Prove that the ideal of $\mathbb{C}G$ generated by a family of elements $\lbrace p_i\rbrace_{i=1}^n$ is equal to $\mathbb{C}G$

Question

Given a finite abelian group $G$ consider the group algebra $\mathbb{C}G$ and a set $\mathcal{P}=\lbrace p_i\rbrace_{i=1}^n$ of elements of $\mathbb{C}G$. Define $I$ to be the ideal of $\mathbb{C}G$ generated by $\mathcal{P}$. I want to find a condition to ensure that $I\neq\mathbb{C}G$.

By representation theory of finite groups, since $G$ is a finite abelian group, every irreducible representation fo $\mathbb{C}G$ has dimension $1$. Hence $\mathbb{C}G$ decomposes as a direct sum: $$\mathbb{C}G\cong\bigoplus_{i=1}^{|G|}\mathbb{C}e_i$$ where $\lbrace e_i\rbrace_{i=1}^{|G|}$ is a set of elements such that $\sum\limits_{i=1}^{|G|}e_i=1$ (for example if $G=C_n$ then it is the set of $n$-th roots of unity)

Now, we know that there is a bijection between the set of characters of $G$ and sums of elements of $e_i$. Hence, if $e$ is defined as the sum of some $e_i$ and $\chi_e$ is its corresponding character then $\forall~\lambda=\sum\limits_{g\in G}\alpha_g g\in\mathbb{C}G$ we have: $$\lambda e=\left(\sum\limits_{g\in G}\alpha_g \chi_e(g)\right)e$$ and, this affects in the decomposition as: $$\lambda\mathbb{C}G=\bigoplus_{\lambda\cdot e_i\neq 0}\mathbb{C}e_i$$ In particular, this means that if there is a character $\chi:G\to\mathbb{C}$ such that $\chi(p_i)=0~\forall~i=1,\dots,n$ then: $$p_i \mathbb{C}G\neq\mathbb{C}G~\forall~i=1,\dots,n$$ This proves the following:

If there exists a character $\chi:G\to\mathbb{C}$ such that $\chi(p_i)=0~\forall~i=1,\dots,n$ then $I\neq\mathbb{C}G$

Is it correct? I've recently started to learn chacter theory, so I'm not sure if this makes any sense. In the case where the proof is incorrect I would like to know whether the final result is true or not. Tanks for your feedback

Answer 1

Yes, this is basically correct.

A character $\chi\colon G\to \mathbb{C}^\times$ [note it should take values in the multiplicative group of non-zero complex numbers] induces an algebra homomorphism $\tilde{\chi}\colon \mathbb{C}G\to \mathbb{C}$ given by $\tilde{\chi}:=\tilde\chi(\sum \alpha_g g)=\sum \alpha_g \chi(g)$. Moreover the kernel of $\tilde{\chi}$, $$ \ker \tilde{\chi}:=\{\lambda\in \mathbb{C}G:\tilde{\chi}(\lambda)=0\}$$ is always a proper ideal of $\mathbb{C}G$, that is not equal to $\mathbb{C}G$, by the rank-nullity theorem, for example.

Your condition that $\tilde{\chi}(p_i)=0$ for all $i=1,\ldots,n$ forces $I=(p_1,\ldots,p_n)$ to be contained in the kernel of $\tilde{\chi}$ and so not equal to $\mathbb{C}G$.

In fact the converse is also true, since every maximal ideal of $\mathbb{C}G$ is the kernel of some $\tilde{\chi}$, and so every proper ideal $I$ of $\mathbb{C}G$ is contained in some $\ker \tilde{\chi}$. Moreover for this choice of $\chi$, $\tilde{\chi}(p)=0$ for all $p\in I$ and so in particular for all $p$ in a generating set for $I$.

Answer 2

Another way to say it is that, if we label so that the character $\chi$ is associated to the primitive idempotent $e$ of the group algebra, then $e$ does not appear in the expression of any $p_{i}$ as a linear combination of the primitive idempotents. Hence $e$ does not appear in the expression of $x$ as a linear combination of primitive idempotents for any $x$ in $I$. Thus, in particular, $e$ lies outside $I$, and $I$ is a proper ideal. This is just another way of saying what @SimonWadsley said.