The question is about the specific case of reflection theorems (copied straight from Franz Lemmermeyer's "Class Groups of Dihedral Extensions"):



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> Let $k^+ = \mathbb{Q}(\sqrt{m})$ with $m\in \mathbb{N}$, and put $k^- = \mathbb{Q}(\sqrt{-3m})$; then the 3-ranks $r\_3^+$ and $r\_3^-$ of $Cl(k^+)$ and $Cl(k^-)$ satisfy the inequalities $r\_3^+ \le r\_3^- \le r\_3^+ + 1$.
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The proofs I have seen either use p-adic arguments or galois actions.


### Question


Is there an explicit surjective map from $Cl(k^-)[3]$ to $Cl(k^+)[3]$ that might, as the theorem suggests, have kernel of size 3?


At the least, an algorithm for such a map?



Explicit map for Scholz reflection principle
One way to think the reflection principle, which is similar to what you are proposing in your question, is a relation between the index 3 subgroups of $Cl(k^{+})$, which I'll call $I\_{3}(m)$, and the subgroups of $Cl(k^{-})$ order 3 which I'll call $S\_{3}(-3m)$. It is not difficult to see that $$|S\_{3}(-3m)|=\frac{3^{r\_{3}^{-}}-1}{2}$$ and that $$|I\_{3}(m)|=\frac{3^{r\_{3}^{+}}-1}{2},$$ hence any injective map $$ \Phi\_{m}: I\_{3}(m) \rightarrow S\_{3}(-3m)$$ would yield to $r\_{3}^{+}\leq r\_{3}^{-}$. It is a result of Hasse that the set $I\_{3}(m)$ is in bijection with the isomorphism classes of cubic fields of discriminant $m$ (notice that here I'm assuming that $m$ is fundamental, i.e., $m=disc(k^{+})$) hence what we are looking is for a map $\Phi$ that takes a cubic field $K$ and produces a subgroup of $Cl(k^{-})$ of order $3$. In other words given a cubic field $K$ of discriminant $m$ we need to associate a primitive, binary quadratic form of discriminant $-3m$ with the extra condition that the form has order 3 under Gauss composition. To shorten the exposition I'll assume $(3,m)=1$ however all that I'm saying can be worked out in full generality. One natural way to define $\Phi\_{m}$ is as follows: Let $O\_{K}^{0}$ be the set of integral elements in $K$ with zero trace. Let $q\_{K}(x):=Tr(x^{2})/2$. Then, one can show that $q\_{K}(x)$ is a primitive, binary quadratic form of discriminant $-3d$. Moreover, as an element of the class group $q\_{K}^{2}$ has order $3$. It is possible to show that the map $\Phi\_{m}$ sending $K$ to the group generated by $q\_{K}^{2}$ is injective, so the result follows. 


All the above results should be appearing at some point soon in ANT, but I can email you a copy of the article if you are curious of the details. 


**Added:** In response to Alex comment I should say that the other inequality can be also derived with the same ideas I explained above. Now, you start with $I\_{3}(-3m)$ and you notice that $$|I\_{3}(-3m)|=\frac{3^{r\_{3}^{-}}-1}{2}$$ Moreover, $S\_{3}(-3(-3m))=S\_{3}(m)$ hence by using the trace you get a map $$ \Phi\_{-3m}: I\_{3}(-3m) \rightarrow S\_{3}(m).$$ The diference here is that this map is not injective, but it can be shown that roughly the map is 3-to-1 hence the other inequality. So summarizing Scholz reflection principle is a relation between index 3 subgroups in one class group and subgroups of order 3 in the other, and one way to make this relation explicit is via the trace form.


One place to see where the difference in the behavior $\Phi\_{m}$ and $\Phi\_{-3m}$ is, as professor Lemmermeyer already pointed out, Bhargava's first paper on Higher composition laws, more specifically Corollary 15. Another place to look at this is J. W. Hoffman and J. Morales, *Arithmetic of binary cubic
forms*, Enseign. Math. (2) **46**, 2000, 61-94. 


