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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.

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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 $$\Phi: Phi_{m}: I_{3}(m) \rightarrow S_{3}(-3m)$$ 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 $Cl(k^{-})$ 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 $\Phi$ \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$\Phi$\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 cubicforms, Enseign. Math. (2) 46, 2000, 61-94.

1

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 $$\Phi: I_{3}(m) \rightarrow S_{3}(-3m)$$ 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 $Cl(k^{-})$ 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 $\Phi$ 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 $\Phi$ 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.