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Let $k=k_1 k_2$ s.t. $(k_1,k_2)=1$ and let $\chi$ be a dirichlet character mod $k$. I'm trying to prove that there exsists $\chi_1,\chi_2$ dirichlet characters mod $k_1,k_2$ respectively, s.t. $\chi(r) = \chi_1(r)\chi_2(r)$ for every $r \in Z$.

I tried to somehow create $\chi_1,\chi_2$ using $\chi$ values but I didnt managed to ge a solution this way. I have a strong feeling that the chineese remainder theorem is somehow related here.

Can someone please help me get towards the solution? Thanks.

Edit: I think the solution is to use corrolary from the chineese remainder that we can look the multiplicative group mod $k$ as a multiplexion of two multiplicative group mod $k_i$.

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    $\begingroup$ Are you familiar with how to think of the CRT as an isomorphism between unit groups? And have you tried an example, e.g. if you start with a Dirichlet character mod 20 can you extract from it a pair of Dirichlet characters mod 4 and mod 5? And conversely, if you start with Dirichlet characters mod 4 and mod 5 can you construct from them a Dirichlet character mod 20 using the CRT? $\endgroup$ – KCd Jun 22 '14 at 0:07
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Here is a second try if $f:G\times H \rightarrow \mathbb{C}$ is a homomorphism, let $g(x)=f(x,e)$ and $h(y)=f(e,y)$ then

$$f(x.y)=f((x,e)(e,y))=g(x)h(y)$$

Note that $U(nm)\cong U(n)\times U(m)$ for $n$ and $m$ relative prime.

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  • $\begingroup$ Although its a great direction I couldnt prove that $\chi_i$ is acctualy a dirichlet char mod $k_i$. $\endgroup$ – Snufsan Jun 22 '14 at 19:04
  • $\begingroup$ I gave another attempt. $\endgroup$ – Rene Schipperus Jun 22 '14 at 20:42
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Remember that if $n=p_1^{k_1}p_2^{k_2}\dots$ then the CRT says that the unit group factorize as $$ (\mathbb{Z}/n\mathbb{Z})^\times\cong (\mathbb{Z}/{p_1^{k_1}}\mathbb{Z})^\times \times (\mathbb{Z}/{p_2^{k_2}}\mathbb{Z})^\times \times (\mathbb{Z}/{p_3^{k_3}}\mathbb{Z})^\times \dots $$ In your case $$ (\mathbb{Z}/k\mathbb{Z})^\times\cong (\mathbb{Z}/{k_1}\mathbb{Z})^\times \times (\mathbb{Z}/{k_2}\mathbb{Z})^\times \;. $$ Now, Dirichlet characters are completely multiplicative functions, so $$ \chi:(\mathbb{Z}/{k_1}\mathbb{Z})^\times \times (\mathbb{Z}/{k_2}\mathbb{Z})^\times \rightarrow \mathbb{C} $$ and the identity you're searching for comes simply restricting to $$ \chi_i:(\mathbb{Z}/{k_i}\mathbb{Z})^\times \rightarrow \mathbb{C} $$ for $i=1,2$.

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  • $\begingroup$ Yeah thats exactly what I suggested in my edit yesterday. $\endgroup$ – Snufsan Jun 23 '14 at 15:34

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