For $p\equiv3\pmod4$, let $h_{\mathbb{Q}(\sqrt{-p})}$ denote the class number of $\mathbb{Q}(\sqrt{-p})$, the class number formula shows $h_{\mathbb{Q}(\sqrt{-p})} < p$. I wonder if there is a pure algebraic proof for $p\nmid h_{\mathbb{Q}(\sqrt{-p})}$, which means avoiding inequality bounds for class numbers.
I've tried the following approach:
By class field theory, the proposition is equivalent to $\mathbb{Q}(\sqrt{-p})$ has no unramified Abel extension with degree $p$. If such extension exists, say $[L:\mathbb{Q}(\sqrt{-p})]=p$, then $\lvert\operatorname{disc}(L)\rvert = p^p$, and Galois theory gives a subfield $K$ with $[K:\mathbb{Q}]=p$, $K$ unramified except $p$.
The complex multiplication shows the Hilbert class field $H=\mathbb{Q}(\sqrt{-p})(j(\frac{-p+\sqrt{-p}}{2}))$, and I know the degree of the characteristic polynomial of $j(\frac{-p+\sqrt{-p}}{2})$ is less than $p$.
By Kummer theory, the proposition is equivalent to the following proposition: For the cyclotomic field $F=\mathbb{Q}(\zeta_p)$ containing $\mathbb{Q}(\sqrt{-p})$, and any elements $a\in\mathbb{Q}(\sqrt{-p}),a\not\in F^{p}$, and $(a)=I^p$ for some ideal, then $F(\sqrt[p]{a})/F$ ramified at $p$. GTM083,Introduction to Cyclotomic Fields(Exercise 9.3) call such number singular primary, and I find a preprint (Queme - Singular integers and p-class group of cyclotomic fields) suggests the existence of such $a$ implies the Vandiver conjecture is false.
Is there anything I've missed?
\midspaces like a binary relation, whereas, I believe,|spaces likemathpunct(can never keep this straight), so you want $p \mid h$p \mid hinstead of $p | h$p | h. But, of course, you actually want non-divisibility, for which $p \not| h$p \not| hbecomes $p \not\mid h$p \not\mid h. This latter can be abbreviated to justp \nmid h. I edited accordingly. $\endgroup$