Let $p$ be an odd prime number and let $\alpha \geq 1$ be an integer. Let $\chi$ be a real, non-principal, primitive Dirichlet character mod $p^{\alpha}$.

How does one show that $\alpha = 1$?

If we choose an integer $g$, not divisible by $p$, such that $g + p^{\alpha}\mathbb{Z}$ generates $(\mathbb{Z}/p^{\alpha}\mathbb{Z})^\times$, then $\chi$ takes the form

$$\chi(n) = \exp{\left(\frac{2 \pi i}{\phi(p^{\alpha})}m\nu(n)\right)} = \exp{\left(\frac{m}{p^{\alpha-1}((p-1)/2)}(\pi i)\nu(n)\right)} \quad \text{for} \quad (n,p^{\alpha}) = 1$$

where $\nu(n)$ is (mod $\phi(p^{\alpha}))$ determined by the congruence $g^{\nu(n)} \equiv n \pmod{p^{\alpha}}$ and where $0 \leq m \leq \phi(p^{\alpha})-1$ is fixed. We have $m \neq 0$ since $\chi$ is non-principal.

Since $\chi$ is real and since a possible value for $\nu(n)$ is $1$, we must have that $m$ is divisible by $p^{\alpha-1}(p-1)/2$. If we had $\alpha \geq 2$, we'd also have that $m$ is divisible by $p$ and $\chi$ could then be rewritten as

$$ \chi(n) = \exp{\left(\frac{2 \pi i}{\phi(p^{\alpha-1})}m'\nu(n)\right)} = \exp{\left(\frac{m'}{p^{\alpha-2}((p-1)/2)}(\pi i)\nu(n)\right)}, $$ which looks like a character mod $p^{\alpha-1}$. But the (or my) problem is that $g$ generated $(\mathbb{Z}/p^{\alpha}\mathbb{Z})^\times$ and need not be a generator for $(\mathbb{Z}/p^{\alpha-1}\mathbb{Z})^\times$.

Could anyone explain why $\chi$ is induced by a character of smaller modulus?


By assumption, since $\chi$ is a real character, we can write $$\chi:(\mathbb Z/p^\alpha\mathbb Z)^\times\to \{\pm1\}.$$

Since $p$ is odd, $(\mathbb Z/p^\alpha\mathbb Z)^\times$ is cyclic. Hence there is just one non-trivial possibility for $\chi$: it must send any generator $g$ to $-1$.

It follows that this $\chi$ must be the character induced from the unique real character of $(\mathbb Z/p\mathbb Z)^\times$, so in order for $\chi$ to be primitive, $\alpha$ must be $1$.


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