Re: Proving a result in algebra
- From: quasi <quasi@xxxxxxxx>
- Date: Sun, 18 Nov 2007 20:01:28 -0500
On Sun, 18 Nov 2007 16:30:39 -0800 (PST), Ginhead
<ginhead.e@xxxxxxxxx> wrote:
Hi all !
Well, I have been trying to prove the result that follows but so far
only succeeded in the first part, so I hope someone out there will be
kind enough to provide the detailed answer to the second part. Here is
the result :
Fix a prime number P and a natural number N (superior or equal to 1),
then there exists a field F of characteristic P such that any
polynomials of F[X] of degree inferior or equal to N has a zero (or a
root) and F is not algebraically closed.
I have been trying to prove the non algebraically ''closedness'' for
about three days now, so I really would like the solution to end my
misery.
Sketch:
Let K be an algebraic closure of Z_p.
Let F be the subfield generated by all subfields of degree at most N
over Z_p. Thus, F is the smallest field containing all the fields of
degree at most N over Z_p.
Argue that each element a of F must be contained in a field generated
over Z_p by finitely many elements of degree at most N.
Thus, let a be an element of Z_p(b_1, ..., b_k) where each b_i has
degree at most N over Z_p.
Form ascending the tower of fields
Z_p < Z_p(b_1) < ... < Z_p(b_1, ..., b_k)
Then the the dimension of Z(b_1,...,b_k) over Z_p is a product of k
positive integers, all at most N.
It follows that degree of a over Z_p cannot have a prime factor
exceeding N.
But K has elements of all degrees over Z_p.
quasi
.
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- Proving a result in algebra
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