How many subspace in F2?

**vincisonfire** · January 11th 2009, 09:22 AM

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?

**ThePerfectHacker** · January 11th 2009, 09:54 AM

Originally Posted by vincisonfire

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.

**vincisonfire** · January 11th 2009, 10:09 AM

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?

**ThePerfectHacker** · January 11th 2009, 10:19 AM

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.

**vincisonfire** · January 11th 2009, 10:32 AM

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!

**Opalg** · January 11th 2009, 11:20 AM

Originally Posted by vincisonfire

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.

**vincisonfire** · January 11th 2009, 11:51 AM

Better now?

**Opalg** · January 11th 2009, 12:01 PM

Originally Posted by vincisonfire

Better now?

That's more like it.

**NonCommAlg** · January 11th 2009, 12:35 PM

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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