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Thread: How many subspace in F2?

  1. #1
    Senior Member vincisonfire's Avatar
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    How many subspace in F2?

    Hi, I have to find the number of subspaces of  \mathbb F^2 where  \mathbb F is a finite field with n elements.
    I know there are the trivial subspaces.
    I thought the subspaces may be the possible lines ( I mean the possible inclinations )
    For example,  \mathbb F_7^2 would have the lines whose parallel vectors are (0,1) (0,2) ... (0,6) (1,0) (1,1) (1,2) (1,3) ... (1,6) (2,0) (2,1) (2,3) (2,5) etc.
    Is it right?
    Last edited by vincisonfire; Jan 11th 2009 at 10:52 AM.
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    Quote Originally Posted by vincisonfire View Post
    Hi, I have to find the number of subspaces of  \mathbb F^2 where  \mathbb F is a finite field with n elements.
    I know there are the trivial subspaces.
    I thought the subspaces may be the possible lines ( I mean the possible inclinations )
    For example,  \mathbb F_7^2 would have the lines whose parallel vectors are (0,1) (0,2) ... (0,6) (1,0) (1,1) (1,2) (1,3) ... (1,6) (2,0) (2,1) (2,3) (2,5) etc.
    Is it right?
    Here is a hint. A subspace of \mathbb{F}^2 must be an additive subgroup of (\mathbb{F}^2,+). If you take for example, \mathbb{F}_7\times \mathbb{F}_7 the additive subgroups include  \{ 0 \}\times \{ 0\}, \mathbb{F}_7\times \{ 0 \}, \{ 0\}\times \mathbb{F}_7, and \mathbb{F}_7\times \mathbb{F}_7. Now you need to see which ones are also closed under scalar multiplication.
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  3. #3
    Senior Member vincisonfire's Avatar
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    I don't see why for example \{ (a,b) : b = 2a \} does not work.
    (0,0) (1,2) (2,4) (3,6) (4,1) (5,3) (6,5) isn't a additive subgroup?
    And isn't it closed under scalar multiplication?
    Last edited by vincisonfire; Jan 11th 2009 at 11:20 AM.
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    I am sorry! This is my second time making this mistake . For some reason I assume the subgroups of G_1\times G_2 must have form H_1\times H_2 where H_1,H_2 are subgroups. But the point still applies above that a subspace must be a subgroup.
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    Senior Member vincisonfire's Avatar
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    Would it be 3 + n because you have the subspaces "generated" by (1,0) (1,1) ... (1,5) (1,6) and (0,1) plus the two trivial subgroups?
    I think it is that. Thanks a lot!
    Last edited by vincisonfire; Jan 11th 2009 at 12:17 PM.
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    Quote Originally Posted by vincisonfire View Post
    Would it be 3 + n because you have the subspaces "generated" by (1,0) (1,1) ... (1,5) (1,6) and (0,1) plus the two trivial subgroups?
    That is correct. If a subspace S is not the zero subspace then it must contain a nonzero element (x,y). If x=0 then S = \{0\}\times\mathbb{F}. If x≠0 then x^{-1}(x,y) = (1,x^{-1}y)\in S, so S contains all the elements (k,kx^{-1}y). If S contains other elements than those, then you should be able to prove by similar arguments that S is the whole of \mathbb{F}\times\mathbb{F}.

    Edit: That Escher avatar is TOO BIG.
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    Senior Member vincisonfire's Avatar
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    Better now?
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    Quote Originally Posted by vincisonfire View Post
    Better now?
    That's more like it.
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  9. #9
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    there is a formula for the general case: (something to think about!)

    suppose \mathbb{F} is a finite field of order n. the number of subspaces of \mathbb{F}^m of dimension 1 \leq k \leq m is: \prod_{j=0}^{k-1} \frac{n^{m-j}-1}{n^{k-j} - 1}. so the number of all subspaces of \mathbb{F}^m is: 1 + \sum_{k=1}^m \prod_{j=0}^{k-1} \frac{n^{m-j}-1}{n^{k-j} - 1}.
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