I need to prove that $\{v=(6,9),w=(7,8)\}\in Z_{p}\times Z_{p}$ is linearly independent when $p=2$ and linearly dependent when $p=3$. The problem is my freshman algebra course did not cover rings and I'm a bit lost.
My approach to the problem has been trying to find $a_1,a_2\in Z$ such that $a_1 6+a_2 7 \equiv 0\mod 2$ and $a_1 9+a_2 8 \equiv 0\mod 2$ ,that is:$a_1 6+a_2 7=2k$ and $a_1 9+a_2 8=2k'$ for some $k,k'\in Z$. But I don't seem to get to anything, is this approach bad? Another thing, is there any way to find all the valors of $p$ for which the set is linearly dependent? Thanks for your help.
For $p=2$, $v=(0,1)$ and $w=(1,0)$ . Then $a.v +b.w =0$ implies $(a,b)=(0,0)$ for which to hold both $a=0$ and $b=0$ must hold.
For $p=3$ , $v=(0,0)$ and $w=(1,2)$. Now , you know that for any vector space , any subset containing the $0$ vector is linearly dependant . So the set $\{(0,0),(1,2)\}$ is linearly dependant. Or more precisely , $0.w + 1.v = (0,0)$ holds for $p=3$.