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 theoory 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

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