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April 17th, 2011, 12:27 AM  #1 
Newbie Joined: Mar 2011 Posts: 14 Thanks: 0  About Field Extension
Let K/k be a field extension. If and ,prove that there are with . thanks 
April 17th, 2011, 12:30 AM  #2 
Newbie Joined: Mar 2011 Posts: 14 Thanks: 0  Re: About Field Extension
Let be a field extension. If and ,prove that there are with .

April 17th, 2011, 01:00 AM  #3 
Newbie Joined: Mar 2011 Posts: 14 Thanks: 0  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. 
April 17th, 2011, 03:43 AM  #4 
Newbie Joined: Apr 2011 From: Milan, Italy Posts: 16 Thanks: 0  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 nontrivial 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] 
April 17th, 2011, 05:58 AM  #5 
Newbie Joined: Mar 2011 Posts: 14 Thanks: 0  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 . 
April 17th, 2011, 07:32 AM  #6 
Newbie Joined: Apr 2011 From: Milan, Italy Posts: 16 Thanks: 0  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

April 18th, 2011, 10:57 PM  #7  
Newbie Joined: Mar 2011 Posts: 14 Thanks: 0  Re: About Field Extension Quote:
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.  
April 20th, 2011, 08:59 AM  #8 
Newbie Joined: Apr 2011 From: Milan, Italy Posts: 16 Thanks: 0  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 ntuple 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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