# Quotient Space over finite field.

I'm looking at a vector space $V = F^3$ where $F = \{0,1,2\}$ the field with three elements. I have a subspace $W = \text{span}(1,2,1)$ and I'm trying to explicitly describe the quotient space $V/W$.

I know the equivalence class for the zero vector is $W$ which is to say $\{(0,0,0),(1,2,1),(2,1,2)\}$. I know I need two other vectors that form a basis with $(1,2,1)$ for V. Take $(1,1,0)$ and $(0,0,1)$. Then to find the equivalence class for these two vectors I look at $E(v) = v + w; \forall w \in W$. So we get $$E(1,1,0) = \{(1,1,0),(2,0,1),(0,1,2)\}$$ and $$E(0,0,1) = \{(0,0,1),(1,2,2),(2,1,0)\}$$

My question is I know dim($V/W$) = dim($V$) - dim($W$) so I'm looking for two vectors in the quotient space that are linearly independent to form a basis but I don't think the above is correct because if I look at $E(1,1,0) + E(0,0,1)$ I should get $E(1,1,1)$ which does not appear in any of my equivalence classes. Have I made a mistake or misunderstood what is supposed to happen?

If someone could point me in the right direction I would greatly appreciate it.

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Did you mean that you're looking at a vector space of dimension three over $\,F\,$ and etc.? Otherwise the question is hard to understand... – DonAntonio Mar 27 '13 at 4:07
The quotient space V/W is of dimension 2, it means you can generate all the equivalence classes with 2 elements. There are more than 2 equivalance classes! (there are 9 equivalence classes). – Long Mai Mar 27 '13 at 4:18
@DonAntonio yes V is a vector space of dimension 3 over the field with three elements. – AvatarOfChronos Mar 27 '13 at 10:03
@LongMai can you expand on your comment please? – AvatarOfChronos Mar 27 '13 at 10:04
@LongMai , It has just occurred to me why there should be 9 equivalence classes and not 2. I seems to have confused some stuff in my class notes. Thanks for your comment. – AvatarOfChronos Mar 27 '13 at 11:20

The two elements you have picked are a basis. I'm not sure how you are calculating $E(1, 1, 0) + E(0, 0, 1)$ but by definition it is $E((1, 1, 0) + (0, 0, 1)) = E(1, 1, 1)$.