Inequality involving the degree of two different field extensions #5054
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I haven't been able to spot any problems in the PR. Seems like you're getting pretty close to the final result! |
avekens
reviewed
Oct 17, 2025
| with integer coefficients of products of elements of ` A ` . | ||
| (Contributed by Thierry Arnoux, 5-Oct-2025.) $) | ||
| elrgspn $p |- ( ph -> ( X e. ( N ` A ) <-> E. g e. F | ||
| X = ( R gsum ( w e. Word A |-> ( ( g ` w ) .x. ( M gsum w ) ) ) ) ) ) $= |
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Why is the new proof so much longer? Only to eliminate the former hypothesis "( ph -> X e. B )"?
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Yes, only to eliminate the hypothesis.
I eliminated the hypothesis by proving both X e. B from both sides of the final biconditional, which requires closure of the sums, both the (infinite) larger sum and the embedded sum over each word.
avekens
approved these changes
Oct 18, 2025
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This is (finally!) the proof of Bourbaki, Algebra 2, Chapter V, Proposition 5 of , as proposed by @savask.
From here, I should be able to loop back to (towers of) quadratic field extensions.