class number of $\mathcal O_D$ is odd for $D\equiv1 \pmod 4$, quadratic imaginary fields Take $\ell$ prime such that $\ell \equiv -1 \pmod 4$ and take $D=-\ell$ or $D=-4\ell$. Then, I have seen as a statement in a paper that the class number of the ring of integers of $K=\mathbb Q(\sqrt{-D})$ is odd. That means $\mathcal O_D=\mathbb Z [\frac{1}{2}(1+\sqrt{-D})]$ and discriminant=$-D$ and we want  to prove $\mathcal O_D$ has odd class number. I am not sure how obvious or hard this is so I wanted to ask: Is there an easy explanation or a reference for that?
 A: I assume that $K=\Bbb Q(\sqrt{D})$ where $D=-\ell$ and $\ell\equiv-1\pmod 4$ is a prime (instead of $\Bbb Q(\sqrt{-D})$ because of the title and the statement about $\mathcal O_K$) (In the case $K=\Bbb Q(\sqrt{-D})$ the same argument almost works, one has to take care of the prime $2$ which ramifies in $K$ in this case, see the comment below). Here is an elementary argument that the class number of $\mathcal O_K$ is odd. Let $\sigma$ be the non-trivial automorphism of $K$.
Lemma: Let $I$ be an ideal. Then $[\sigma(I)]=[I]^{-1}$.
Proof: It suffices to prove this in the case $I=P$ prime. Let $p$ be the rational prime lying under $P$. By considering the different possible factorizations of $p$ in $\mathcal O_K$ one easily sees the claim (note that if $p\mathcal O_K=PQ$, then $\sigma(P)=Q$).
Now let $[I]$ be any ideal class with $[I]^2=1$. By the lemma we have $I\sigma(I)^{-1}=\alpha\mathcal O_K$ for some $\alpha\in K$. Note that $N(\alpha)=1$ so that there is a $\beta\in K$ such that $\alpha=\sigma(\beta)/\beta$ by Hilbert 90. Thus, $J:=I\beta$ satisfies $[J]=[I]$ and $J=\sigma(J)$. Thus, $J^2=J\sigma(J)=(N(J))$ and $N(J)\in\Bbb Q$. Write $N(J)=p_1^{a_1}\cdots p_n^{a_n}\ell^b$ for distinct rational primes $p_1,\dots,p_n\ne\ell$ and $a_1,\dots,a_n,b\in\Bbb Z$. Since only $\ell$ is ramified in $K$, each $p_i$ either stays prime oder splits into two distinct prime ideals in $\mathcal O_K$. Hence, by unique factorization and $J=\sigma(J)$ we get that the $a_i$ are even and $J=(p_1^{a_1/2}\cdots p_n^{a_n/2}\sqrt{D}^{k})$ (note that $(\sqrt{D})$ is the prime lying above $\ell$). Thus $1=[J]=[I]$. Therefore the class group of $\mathcal O_K$ contains no element of order $2$ and hence has odd order.
(the idea comes from this answer.)
Something more general is true:
Theorem: Let $D\equiv1\pmod 4$ be negative and $r$ the number of (odd) primes dividing $D$. Then the class group of integer binary quadratic forms with discriminant $D$ has exactly $2^{r-1}$ elements of order $\leq2$.
(Cox "Primes of the form $x^2+ny^2$", p. 52)
Theorem: $D$ as in the previous theorem. The class group of integer binary quadratic forms with discriminant $D$ is isomorphic to the class group of $\Bbb Q(\sqrt{D})$.
(Cox, p. 113)
In your case $D=-\ell$ we have $r=1$, so there is no element of order $2$ so that the class group has odd order.
