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Question Number 169989 by udaythool last updated on 13/May/22

  Prove or disprove:  For any extension E of a field F,  F(u)=F[u]    ∀ u∈E.  Where F(u) is a smallest subfield  of E containing F and u and  F[u]={f(u)∣f(x)∈F[x]}, F[x] is a  polynomial ring over F.

$$ \\ $$$$\mathrm{Prove}\:\mathrm{or}\:\mathrm{disprove}: \\ $$$$\mathrm{For}\:\mathrm{any}\:\mathrm{extension}\:{E}\:\mathrm{of}\:\mathrm{a}\:\mathrm{field}\:{F}, \\ $$$${F}\left({u}\right)={F}\left[{u}\right]\:\:\:\:\forall\:{u}\in{E}. \\ $$$$\mathrm{Where}\:{F}\left({u}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{smallest}\:\mathrm{subfield} \\ $$$$\mathrm{of}\:{E}\:\mathrm{containing}\:{F}\:\mathrm{and}\:{u}\:\mathrm{and} \\ $$$${F}\left[{u}\right]=\left\{{f}\left({u}\right)\mid{f}\left({x}\right)\in{F}\left[{x}\right]\right\},\:{F}\left[{x}\right]\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{polynomial}\:\mathrm{ring}\:\mathrm{over}\:{F}. \\ $$

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