Let K be a field extension of F and let a∈K. Show that [F(a):F(a3)]≤3. Find examples to illustrate that [F(a):F(a3)] can be 1,2 or 3.
Attempt: F⊂F(a3)⊆F(a)
The minimal polynomial for a3 over F is x−a3=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 E1⋂E2=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 x3−a3, 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)|(x3−a3)degma(x)≤deg(x3−a3)=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 F⊆E′⊆E1, 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
F⊆E1∩E2⊆E1⟹E1∩E2=F or E1
If it's F, we're done, if not then by using the same condition on E2 we see that E1∩E2=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 E1∩E2 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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