Let $K/\mathbb{Q}$ be an imaginary quadratic extension of discriminant $-D_K < -3$ and fix a prime $p > 3$ that is split in $K$ as $p\mathcal{O}_K = \mathfrak{p}\overline{\mathfrak{p}}$. Consider the character $$ \psi: I_K^{(p)} \rightarrow \{\pm 1\}$$ $$ \mathfrak{q} \mapsto \left(\frac{N(\mathfrak{q})}{p}\right)$$
where $N(\cdot)$ is the norm map, and $\left(\frac{\cdot}{p}\right)$ is the Legendre symbol. As far as I can see, $P_{K, 1}^{(p)} = \{x \in K \mid x - 1 \in p \mathcal{O}_K\}$ is in the kernel of this map. So this character defines a degree-$2$ extension of $K$, which we denote by $L$. The reason this is not the trivial character is because there exists a prime $q$ split in $K$ such that $(\frac{q}{p}) = -1$ (otherwise we would have $\mathbb{Q}(\sqrt{(-1)^{(p - 1)/2}p})$ be contained in $K$).
Here is an interesting property of $L$: $q$ is split in $L$ for every rational prime $q$ that is inert in $K$, because $\psi(q) = \left(\frac{N(q)}{p}\right) = \left(\frac{q^2}{p}\right) = 1$. So $L$ is contained in $H_K$, the Hilbert class field of $K$. But that's saying that the class number of $K$ must be even. Clearly, there are imaginary quadratic fields with odd class number. So, I would love to see where my mistake could have come from.
The reason I am interested in this character is, I was hoping that if I look at it as a Hecke character than the restriction to $\mathcal{O}_{K, \mathfrak{p}}^{\times}$ would also give me the (lift of) the Legendre symbol, which would be helpful for a problem that I am looking at.
