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It is not difficult to see that a group ring $K[G]$ ($K$ a domain) has non-trivial zero-divisors whenever there exists a non-trivial torsion element $g\in G$. [In fact, in this case, $1-g$ is such a non-trivial zero-divisor]

My question is the following: is it possible that a crossed product $K*G$ is a domain even if $G$ has non-trivial torsion (e.g., what happens when $G$ is finite)?

My guess is that $K*G$ may very well be a domain. In fact, if $G'$ is a group such that $K*G'$ is a domain and $G=G'/H$ is quotient of $G'$ (I'm thinking for example to the case $G'=\mathbb Z$ and $G=\mathbb Z(p)$), then $D=K*H$ is a domain and so $K*G'=D*(G'/H)=D*G$ cannot have non-trivial zero divisors. It seems that this example answers my question but in a very indirect way... do you have any straightforward reason for which this phenomenon may happen? Thank you in advance.

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well, it is not really an answer but an illuminating example: the ring of complex numbers $\mathbb C$ can be viewed as a crossed product ring $\mathbb R*C_2$. In fact, the elements of $\mathbb C$ can be viewed as pairs $(r,s)$, with $r,s\in \mathbb R$, with the sum defined component-wise and multiplication as follows: $$(r_1,s_1)(r_2,s_2)=(r_1r_2-s_1s_2, r_1s_2+s_1r_2)\,.$$ The unique difference with the multiplication in the group ring $\mathbb R[C_2]$ consist in a $-1$, in fact, the multiplication in this second ring is $$(r_1,s_1)(r_2,s_2)=(r_1r_2+s_1s_2, r_1s_2+s_1r_2)\,.$$ Anyway, from the point of view of the zero divisors this makes an enormous difference, $\mathbb R[C_2]$ has (infinitely many) zero divisors, while $\mathbb C$ is a field: $$\text{In }\mathbb R[C_2]:\ \ \ \ (1,1)(1,-1)=(1-1,1-1)=(0,0)\,,$$ $$\text{In }\mathbb C:\ \ \ (1,1)(1,-1)=(1+1,1-1)=(2,0)\neq 0\,.$$

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