Let $\mathbb{K}$ be a finite field of order $q$ and with $p = char(\mathbb{K})$. Let $\mathbb{K} = \{\lambda_1 , \dots , \lambda_q\}$. I know that
$x^q - x = \prod_{i=1}^{q}{(x-\lambda_i)}$
So if I expand the RHS I get
$$x^q - x = \sum_{k=0}^{q}{(-1)^{q-k} x^k e_{q-k}(\lambda_0 , \dots , \lambda_q)}$$
Where $e_k$ is the $k$-th elementary symmetric polynomial in $q$-variables
Since the equation holds componentwise I get
$$ e_{k}(\lambda_1 , \dots , \lambda_q) = \sum_{i_1 < \dots < i_k }{ \lambda_{i_1} \cdot \dots \cdot \lambda_{i_k} } = (-1)^{k+1} \delta^{k}_{q-1}$$
Where $\delta^{i}_j$ denotes the Kronecker delta symbol.
I consider now the ring $R$ of the polynomial in $q$-variables over the ring $\mathbb{Z}$, so $$R := \mathbb{Z}[x_1 , \dots , x_q]$$
I consider the homomorphism map $\pi \, : \, R \to \mathbb{K}$ such that $\pi(x_i) = \lambda_i$ (and $\pi(1) = 1$ )
Clearly $\frac{\mathbb{Z}[x_1 , \dots , x_q]}{ker(\pi)} \cong \mathbb{K}$
From the previous discussion, I know that
$e_k(x_1, \dots , x_q) + (-1)^k \delta^k_{q-1} \in ker(\pi)$
Also, since $p =char(\mathbb{K})$, I know that $p \in ker(\pi)$
My question is, if I let $I$ be the ideal generated by $p$ and the elements $e_k(x_1 , \dots , x_q) + (-1)^k \delta^k_{q-1}$, is it true that $I = ker(\pi)$ ?