Let $K/F$ is finite separable extension of fields with Galois closure $L$ and let let $G= \Gal(L/F)$. Then any subfield of $L$ is a resolvent field for $K/F$. These are in bijection with the subgroups of $G$.

One way to group these fields is to collect together subfields with the same Galois closure since given one such field, the other fields are in a collection are visible as siblings of the field. A natural representative of such a collection would be a field of minimal degree over $F$ with ties broken by taking a field whose Galois closure has Galois group $nTj$ with minimal $j$.

In group theoretic terms, $G\leq S_n$ where $n=[K:F]$. The **resolvents** of $G$ correspond to quotients $G/N$ where $N \unlhd G$. For each normal subgroup $N$, we then look for $H\leq G$ such that the core of $H$ equals $N$, and then pick the subgroup $H$ of minimal index in $G$ so that when $G$ acts transitively on the $m$ left cosets of $H$ by left translation, the T-number of the permutation reputation is minimized.

The fixed field of $H$ is then a stem field for $L^N$, the fixed field of $N$ over $F$. The Galois group for $L^N/F$ has is a subgroup of $S_m$, up to conjugation.

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- Review status: reviewed
- Last edited by John Jones on 2019-09-15 18:46:57

**History:**(expand/hide all)

- 2019-09-15 18:46:57 by John Jones (Reviewed)
- 2019-09-04 20:35:24 by John Jones
- 2019-09-04 20:31:49 by John Jones
- 2019-05-22 17:46:37 by John Jones
- 2019-05-03 16:37:43 by Kiran S. Kedlaya (Reviewed)
- 2019-05-03 00:20:04 by John Jones
- 2018-07-07 21:31:09 by John Jones

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