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Let $a\not\equiv 3\pmod 4$ and $a\ne 0$. How do you show that $\chi(n):=\left(\frac{a}{n}\right)$ (where $\left(\frac{\cdot}{\cdot}\right)$ denotes the Kronecker symbol) defines a character of modulus $m:=4|a|$ if $a\equiv 2 \pmod 4$ and $m:=|a|$ otherwise?

What I have to show is that we have $\left(\frac{a}{n}\right)=\left(\frac{a}{m+n}\right)$. The Kronecker symbol is defined using the prime factorisation of the lower argument. Unfortunately the prime factorisation of $n$ and $n+m$ can be totally different. So I don't have a clue how to prove that identity.

principal-ideal-domain
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1 Answers1

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The Kronecker symbol defines a character, because it is completely multiplicative by definition, i.e., because the Jacobi symbol is multiplicative. So we have $\chi(nm)=\left(\frac{a}{nm}\right)=\left(\frac{a}{n}\right)\left(\frac{a}{m}\right)=\chi(n)\chi(m)$, unless $a=-1$, one of $m,n$ is zero and the other one has odd part congruent to $3\bmod4$. So the only problem is, to be a bit careful with the different cases of the definition - see here.
It also follows from the properties of the Jacobi symbol that for $a\not\equiv3\pmod 4$, $a\ne 0$, we have $\left(\tfrac am\right)=\left(\tfrac an\right)$ whenever $$m\equiv n\bmod\begin{cases}4|a|,&a\equiv2\pmod 4,\\|a|&\text{otherwise.}\end{cases}$$

Dietrich Burde
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  • The statement is not only that it defines a character but a character of modulus $m$ with $m$ defined as in my question. So we have also to show that $\left(\frac{a}{n}\right)=\left(\frac{a}{m+n}\right)$ for all integers $n$. That's the part I have a problem with. – principal-ideal-domain Aug 14 '15 at 08:55
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    All right. This can be found, e.g. in the book of Montgomery, Hugh L; Vaughan, Robert C. (2007). Multiplicative number theory – Dietrich Burde Aug 14 '15 at 11:11