains), but generally it fails in non-domains like $\,\Bbb Z_8,\,$ where $\,4\cdot 2 = 0.\,$ Indeed, as explained here we can deduce that $\,f(a) = 0 = f(b)$ $\Rightarrow f(x) = (x-a)(x-b)g(x)\,$ only when $\,a-b\,$ is cancellable.

Let's see how it breaks. $\,f(0) = 0\,\Rightarrow\, f = x\, g.\,$ $\,0 = f(1) = 1\cdot g(1)\, $ so $\, g= (x\!-\!1)h,\,$ so $\,f = x(x\!-\!1)h.\,$ $\,0 = f(2) = 2(1)h(2),\,$ but we can't conclude $\,h(2)=0\,$ because $2$ is not cancellable in $\,\Bbb Z_8$. But we can infer $\,8\mid 2h(2)\,\Rightarrow\,4\mid h(2).\,$ Taking $\,h = 4\,$ does the job, i.e. $\,8\mid 4x(x\!-\!1)\,$ by $\,2\mid x(x\!-\!1),\,$ for all $\,x.$