My Math Forum About Field Extension

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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 $A \subseteq K$ and $u \in k(A)$,prove that there are $a_1 ,..., a_n \in A$ with $u \in k(a_1 ,..., a_n)$. thanks
 April 17th, 2011, 12:30 AM #2 Newbie   Joined: Mar 2011 Posts: 14 Thanks: 0 Re: About Field Extension Let $K/k$ be a field extension. If $A \subseteq K$ and $u \in k(A)$,prove that there are $a_1 ,..., a_n \in A$ with $u \in k(a_1 ,..., a_n)$.
April 17th, 2011, 01:00 AM   #3
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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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 ??.JPG (25.8 KB, 613 views)

 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 $[F(S):F]$ 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 $[F(S):F]=n$, 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 $S_0 := S$, or rather define $S_0$ as the set of the linearly independent elements of S. On the other hand, if S is not finite, we can write $S=\{s_1, \dots, s_k, \dots \}$. Now consider the extension field $F(s_1)$, generated by $s_1$ over F; since S is a subset of E\F, this extension is not trivial, and so $[F(s_1):F]=k_1$, for some suitable integer $k_1>1$. If $k_1=n$, we have finished. If not, we take the second element of $s_2$ and consider the field extension $F(s_1,s_2)$; there are two possibilities: $[F(s_1,s_2):F(s_1)]$ may be 1 or $k_2>1$, for some suitable positive integer. In the first case, we have $s_2 \in F(s_1)$ and consequently $F(s_1,s_2)=F(s_1)$. In the second one, we obtain a non-trivial extension field, $F(s_1,s_2)$, such that $[F(s_1,s_2):F(s_1)]=k_2$. Multiplicativity of degrees leads to $[F(s_1,s_2):F]= k_1 k_2$. If $n=k_1 k_2$, 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
 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
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Quote:
 Originally Posted by yuesining Let K/k be a field extension. If $A \subseteq K$ and $u \in k(A)$,prove that there are $a_1 ,..., a_n \in A$ with $u \in k(a_1 ,..., a_n)$. 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 .

 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 $u \in k(A)$ there exist a n-tuple of element $a_1, \dots , a_n \in A$ s.t. $u$ is a linear combination of these one with coefficients in $k$ (without lost of generality can be supposed $A \cap k= void$). It's clear that each element may have a different sistem of generators. Instead, we have proved that each element of $k(A)$ can be expanded in terms of the same set of generators!!A stronger statement, indeed... Bob...[ITA]

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