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Consider the field $\mathbb{F}_p$ for $p$ prime. Find the number of subspaces of dimension 1 and 2 of the three dimenstional vector space $\mathbb{F}_p^3$. I know that $\mathbb{F}_p$ is generated by all elements relatively prime to $p$, but two subspaces are supposed to be equal if their bases can be expressed as linear combinations of each other. So aren't all 1 dimensional subspaces of $\mathbb{F}_p^3$ isomorphic to each other and in particular $\mathbb{F}_p$? Similarly for subspaces of dimension 2, you need two elements to generate a subset but all are linear combinations of one another so there can only be one.

This doesn't sound right, but I'm not sure what to do. I can't seem to find any information that would help figure this out.

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Yes, all subspaces of a given dimension are isomorphic to each other, but the question does not ask you to count subspaces up to isomorphism, it just asks you to count subspaces. So e.g. the subspace spanned by $(0,1,1)$ and the subspace spanned by $(1,0,1)$ are different. On the other hand, the subspace spanned by $(0,1,1)$ and the subspace spanned by $(0,2,2)$ (if $p \ne 2$) are the same.

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Thanks. I guess I was making it a little more complicated than it needed to be. –  anon Sep 18 '11 at 17:29
    
Just to make sure I understand this, subspaces of one dimension can only have 1 coordinate non-zero, and since in $\mathbb{F}_p$ all elements are linear combinations of one another, then there are only 3 subspaces. Then in your example, the subspace spanned by $(0,1,1)$ and the subspace spanned by $(0,1,2)$ would not be equal. –  anon Sep 18 '11 at 17:48
    
I think you're still confused. What do you mean by "only have 1 coordinate non-zero"? –  Robert Israel Sep 18 '11 at 19:06
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You're right, I am. Fortunately, I managed to find a good section in Dummit and Foote that is working a lot better than the notes provided by my professor. I think I meant that a subspace spanned by (0,1,1) is not one-dimensional since there are two coordinates that are non-zero, which is definitely not correct. The dimension of a (finite) subspace is equal to the order of the basis, so if the basis is generated by a single element, then the subspace generated by that basis is one-dimensional. –  anon Sep 18 '11 at 19:13

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