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April 17th, 2011, 12:27 AM   #1
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About Field Extension

Let K/k be a field extension. If and ,prove that there are with .

thanks
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April 17th, 2011, 12:30 AM   #2
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Re: About Field Extension

Let be a field extension. If and ,prove that there are with .
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April 17th, 2011, 01:00 AM   #3
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Re: About Field Extension

maybe that problem can also be describe like this:
[attachment=0:27hsfxwr]??.JPG[/attachment:27hsfxwr]
that's a problem in a textbook,too.
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File Type: jpg ??.JPG (25.8 KB, 613 views)
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April 17th, 2011, 03:43 AM   #4
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Re: About Field Extension

hi yuesining,

that's my idea:

We deal with a finite extension field F(S) over F. We denote the degree of F(S) seen as vectorial space over the field F, and it is also the degree of the extension. Since F(S) is finite over F, we can suppose , where n is a positive integer >1. Furthermore, we can assume that S is a subset of E\F, in case not we can drop the elements in F (they are unuseful as generators over F). Now, if S is finite, we can define , or rather define as the set of the linearly independent elements of S. On the other hand, if S is not finite, we can write . Now consider the extension field , generated by over F; since S is a subset of E\F, this extension is not trivial, and so , for some suitable integer . If , we have finished. If not, we take the second element of and consider the field extension ; there are two possibilities: may be 1 or , for some suitable positive integer. In the first case, we have and consequently . In the second one, we obtain a non-trivial extension field, , such that . Multiplicativity of degrees leads to . If , we have finished; if not, we continue with the next step. We are in front of a inductive process that must terminate. In fact suppose ad absurdum that the process doesn't finish. This means that adding all elements of S, one by one, we obtain an extension field of degree <n, impossible! Thus the inductive process terminates after adding a finite number of elements, say . This is the set you were searching for.

If there is any problem, I'm at your service...

Bob...[ITA]
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April 17th, 2011, 05:58 AM   #5
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Re: About Field Extension

I deeply appreciate your help.In the textbook that I'm reading these days ,this proposition is regarded as "facts" so that the textbook provide no proof .But when need prove a proposition on separable extension , it's useful .Thank you very much indeed .
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April 17th, 2011, 07:32 AM   #6
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Re: About Field Extension

you're welcome...i've just added you as a friend..take my msn address, if you want...you never know...XD
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April 18th, 2011, 10:57 PM   #7
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Re: About Field Extension

Quote:
Originally Posted by yuesining
Let K/k be a field extension. If and ,prove that there are with .

thanks
I would say that there maybe some differences between the proposition in layer 2 and layer 1.Then ,how to prove the proposition in layer 1 .
And to roberto.laface
Thank you , and I also have added you as my friend.
In addition,I don't have a MSN account , and prefer to use gmail .
This is my gmail address :yuesiningmath@gmail.com,and I'm glad to be your friend.
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April 20th, 2011, 08:59 AM   #8
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Re: About Field Extension

well...

it's not so far from what I've proved last time...In fact I proved a more general proposition: in layer 1, your request was to show that for each element in there exist a n-tuple of element s.t. is a linear combination of these one with coefficients in (without lost of generality can be supposed ). It's clear that each element may have a different sistem of generators. Instead, we have proved that each element of can be expanded in terms of the same set of generators!!A stronger statement, indeed...

Bob...[ITA]
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