3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Sammy Black Nov 24 '13 at 18:14
  • $\begingroup$ The maximal ideal has codimension $1$. (By the way, the kernel is usually called the augmentation ideal of the group ring.) $\endgroup$ – Sammy Black Nov 24 '13 at 18:18
  • $\begingroup$ typo fixed. I haven't studied the term codimension. I do know what a dim of a vector space is. $\endgroup$ – jreing Nov 24 '13 at 18:26
  • $\begingroup$ If $K$ is a subspace of $V$, then $\operatorname{codim} K = \dim V - \dim K$. For any linear map, $\operatorname{codim} \operatorname{Ker}f = \dim \operatorname{Im} f$. $\endgroup$ – Sammy Black Nov 24 '13 at 18:46
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.