Basis for the annihilator Let the field $F$ be given by $F=\mathbb{Z}/5\mathbb{Z}$ and let $V=F^3$. Let $S$ be a subspace of $V$ spanned by $v_1=(1,2,3),v_2=(2,1,2)$. I'm now asked to find a basis for $S^0.$
I've tried solving this and I get $(-1/3,-4/3,1)$. Is it right that the $S^0$ has dimension $1$?

Since $(v_1,v_2)$ are linearly independent vectors, we have $\dim S = 2$ and indeed
$$ \dim S^{0} = \dim V - \dim S = 3 - 2 = 1. $$
To check your answer, note that we indeed have
$$ \left( -\frac{1}{3}, -\frac{4}{3}, 1 \right) \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = -\frac{1}{3} - \frac{8}{3} + 3 = 0, \\
\left( -\frac{1}{3}, -\frac{4}{3}, 1 \right) \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} = -\frac{2}{3} -\frac{4}{3} + 2 = 0 $$
and so your vector belongs to $S^0$ and spans it by dimensional reasons. 
BTW, since you work over $\mathbb{Z}/5\mathbb{Z}$, the $\frac{1}{3}$ stands for the multiplicative inverse of $3$ which is (or, more precisely, can be identified with the equivalence class of) $2$ and so $-\frac{1}{3}$ is identified with $3$. Thus, your spanning vector of $S^0$ can also be written as
$$ (3, 2, 1). $$
