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This is my first post on StackExchange. I had a quick question about notation (appearing in research literature) that I was unable to find by repeated searches, and I was hoping that someone would be able to help.

I am dealing with finite fields $\mathbb{F}_q$, where $q$ is a power of a prime $p$, and a multiplicative character $\chi: \mathbb{F}_q^* \to \mathbb{C}^*$. I want to consider a character satisfying $\chi(\mathbb{F}_q^*)=1$, notation I am not familiar with. If anyone could help me define this or direct me to a source that can, it would be much appreciated.

Thank you very much!

Edit: Please see the comments below for the context of the notation.

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    Unless this is a horrible clash of notation, this just means that the character is trivial, no? Welcome, by the way.2012-05-23
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    I think it's just slight abuse of notation saying that the image of the domain under $\chi$ is $\{1\}$, ie the character is trivial as Dylan says.2012-05-23
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    Given the lack of context, would this article be of any help? http://en.wikipedia.org/wiki/Dirichlet_character2012-05-23
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    That's also what I thought, and on closer inspection, this now makes sense. Just to confirm, this is the way it is used in context: "Let $A=(\mathbb{F}_q[x]/(h(x)))^*$ and $\hat{A}$ denote the group of all characters of $A$. [$h(x)$ is a generic polynomial in $\mathbb{F}_q[x]$ of degree $d$.] Let $\hat{B}=\{\chi \in \hat{A}\ |\ \chi(\mathbb{F}_q^*)=1\}$, an abelian subgroup of order $\leq q^d$." So what this is saying is that $\hat{B}$ is the set of characters that is trivial over the base field?2012-05-23
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    *So what this is saying is that* $\hat{B}$ *is the set of characters that is trivial over the base field?* Yep. (Which of course forms an abelian group under pointwise multiplication...)2012-05-23

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To summarize: the notation $\chi(\Bbb F_q^*)=1$ means that the image of $\Bbb F_q^*$ (the base field, minus $0$ of course) is simply $\{1\}$. The set of these $\chi$s, denoted $\hat{B}$, forms an abelian group under pointwise multiplication.