Is this set a basis of $F_{3}^{1*3}$ or $F_{2}^{1*3}$?

Question

Is this set a basis of $F_{3}^{1*3}$ or $F_{2}^{1*3}?$:

{0,1,1),(1,0,1),(1,1,0)}

This was a question on my last quiz and I decided that it was a basis for $F_{2}^{1*3}$, because there are only 1 and 0. But the answer had to be $F_{3}^{1*3}$ Can someone help me understand why it was a basis of residue field 3 instead of 2?

Thanks in advance!

Answer 1

Even though {$(0,1,1),(1,0,1),(1,1,0)$} could be a collection of vectors over $F_2$,

it is not a basis for a three-dimensional space over $F_2,$

because these three vectors are not linearly independent over $F_2$,

because, over $F_2, (1,1,0)=(0,1,1)+(1,0,1).$

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They are linearly independent over $F_3$, because if $a_1(0,1,1)+a_2(1,0,1)+a_3(1,1,0)=(0,0,0),$

where $a_1,a_2,a_3\in F_3,$ then $(a_2+a_3, a_1+a_3, a_1+a_2)=(0,0,0)$.

The difference of the third and first components implies $a_1-a_3=0$;

combining this with the second component $a_1+a_3=0$ means $2a_1=0$,

which implies $a_1=0$ in fields not of characteristic $2$ such as $F_3$.

From $a_1=0$ and $a_1+a_3=a_1+a_2=0,$ it follows that $a_2=a_3=0$ too.