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Fixed two typos, updated a reference, plus a few other minor tweaks

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edited Oct 31, 2020 at 11:51

It follows from the Cohen-Lenstra heuristic that every finite abelian group is expected to be isomorphic to infinitely many class groups of real quadratic fields (even to a positive proportion of real quadratics, which is stronger), but nothing like this is known.

However, if you take the Galois action into account, then things get interesting: there are Galois modules that are not isomorphic to any class group of a Galois number field with the respective Galois group. See Corollary 4.812 and the bottom ofdiscussion following it on page 1720 in this paper of mine with Lenstra: https://arxiv.org/abs/1803.06903v2 https://arxiv.org/abs/1803.06903v4. You have to read a bit between the lines, but it is shownWe show there that if $G$ is a cyclic group of order degree $58$, then there are finite $\mathbb{Q}[G]$$\mathbb{Z}[G]$-modules that cannot be realised as the class group of a $G$-extension (the "almost all" in the last line of page 17 can be strengthened to "all but one, namely the one for the field $\mathbb{Q}_{\zeta_{59}}$").

Of course, there are cheap ways of doing that, by demanding that theconsidering modules whose fixed submodule is something silly, contradicting the fact that the class group of $\mathbb{Q}$ is trivial, but that is not what is happening in our paper. For example our obstruction cannot be seen by looking at any particular $p$-Sylow of the class group.

It follows from the Cohen-Lenstra heuristic that every finite abelian group is isomorphic to infinitely many class groups of real quadratic fields (even to a positive proportion of real quadratics, which is stronger), but nothing like this is known.

However, if you take the Galois action into account, then things get interesting: there are Galois modules that are not isomorphic to any class group of a Galois number field with the respective Galois group. See Corollary 4.8 and the bottom of page 17 in this paper of mine with Lenstra: https://arxiv.org/abs/1803.06903v2. You have to read a bit between the lines, but it is shown there that if $G$ is a cyclic group of order degree $58$, then there are finite $\mathbb{Q}[G]$-modules that cannot be realised as the class group of a $G$-extension (the "almost all" in the last line of page 17 can be strengthened to "all but one, namely the one for the field $\mathbb{Q}_{\zeta_{59}}$").

Of course, there are cheap ways of doing that, by demanding that the fixed submodule is something silly, contradicting the fact that the class group of $\mathbb{Q}$ is trivial, but that is not what is happening in our paper. For example our obstruction cannot be seen by looking at any particular $p$-Sylow of the class group.

Of course, there are cheap ways of doing that, by considering modules whose fixed submodule is something silly, contradicting the fact that the class group of $\mathbb{Q}$ is trivial, but that is not what is happening in our paper. For example our obstruction cannot be seen by looking at any particular $p$-Sylow of the class group.

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created Jul 16, 2019 at 18:13

Alex B.

However, if you take the Galois action into account, then things get interesting: there are Galois modules that are not isomorphic to any class group of a Galois number field with the respective Galois group. See Corollary 4.8 and the bottom of page 17 in this paper of mine with Lenstra: https://arxiv.org/abs/1803.06903v2. You have to read a bit between the lines, but it is shown there that if $G$ is a cyclic group of order degree $58$, then there are finite $\mathbb{Q}[G]$-modules that cannot be realised as the class group of a $G$-extension (the "almost all" in the last line of page 17 can be strengthened to "all but one, namely the one for the field $\mathbb{Q}_{\zeta_{59}}$").