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I'm having difficulty understanding the proof that Fueter provides in section 5 of his "Ueber kubische diophantische Gleichungen" (published 1930), specifically the material in (case) B.I. I've tried looking but haven't been able to find an English translation of his work.

Some background: The theorem involves the insolvability of $x^3-y^2 = D$ over $mathbbQ$ for some integer $D$. He requires that $D equiv 7 pmod9$, $D notequiv 1 pmod4$ and $D not equiv 4 pmod16$, and 3 does not divide the class number of $mathbbQ(sqrt-D)$.

In B.I., we have the equation $$t_2^3=r^2-27Dt_1^6$$ (which is labeled equation (10') in the text.) This cubic Diophantine equation is associated to the binary cubic form: $v^3 - 3t_2v-2r=0$, and we have $$v = sqrt[3]r+3t_1^3sqrt3D + sqrt[3]r-3t_1^3sqrt3D, quad r notequiv 0 pmod3.$$

Case B.I. is when the equation for $v$ is irreducible over $mathbbQ(sqrt-D)$. As far as I understand it, Fueter proceeds to show that the field extension $K = mathbbQ(sqrt-D)(v)$ is an unramified extension. Applying results using the Hilbert class field then contradicts the condition that $3nmid h_mathbbQ(sqrt-D)$. What I do not understand is how he concludes that $K$ is an unramified extension.

The relevant passage follows:

v legt einen zu $mathbbQ(sqrt-D)$ relativabelschen Korper $K$ fest. Die Relativdiskriminante von $v$ in bezug auf $mathbbQ(sqrt-D)$ ist $d = -(2cdot 3^3 cdot t_1^3) 2 cdot D$. Wegen (10') kann die Relativdiskriminante von $K$ zu $mathbbQ(sqrt-D)$ kein zu $(3)$ teilerfremdes Primideal enthalten. Denn die Radikanden der dritten Wurzeln von $v$ sind dritte Potenzen von Idealen (Beweis die fruher). $(3)$ ist wegen (8) Primideal in $mathbbQ(sqrt-D)$. Es kann ebenfalls nicht in der Relativdiskriminante auftreten. Denn die Radikanden der dritten Wurzeln sind (mod $3sqrt3$) dem Kubus einer Zahl des Korpers $mathbbQ(sqrt3D)$ kongruent, wegen (10'): $$r+3t_1^3sqrt3D equiv r equiv (pm 1)^3 pmod3sqrt3.$$ Die Relativdiskriminante von $K$ zu $mathbbQ(sqrt-D)$ is daher eins, und die Klassenzahl von $mathbbQ(sqrt-D)$ ist durch drei teilbar.

Equation 8 refers to the congruence conditions imposed on $D$. I translate the passage above as follows:

Write $K = mathbbQ(sqrt-D)(v)$. The relative discriminant of $v$ in $mathbbQ(sqrt-D)$ is given by $d$. By equation (10'), the relative discriminant of $K$ to $mathbbQ(sqrt-D)$ cannot be contained in any ideals dividing (that is, lying above) $(3)$. The radicands in $v$ are thus cubes of ideals (proof as before). By (8), $(3)$ is a prime ideal in $mathbbQ(sqrt-D)$. It also cannot occur in the relative discriminant (of $K$ to $mathbbQ(sqrt-D)$). Then the radicands are (modulo $3sqrt3$) perfect cubes of integers in $mathbbQ(sqrt3D)$ and by (10') we have $$r+3t_1^3sqrt3D equiv r equiv (pm 1)^3 pmod3sqrt3.$$ The relative discriminant of $K$ over $mathbbQ(sqrt-D)$ is thus one and the class number of $mathbbQ(sqrt-D)$ is divisible by 3.

Why does Fueter study the irreducibility over $mathbbQ(sqrt-D)$ instead of over $mathbbQ(sqrt3D)$? What does equation (10') have to do with the divisibility of the ideal generated by $d$? What is the relevance of the radicands being perfect cubes of integers in $mathbbQ(sqrt3D)$?