Why the group ring $\mathbb{F}_2[Q_8]$ is not Armendariz? Any help? Actually, I want to find the zero divisors in its polynomial ring. The group ring contains the $\mathbb{F}_2$ linear combination of the group $Q_8$.
Armendariz ring: A ring $R$ is said to be Armendariz if whenever polynomials $f(x),g(x)\in R[x]$ satisfy $f(x)g(x)=0$ then $a_ib_j=0$ where $f(x)=a_0+a_1x+...+a_nx^n$ and $g(x)= b_0+b_1x+...+b_mx^m$.
I want to find two polynomials in the polynomial ring of $\mathbb{F}_2[Q_8]$ such that their product is zero while one of the product of the coefficients is not zero.
Using a bit of code I found a solution. I will use the notation $Q_8 = \{e,\bar{e},i,\bar{i},j,\bar{j},k,\bar{k}\}$. Let $a = e + k$, $b = \bar{j} + \bar{k}$, $c = e + \bar{e} + k + \bar{k}$, and $d = j + \bar{j} + k + \bar{k}$. Then $ac = bd = 0$ but $ad = bc = \sum_{s \in Q_8} s$. Hence $f = a + bx$ and $g = c + dx$ are two polynomials satisfying your condition.
Have you determined the zero divisors in $\mathbf{F}_2[Q_8]$? Surely you should do this before trying to find all zero divisors in the polynomial ring.