# Showing the augmentation ideal of $\mathbb{F}[G]$ is a maximal ideal

Given a basis of $\Bbb R^n,\ G:=\{e_0,...,e_{n-1}\}$, we define multiplication on the elements of the basis by $e_i\cdot e_j=e_{i+j}$ (where $i+j$ is calculated modulo $n$).

For a field $\Bbb F$ we define the ring $\Bbb F[G]=\{\sum_{j=0}^{n-1}a_je_j :a_j\in \Bbb F\}$ with the natural addition and multiplication.

We define a map $f:\Bbb F[G] \to \Bbb F$ by $f(\sum_{j=0}^{n-1}a_je_j)=\sum_{j=0}^{n-1}a_j$.

1. Prove this is a homomorphism
2. Prove that $\ker(f)$ is a maximal ideal.

What I did:

1. I succeeded showing it.

2. I think that $\ker(f)$ is only $0$, because otherwise the image would have been different than $0$. So actually I need to show that the only ideal in this field is $(0)$. I was wondering if this is a good direction?

Thanks.

• Look again at your equation $e_i + e_i = e_{i + j}$. I think that you want multiplication, otherwise the set is not linearly independent. – Sammy Black Nov 24 '13 at 18:14
• The maximal ideal has codimension $1$. (By the way, the kernel is usually called the augmentation ideal of the group ring.) – Sammy Black Nov 24 '13 at 18:18
• typo fixed. I haven't studied the term codimension. I do know what a dim of a vector space is. – jreing Nov 24 '13 at 18:26
• If $K$ is a subspace of $V$, then $\operatorname{codim} K = \dim V - \dim K$. For any linear map, $\operatorname{codim} \operatorname{Ker}f = \dim \operatorname{Im} f$. – Sammy Black Nov 24 '13 at 18:46

As $f$ is a homomorphism of rings, $\ker f$ is an ideal. By the first isomorphism theorem, $\mathbb{F}[G]/\ker f \cong \operatorname{im} f$. For any $\alpha \in \mathbb{F}$, $\alpha e_0 \mapsto \alpha$ so $f$ is surjective and therefore $\mathbb{F}[G]/\ker f \cong \mathbb{F}$. As the quotient by $\ker f$ is a field, $\ker f$ is a maximal ideal.