# Thread: How many subspace in F2?

1. ## 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?

2. 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.

3. 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?

4. 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.

5. 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!

6. 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.

7. Better now?

8. Originally Posted by vincisonfire
Better now?
That's more like it.

9. 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: $1 + \sum_{k=1}^m \prod_{j=0}^{k-1} \frac{n^{m-j}-1}{n^{k-j} - 1}.$