When $p$ is a prime $\equiv9\bmod16$, the class number, $h$, of $\mathbb Q(p^{1/4})$ is known to be even. In 


[[Charles J. Parry,
A genus theory for quartic fields.
*Crelle's Journal* **314** (1980), 40--71]](http://dx.doi.org/10.1515/crll.1980.314.40)


it is shown that $h/2$ is odd when 2 is not a fourth power in $\mathbb Z/p\mathbb Z$. Does this still hold
when 2 is a fourth power?


Some years ago I gave an (unpublished) proof that this is true
provided the elliptic curve $y^2=x^3-px$ has positive rank, and in particular that it is
true on the B. Sw.-D. hypothesis. It's known that the above curve has positive rank for
primes that $\equiv5$ or $7\bmod16$, but to my knowledge $p\equiv9\bmod16$ remains untouched. But perhaps
there's an elliptic-curve free approach to my question?



If p is a prime congruent to 9 mod 16, can 4 divide the class number of Q(p^(1/4))?
Let $p \equiv 1 \bmod 8$ be a prime number, let $K = {\mathbb Q}(\sqrt[4]{p})$, and let $F$ be the quartic subfield of the field of $p$-th roots of unity. An easy exercise involving Abhyankar's Lemma shows that $FK/K$ is an unramified quadratic extension, hence the class number of $K$ is always even.


The field $KF$ has the quartic subfield $L = {\mathbb Q}(\sqrt{u})$, where $u$ is the fundamental unit of $k = {\mathbb Q}(\sqrt{p})$. An routine application of the ambiguous class number formula to $L/k$ shows that $L$ has odd class number (there are two ramified primes, one infinite and the other one above $2$; clearly $-1$ is not a norm residue at the infinite prime). 


Now I claim that if $p \equiv 9 \bmod 16$, the class number of $KF$ is odd. By class field theory, this implies that the $2$-class number of $K$ must be $2$. An application of the ambiguous class number formula to $KF/L$ shows that the $2$-part of the ambiguous class group has order
$$ h = \frac{2}{(E:H)}, $$
where $E$ is the unit group of $L$ and $H$ its subgroup of units that are norms from all completions of $KF$: in fact, only the two prime ideals above $p$ are ramified in $KF/L$. Thus it is sufficient to show that $E \ne H$. I will show that $\sqrt{u}$ is a quadratic nonresidue modulo the primes $\mathfrak p$ above $p$. But if $u = T + U \sqrt{p}$ (replace $u$ by $u^3$ in order to guarantee that $T$ and $U$ are integers), then $(\sqrt{u}/{\mathfrak p})\_2 = (u/\mathfrak p)\_4 = (T/p)\_4 = (T^2/p)\_8 = (-1/p)\_8 = -1$
because $p \equiv 9 \bmod 16$; here we have used the congruence $T^2 \equiv -1 \bmod p$.


The reason why the case $(2/p)\_4 = -1$ is easier is because in this case, the ideal above
$2$ ramified in $K$ generates a class of order $2$ in the $2$-class group, whereas this prime generates a class with odd order if $(2/p)\_4 = +1$, which means that there is no strongly ambiguous ideal class in this case.


**Edit.** Paul Monsky has kindly written up this argument, filled in all the details, and made it available [here](http://arxiv.org/abs/1009.3990). Thanks!


