Group ring zero divisor example.

Question

Here is the question I am trying to understand its solution:

Give an example of a zero divisor in the group ring $\mathbb Z(C_3)$ where $C_3$ is the cyclic group with 3 elements.

My professor gave us the following solution:

(g^2 + g+ 1)(g-1)

I do not know how can I get another simple solution to this question, could someone clarify this to me please?

Answer 1

$\mathbb Z(C_3)\simeq \mathbb Z[x]/(x^3-1)$, and if we have in the quotient ring $\bar p(x)\bar q(x)=0$, then $x^3-1 \mid pq$, as $x-1$ and $x^2+x+1$ are irreducible, there must be $$x-1\mid p \text{ or }x-1\mid q$$

$$x^2+x+1\mid p \text{ or } x^2+x+1 \mid q$$

As neither $\bar p$ nor $\bar q$ is $0$, we must have one of $p,q$ has $x-1$ as a factor while the other has $x^2+x+1$ as a factor, so there is essentially no other "simple" example. All we can do is to scale $x-1$ and $x^2+x+1$. Note that $g^2+g+1$ is $C_3$-invariant, hence all of its multiples are $\mathbb Z$-multiples. But $g-1$ allows slightly more, such as $(g+1)(g-1)=g^2-1$, etc. So we may take take $$(g^2-1)(g^2+g+1)=0$$

Alternatively we may apply the Chinese Remainder Theorem to get $$\mathbb Z[x]/(x^3-1)\simeq \mathbb Z[x]/(x-1)\times \mathbb Z[x]/(x^2+x+1)\simeq \mathbb Z\times \mathbb Z[\zeta_3]$$

As both of $\mathbb Z$ an $\mathbb Z[\zeta_3]$ are integral domains, it should be clear what the zero divisors are.

As pointed out in the comment, we cannot apply CRT here, but we do have a homomorphism: $\mathbb Z[x]/(x^3-1)\rightarrow \mathbb Z\times\mathbb Z[\zeta_3]$ by $p(x)\mapsto (p(1), p(\zeta_3))$. This is injective but not surjective due to the failure of the CRT condition. The image of the map is $$\{(a,b+c\zeta_2): a,b,c\in\mathbb Z, a\equiv b+c\text{ mod } 3 \}$$

We can still read the zero divisors from the image as a subring, but this is probably too much trouble for the simple question.

Answer 2

If $G$ is a finite group acting on an abelian group $A$, or equivalently, $A$ is a $\mathbb Z[G]$-module, you can always "average" over $G$ to get elements of $A$ which are invariant under the action of the group: Let $s_G = \sum_{g \in G} g \in \mathbb Z[G]$. Then if $a\in A, g_1 \in G$ we have $$ g_1(s_G(a)) = g_1\left(\sum_{g \in G} g.a \right)= \sum_{g \in G}g_1(g(a)) = \sum_{g \in G} (g_1.g)a = \sum_{k \in G} k.a = s_G(a), $$ where in the third equality we have set $k=g_1g$, and used the fact that as $g$ runs over all of $G$, so does $k$, since left-multiplication by $g_1$ is a bijective map from $G$ to itself.

But now since right-multiplication by $h\in G$ is also a bijective map from $G$ to itself, the same argument shows that, in the group ring $\mathbb Z[G]$, we have $s_G.h=s_G=h.s_G$ for all $h \in G$. But then $h.s_G = 1.s_G$ for all $h \in G$, so that $(h-1)s_G=0$ in $\mathbb Z[G]$ and hence a group ring always contains zero divisors as soon as the group $G$ is nontrivial.