Degree of the maximal $p$-extension inside the ring class field of conductor $p^{\alpha+1}$ Fix a prime $p$. Let $K$ be an imaginary quadratic extension of $\mathbb{Q}$ and suppose that the class number of $K$ is prime to $p$. Let $H_{p^{\alpha+1}}$ be the ring class field of conductor $p^{\alpha+1}$, where $\alpha\ge 0$. Call $K_\alpha$ the maximal $p$-extension of $K$ contained in $H_{p^{\alpha+1}}$.
It seems that $[K_\alpha:K]=p^\alpha$. Why?
 A: Suppose $p\ge 5$ and  $p\nmid d_K$, where $d_K$ is the discriminant of $K$ over $\mathbb{Q}$.
Theorem 7.24 of Cox,  "Primes of the form $x^2+ny^2$" shows that
$h(\mathcal{O})=\frac{h(\mathcal{O}_K) f}{[\mathcal{O}_K^\times:\mathcal{O}^\times]}\prod_{l|f}(1-\Big(\frac{d_K}{l}\Big)\frac{1}{l})$
where $\mathcal{O}_K$ is the ring of integers of $K$, $\mathcal{O}$ is the order of $K$ of conductor $f$, $h(-)$ is the class number, $\Big(\frac{d_K}{l}\Big)$ is the Legendre symbol and the product is among the primes $l$ dividing the conductor $f$.
In our case, $f=p^{\alpha+1}$, hence the formula becomes
$h(\mathcal{O})=\frac{h(\mathcal{O}_K) p^{\alpha+1}}{[\mathcal{O}_K^\times:\mathcal{O}^\times]}(1-\Big(\frac{d_K}{p}\Big)\frac{1}{p})$.
Compute now the $p$-part of this equation. By hypothesis, $h(\mathcal{O}_K)$ is not divisible by $p$, and it is known that $[\mathcal{O}_K^\times:\mathcal{O}^\times]\in\{1,2,3\}$, hence also this factor is not divisible by $p$. Also, we have that $(1-\Big(\frac{d_K}{p}\Big)\frac{1}{p})=\frac{p \pm1}{p}$. This implies that the $p$-part of $h(\mathcal{O})$ is $p^\alpha$.
It seems that when $p=2,3$ or $p| d_K$, this argument fails.
In particular, I want to point out that if $p| d_K$ and $p\ge 5$, we have that the $p$-part of $h(\mathcal{O})$ is $p^{\alpha+1}$.
