enfocarteu7z

2022-02-16

Lagrange's rational function theorem states that if one has two rational functions in multiple variables $f({x}_{1},{x}_{2},\dots {x}_{n})$ and $g({x}_{1},{x}_{2},\dots {x}_{n})$ then one can express f as a rational function in g if and only if the set of permutations that keep g unchanged is a subset of the set of permutations that preserve f.

Is anyone familiar with the proof of this theorem? While it is fairly clear that if f can be expressed in terms of g the set of permutations that keep g unchanged has to be the subset of those that keep f unchanged, the converse is far from obvious.

Is anyone familiar with the proof of this theorem? While it is fairly clear that if f can be expressed in terms of g the set of permutations that keep g unchanged has to be the subset of those that keep f unchanged, the converse is far from obvious.

shotokan0758s

Beginner2022-02-17Added 8 answers

Consider the field $K=\mathbb{Q}({x}_{1},{x}_{2},\dots {x}_{n})$ , and consider ${K}_{g}\subset K$ to be the subfield generated by g. Define

$H}_{g}=Aut(\frac{K}{{K}_{g}}=\{\sigma \u03f5Aut\left(K\right):\sigma \left(\alpha \right)=\alpha \mathrm{\forall}\alpha \u03f5{K}_{g}\$

Similarly, define$K}_{f$ and $H}_{f$ . Then you want to show that

$K}_{f}\subset {K}_{g}\iff {H}_{g}\subset {H}_{f$

If all the hypotheses are satisfied, this is merely the Fundamental Theorem of Galois Theory

Similarly, define

If all the hypotheses are satisfied, this is merely the Fundamental Theorem of Galois Theory