This field has Galois group \(\mathrm{GL}(1,\mathbb{Z}/20) \cong \mathrm{GL}(1,\mathbb{Z}/4) \times \mathrm{GL}(1,\mathbb{Z}/5)\):
so its poset of subfields is isomorphic to the opposite of the poset of subgroups of \(\mathrm{GL}(1,\mathbb{Z}/4) \times \mathrm{GL}(1,\mathbb{Z}/5)\):
(Pictures by Simon Burton.)
The subfields of the 20th cyclotomic field correspond to the subgroups of its Galois group, and we can draw the lattice of these subfields, and indicate which steps up this lattice involve ramification at 2, 5, or the real prime.
(Pictures by Simon Burton.)
There are 30 such subgroups, and we construct a chart of 12 of them mimicking the method used on February 16th, 2023 for the 20th cyclotomic field.
The theory works most simply when \(k\) contains (or in our conversation is) the Eisenstein field: the rationals with a primitive cube root of \(1\) adjoined. Then \(\mathbb{Z}/3\) splits over \(k\), so one can show that a \(\mathbb{Z}/3\)-torsor over \(K\) amounts to a line object \(L\) in the category of \(k\)-vector spaces together with a trivialization of its tensor cube: $$ F \colon 1_{\otimes} \to L^{\otimes 3} $$ where the tensor unit \(1_{\otimes} \in \mathrm{Vect}\) is just the field \(k\). Even more concretely, we get a \(\mathbb{Z}/3\)-torsor over \(k\) from a nonzero element \(A \in k\): to do this, we take \(L = k\) and take \(F\) to be multiplication by \(A\). Then two choices of \(A\), say \(A\) and \(A'\), give isomorphic \(\mathbb{Z}/3\)-torsors iff \(A' = B^3A\) for some nonzero \(B \in k\).
When \(k\) is the Eisenstein field, any cubic extension \(K\) is a 6th-degree extension of \(\mathbb{Q}\). We erroneously claim that this is always a Galois extension of \(\mathbb{Q}\). When it actually is, the Galois group of \(K\) over \(\mathbb{Q}\) can thus be either \(\mathbb{Z}/6\), which is abelian, or \(S_3\), which is not. We thus consider this challenge: given \(A \ne 0\) in the Eisenstein field \(k\), determine whether the Galois group of \(K = \sqrt[3]{A}\) over \(\mathbb{Q}\) is abelian or not.
Next, what if \(k\) does not contain the Eisenstein field? Then we can redo our previous discussion using a formal 'Eisenstein object' \(E\) replacing the tensor unit in \(\mathsf{Vect}\), and an 'Eisenstein line object' \(L_E\) replacing \(L\), with a trivialization of its tensor cube: $$ F \colon E \to L_E^{\otimes_E 3} $$
The theory works most simply when \(k\) contains (or is) the Eisenstein field: the rationals with a primitive cube root of 1 adjoined. Then \(\mathbb{Z}/3\) splits over \(k\), so one can show that a \(\mathbb{Z}/3\)-torsor over \(K\) amounts to a line object \(L\) in the category of \(k\)-vector spaces together with a trivialization of its tensor cube. Even more concretely, we get a \(\mathbb{Z}/3\)-torsor over \(k\) from a cube root of nonzero element \(A \in k\). Then two choices of \(A\), say \(A\) and \(A'\), give isomorphic \(\mathbb{Z}/3\)-torsors iff \(A' = B^3A\) for some nonzero \(B \in k\).
When \(k\) is the Eisenstein field, any cubic extension \(K\) is a 6th-degree extension of \(\mathbb{Q}\), and in fact (we claim) a Galois extension. The Galois group of \(K\) over \(\mathbb{Q}\) can thus be either \(\mathbb{Z}/6\), which is abelian, or \(S_3\), which is not. We thus consider this challenge: given \(A \ne 0\) in the Eisenstein field \(k\), determine whether the Galois group of \(K\) over \(\mathbb{Q}\) is abelian or not.
Next, what if \(k\) does not contain the Eisenstein field? Then we can redo our previous discussion using a formal 'Eisenstein object' \(E\) replacing the tensor unit in \(\mathsf{Vect}\), and an 'Eisenstein line object' replacing \(L\).
As already discussed in our April 20, 2023 talk, the 'theory of \(\mathbb{Z}/3\) torsors over \(k\)' is another way of talking about the category of representations of \(\mathbb{Z}/3\) on rational vector spaces. 'Theories' here are symmetric monoidal locally presentable categories, and we can also describe this theory by saying it's the free symmetric monoidal locally presentable category on an Eisenstein line object \(L\) with a trivialization of its tensor cube. We can further constrain this theory by saying that the Eisenstein conjugate of \(L\) is isomorphic to its tensor square (in a way compatible with the above trivialization). This extra constraint makes it so models of this theory in the category of vector spaces over \(k\) correspond to abelian cubic extensions of the number field \(k\).
We can also state this all more concretely, 'gauge-fixing' by taking \(L\) to be the 'standard' Eistenstein line object \(E\) which can be taken to be the sum of two copies of the unit for the tensor product in our theory. We can actually put this into the theory, equipping our theory an isomorphism \(G \colon L \to E\). Then the models of the theory are given by solutions to some polynomial equations, which we outline.
\(\mathrm{Rep}(G)\) is generated, as a 2-rig, by the regular representation of G, which we could call \(r(G)\). But \(\mathrm{Rep}(G)\) also contains another special object \(k^G\): the algebra of functions on the finite set \(G\), treated as a trivial representation of \(G\). This is a coalgebra in \(\mathrm{Rep}(G)\) whose comultiplication encodes the multiplication in \(G\). This coalgebra \(k^G\) has a coaction on \(r(G)\), \(r(G) \to r(G) \otimes k^G\). This coaction makes \(r(G)\) into what James calls a '\(G\)-torsor in \(\mathrm{Rep}(G)\)', meaning that there is also a 'codivision' map \(k^G \to r(G) \otimes r(G)\) obeying a certain equation. All this is dual to how a nonempty set \(X\) is a torsor of \(G\) if it has an action \(G \times X \to X\) together with a division map \(X \times X \to G\) obeying a certain equation.
Indeed, any 2-rig \(R\) contains a Hopf object that we could call \(k^G\): it's the direct sum of \(G\) copies of the unit object, with a comultiplication coming from multiplication in \(G\). We can use this to define a concept of a '\(G\)-torsor in R', copying the definition just given in \(\mathrm{Rep}(G)\).
Such a \(G\)-torsor in a 2-rig \(R\) works out to be the secretly the same thing as a 2-rig map \(F \colon \mathrm{Rep}(G) \to R\) since we can determine such an \(F\) by choosing any \(G\)-torsor in \(R\), and vice versa. Thus, James usually shortcuts the whole process and defines a \(G\)-torsor in a 2-rig \(R\) simply to be a 2-rig map \(F \colon \mathrm{Rep}(G) \to R\).
Finally, suppose \(L\) is any field extending \(k\) with Galois group \(G\). Then \(L\) gives an algebra object in \(\mathrm{Vect}\) which is also a \(G\)-torsor in \(\mathrm{Vect}\).