I'm struggling with Ireland and Rosen, chapter 8, exercise 7.
Suppose that $p=1\bmod 4$ and that $\chi$ is a character of order 4. Write $\chi^2=\rho$ a character of order 2. Show that $J(\chi,\chi)=\chi(-1)J(\chi,\rho)$.
hint: evaluate $g(\chi)^4$ in two ways.
Notions:
- $g(\chi)$ is the gauss sum: $g(\chi)\doteq\sum_t\chi(t)\zeta^t$ with $\zeta$ a $p$th primitive root of unity;
- $J(\chi,\rho)$ is the Jacobi sum: $J(\chi,\rho)\doteq\sum_{x+y=1}\chi(x)\rho(y)$
I don't know how to start neither why $p=1\bmod4$ is useful here…
Answer: since $p=1\bmod 4$, $$ g^4(\chi)=p\chi(-1)J(\chi,\chi)J(\chi,\chi^2) $$
Using properties of the Jacobi sum, $$ g^4(\chi) = (g(\chi)g(\chi))^2 = (J(\chi,\chi)g(\chi^2))^2$$ But since $\chi$ is of order 4, $\chi^2=(\chi^2)^{-1}$ and so $g(\chi^2)^2=g(\chi^2)g((\chi^2)^{-1})=g(\chi^2)\overline{g(\chi^2)}=|g(\chi^2)|^2=p$
Hence, $$ p\chi(-1)J(\chi,\chi)J(\chi,\rho)=pJ(\chi,\chi)^2 $$ which proves the statement.