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Let $p$ be odd and $\mathbb Z_{p^n}$ denote the ring of integers modulo $p^n$. Let the quadratic character, $\eta$, be the function defined on $\mathbb Z_{p^n}^*$ (multiplicative group of $\mathbb Z_{p^n}$) with $\eta(c)=1$ if $c$ is the square of an element of $\mathbb Z_{p^n}^*$ and $\eta(c)=-1$ otherwise.

Let $G$ be a finite abelian group. A character of $G$ is simply a homomorphism $\chi$ from $G$ into the multiplicative group $\mathbb{C}^*$ of complex numbers. That is, $\chi(g_1g_2)=\chi(g_1)\chi(g_2)$ for all $g_1,g_2 \in G$.

We call a character of $\mathbb Z_{p^n}^*$ as multiplicative character of $\mathbb Z_{p^n}$.

Can we say that, $\eta$ is a multiplicative character for $\mathbb Z_{p^n}$ and all multiplictive characters are obtained in this way?

Thank you.

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As $p$ is odd, the group $G=\mathbb{Z}^*_{p^n}$ is cyclic of order $p^{n-1}(p-1)$. So if $g$ is a generator of $G$, and $\omega$ is any complex number that satisfies the equation $\omega^{p^{n-1}(p-1)}=1,$ then we get a homomorphism $\chi=\chi_\omega$ from $G$ to $\mathbb{C}^*$ from the recipe $$ \chi_\omega(g^k)=\omega^k $$ for all $k=0,1,2,\ldots,p^{n-1}(p-1)-1$.

There are $p^{n-1}(p-1)$ choices for $\omega$, and each choice gives you a different character. It is not too hard to see that there are no others.

If you are only interested in characters that take values $\pm1$ only, then you must pick $\omega=\pm1$. The choice $\omega=+1$ gives you the trivial character (everything is mapped to one), and the choice $\omega=-1$ gives the character $\eta=\chi_{-1}$ that you described:
$$ \eta(g^k)=(-1)^k=\begin{cases}+1,\ \text{if $k$ is even,}\\-1,\ \text{if $k$ is odd}.\end{cases} $$ So your $\eta$ is a multiplicative character all right (the square are exactly the even powers of $g$, as the order of $g$, $p^{n-1}(p-1)$, is an even number.

For example, when $p^n=3^2=9$, then $|G|=3^{2-1}(3-1)=6$, and it is easy to see that we can use $g=2$. If we select $\omega$ to be a primitive sixth root of unity, say $\omega=e^{\pi i/3}=(\sqrt3+i)/2$, then we get the character $\chi=\chi_\omega$: $$ \begin{eqnarray*} \chi(1)=\chi(2^0)&=\omega^0=1,\\ \chi(2)=\chi(2^1)&=\omega^1=\omega,\\ \chi(4)=\chi(2^2)&=\omega^2,\\ \chi(8)=\chi(2^3)&=\omega^3=-1,\\ \chi(7)=\chi(2^4)&=\omega^4=-\omega,\\ \chi(5)=\chi(2^5)&=\omega^5=\overline{\omega}. \end{eqnarray*} $$ This time $\chi(0),\chi(3)$ and $\chi(6)$ are left undefined as those numbers are divisible by three. Sometimes a convention is used, when those are defined to be zero. Those three residue classes do not belong to the group $G$, but number theorists sometimes want to define these functions outside of $G$ as well. This time you get the quadratic character $\eta$ as the third power $\chi^3$.

Hope this helps. Ask for clarifications if/where needed.

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