Reducedness of a factor ring in regard of group rings

Question

Let $G$ be the group of order 2, and $K$ be a field of characteristic 2. Assume the group ring $R=KG$, and take the ideal $S=K.\sum _{g\in g}g$. Is it true that the factor ring $R/S$ is reduced?

Thanks for any help to reach the answer.

Answer 1

In this case, the ideal you’re talking about is obviously equal to the augmentation ideal, and the quotient by that here is $K$, so yes.

(The augmentation ideal is generated by $1-g$ for $g\in G$, and with characteristic $2$, $1-g=1+g$.)

Answer 2

According to this widipedia entry on reduced rings, a ring $\tilde R$ is reduced if and only if it has no non-zero nilpotent elements; this means that if $x \in \tilde R$ and there is a positive integer $n$ such that

$x^n = 0, \tag 1$

then

$x = 0; \tag 2$

in fact, we only need to consider the case $n = 2$ here, for if $n \ge 2$ is the smallest positive integer for which (1) binds, then we have

$x^{n - 1} \ne 0, \tag 3$

but

$(x^{n - 1})^2 = x^{2n - 2} = x^n x^{n - 2} = 0; \tag 4$

this shows that if there are no non-zero $y \in \tilde R$ with $y^2 = 0$, then there are no non-zero $x$ with $x^n = 0$, eliminating the need to separately cover the cases $n \ge 3$.

Since $G$ is the group of order $2$, we have

$G = \{e, g\}, \; g^2 = e , \tag 5$

$e$ being the identity element of $g$; then the elements of $R = KG$ are all of the form

$r = \alpha e + \beta g, \; \alpha, \beta \in K, \tag 6$

so

$r^2 = (\alpha e + \beta g)^2 = \alpha^2 e^2 + 2\alpha \beta eg + \beta^2 g^2$ $= \alpha^2 e^2 + \beta^2 g^2 = \alpha^2 e + \beta^2 e = (\alpha^2 + \beta^2)e, \tag 7$

where we have used $\text{char} K = 2$ and $g^2 = e$ is performing this calculation.

Now the elements of $S$ are all of the form $\gamma (e + g)$ where $\gamma \in K$, so if $(r + S)^2 = 0$ in $R/S$, we have

$r^2 + S = (r + S)^2 = 0, \tag 8$

or

$r^2 \in S; \tag 9$

using (7) we thus see that

$(\alpha^2 + \beta^2)e = \gamma(e + g) = \gamma e + \gamma g \tag{10}$

for some $\gamma \in K$. Since $e$ and $g$ are by definition linearly independent over $K$, this shows we must have

$\gamma = 0, \tag{11}$

which in turn forces

$\alpha^2 + \beta^2 = 0; \tag{12}$

also, by $\text{char K} = 2$,

$(\alpha + \beta)^2 = \alpha^2 + 2 \alpha \beta + \beta^2 = \alpha^2 + \beta^2 = 0, \tag{13}$

which forces

$\alpha + \beta = 0 \tag{14}$

since $K$ is a field; again by $\text{char K} = 2$,

$\alpha = \beta, \tag{15}$

whence

$r = \alpha(e + g) \in S; \tag{16}$

thus

$r + S = 0 \tag{17}$

in $R/S$, and we conclude that $R/S$ is reduced.