The classical Gauss sum This is a problem from dummit and foote 14.7.11

I solved everything except b). I tried hard but couldn't solve it. I'm trying to using sum of pth root of unity is 0. But there are $p$ terms in lagrange resolvant. So this won't work. 

Please help me with b) (why lagrange resolvant $(\zeta_p,1)=-1$).
I think the definition in Dummit and Foote is confusing. The following general definition makes more sense:

Definition Let $L/K$ be a Galois extension with cyclic Galois group $G$. Define the Lagrange Resolvent of an element $\theta \in L$ and a character $\chi : G \to K^\times$ to be $$R(\theta,\chi) = \sum_{g \in G} \chi(g) g(\theta).$$

Now we can do a worked example of the problem for $p = 7$. The squares mod 7 are 1,2 and 4.

I believe the characters in question are the principal character $1$ and the character "$-1$" that is $1$ for squares and $-1$ for nonsquares:


$\eta_0 = \zeta_7^1 + \zeta_7^2 + \zeta_7^4$ = -0.5 + 1.32287i
$\eta_1 = \zeta_7^3 + \zeta_7^5 + \zeta_7^6$ = -0.5 - 1.32287i
$R(\zeta_7,1) = \zeta_7^1 + \zeta_7^2 + \zeta_7^3 + \zeta_7^4 + \zeta_7^5 + \zeta_7^6$ = -1
$R(\zeta_7,-1) = -\zeta_7^1 - \zeta_7^2 + \zeta_7^3 - \zeta_7^4 + \zeta_7^5 + \zeta_7^6$ = 2.64575i


and $R(\zeta_7,1) + R(\zeta_7,-1) = 2 \eta_0$.



Update: We can relate back to the definition given in Dummit and Foote:

$$(\alpha, \omega) = \alpha + \omega \sigma(\alpha) + \omega^2 \sigma^2(\alpha) + \cdots + \omega^{n-1} \sigma^{n-1}(\alpha)$$

Now $\sigma$ is a generator of the cyclic Galois group, so in the case of 7 for example we will pick a primitive root of 7, 2 will work and map $\zeta_p^i \mapsto \zeta_p^{2i}$. Not only does this generate the Galois group but you may notice that every second iteration of $\sigma$ gives you a square. So $(\zeta_7, -1)$ agrees with $R(\zeta_7, -1)$ above.
I think the definition in Dummit and Foote is confusing. The following general definition makes more sense:

Definition Let $L/K$ be a Galois extension with cyclic Galois group $G$. Define the Lagrange Resolvent of an element $\theta \in L$ and a character $\chi : G \to K^\times$ to be $$R(\theta,\chi) = \sum_{g \in G} \chi(g) g(\theta).$$

Now we can do a worked example of the problem for $p = 7$. The squares mod 7 are 1,2 and 4.

I believe the characters in question are the principal character $1$ and the character "$-1$" that is $1$ for squares and $-1$ for nonsquares:


$\eta_0 = \zeta_7^1 + \zeta_7^2 + \zeta_7^4$ = -0.5 + 1.32287i
$\eta_1 = \zeta_7^3 + \zeta_7^5 + \zeta_7^6$ = -0.5 - 1.32287i
$R(\zeta_7,1) = \zeta_7^1 + \zeta_7^2 + \zeta_7^3 + \zeta_7^4 + \zeta_7^5 + \zeta_7^6$ = -1
$R(\zeta_7,-1) = -\zeta_7^1 - \zeta_7^2 + \zeta_7^3 - \zeta_7^4 + \zeta_7^5 + \zeta_7^6$ = 2.64575i


and $R(\zeta_7,1) + R(\zeta_7,-1) = 2 \eta_0$.



Update: We can relate back to the definition given in Dummit and Foote:

$$(\alpha, \omega) = \alpha + \omega \sigma(\alpha) + \omega^2 \sigma^2(\alpha) + \cdots + \omega^{n-1} \sigma^{n-1}(\alpha)$$

Now $\sigma$ is a generator of the cyclic Galois group, so in the case of 7 for example we will pick a primitive root of 7, 2 will work and map $\zeta_p^i \mapsto \zeta_p^{2i}$. Not only does this generate the Galois group but you may notice that every second iteration of $\sigma$ gives you a square. So $(\zeta_7, -1)$ agrees with $R(\zeta_7, -1)$ above.