Thursday, August 27, 2015

abstract algebra - Let K be a field extension of F and let ainK. Show that [F(a):F(a3)]leq3



Let K be a field extension of F and let aK. Show that [F(a):F(a3)]3. Find examples to illustrate that [F(a):F(a3)] can be 1,2 or 3.



Attempt: FF(a3)F(a)


The minimal polynomial for a3 over F is xa3=0


I, unfortunately, don't have much idea than this on this problem. Could you please tell me how to move ahead?




Let K be an extension of F. Suppose that E1 and E2 are contained in K and are extensions of F. If [E1:F] and [E2:F] are both prime, show that E1=E2 or E1E2=F



Attempt: [K:F]=[K:E1][E1:F]=[K:E2][E2:F]


Since, [E1:F] and [E2:F] are both prime [E2:F] divides [K:E1] and [E1:F] divides [K:E2]


How do i move ahead?


Thank you for your help.


Answer



Consider x3a3, where here a3 is the number which generates F(a3)/F. Then a is a root of this polynomial, and so we will let ma(x) denote the minimal polynomial for a over F(a3). We know




{ma(x)|(x3a3)degma(x)deg(x3a3)=3.



But since


F(a)F(a3)[x]/(ma(x))


we know



[F(a):F(a3)]=deg F(a3)F(a)=deg ma(x)3



For the second question we proceed similarly.



Let FEE1, then by the tower law [E:F]|[E1:F] since [E1:F] is prime, we have [E:F]{1,[E1:F]}, so either E=F or E=E1, and similarly for any subfield of E2.



But then



FE1E2E1E1E2=F or E1



If it's F, we're done, if not then by using the same condition on E2 we see that E1E2=E2 as well (since it's not F). Hence E1=E2.



Note: Why don't we see K in this computation? What is it there for? The answer is that E1E2 doesn't make sense unless they are both subsets of a common set, K is there because of a set theory restriction, but isn't really essential to the proof other than that, unless we're being very pedantic.


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