Re: Simple extension of fields
- From: quasi <quasi@xxxxxxxx>
- Date: Fri, 16 Jan 2009 16:25:18 -0500
On Thu, 15 Jan 2009 21:21:11 +0100, "Jean Pierre MERX"
<firstname.name@xxxxxxxxxxxxxxxxx> wrote:
How to prove following result:
If k \subset K is a field extension and if for every intermediate fields M,
N we have M \subset N or N \subset M then K is a simple extension of k.
I try to use a transcendance basis in order to reduce the case to the one
where K is an algebraic extension of k. But I'm not able to get a conclusion
for this "simpler case"
Firstly, it's easy to show that K must be algebraic over k, otherwise,
if K contains an element x say, where x is transcendental over k, then
the fact that neither of the fields k(x^2) and k(x^3) contains the
other contradicts the hypothesis.
Next consider the case where [K : k] is finite. Choose a in K such
that [k(a): k] is greatest. If K = k(a) we are done, hence assume k(a)
is a proper subfield of K. Choose b in K, b not in k(a). But then k(b)
is not a subfield of k(a) (since b is not in k(a)), hence, by
hypothesis, k(a) must be a proper subfield of k(b), which implies
[k(b) : k] > [k(a) : k], contrary to our choice of a.
It remains to consider the case where K is algebraic over k but [K :
k] is infinite. But for this case, I'm not sure the claimed result
holds. As a proposed counterexample, consider
k = Q and K = Q(2^(1/2), 2^(1/4), 2^(1/8), ...)
Unless I'm missing something simple (no pun intended), the hypothesis
is satisfied, but K is not a simple extension of k.
quasi
.
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- From: Jean Pierre MERX
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