Do we have [E:K]i=[Er:K] where Er is purely inseparable closure? Let EK be an algebraic extension. Let Es (resp. Er) be the set of elements of E which are separable (resp. purely inseparable) over K. Let [E:K]i=[E:Es]=[E:K][E:K]s, be the inseparable degree. Do we have [E:K]i=[Er:K]?

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Do we have [E:K]i=[Er:K] where Er is purely inseparable closure?  Let EK be an algebraic extension. Let Es (resp. Er) be the set of elements of E which are separable (resp. purely inseparable) over K. Let [E:K]i=[E:Es]=[E:K][E:K]s, be the inseparable degree. Do we have [E:K]i=[Er:K]?

liiipstick0j2 · Accepted Answer

k=F3,K=k(x6,y6),R=K(x+y2),L=K(x3)=K(x3+y6)  LK is a separable quadratic extension, EL is purely inseparable of degree 3 and [E:K]=6.  If there is E/F/K with F/K purely inseparable then [F:K]=3,[E:F]=2.  Let f∈F[T] be the degree 2 minimal polynomial of x+y2.  f stays irreducible over k(x2,y2)⊂F so in fact f=(T−x−y2)(T+x−y2)=T2−2y2T−x2+y4 when y2,x2∈F which is a contradiction as k(x2,y2) has degree 32 over K.

Do we have [E:K]_{i}=[E_{r}:K] where E_{r} is purely inseparable closure? Let

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