$\mathfrak{B}=((1,2,0)^t ,(2,1,2)^t ,(3,1,1)^t)$ is a basis of $\mathbb{R}^3$. For what prime numbers p is $\mathfrak{B}$ a basis of $\mathbb{F}^3_p$? I'm not quite sure, but it seems to me that $\mathbb{F}^3_p$ is the finite field in three dimensions with $p$ elements inside it, where $p$ is the prime number(s) we are looking for. A finite field is simply a field with a finite number of elements, the finite number of elements must always be a prime number
So they are asking for which specific prime numbers the set $\mathfrak{B}$ is still a basis. To be a basis, the elements inside the set $\mathfrak{B}$ need to span the entire vector space, that is
$a_1v_1 + a_2v_2 + a_3v_3 = x$ for any $x$ element in the space
And the base elements need to be linearly independent, that is
$a_1v_1 + a_2v_2 + a_3v_3 = 0$ implies $a_1 = a_2 = a_3 = 0$
In our case $v_1 = (1,2,0)^t$, $v_2 = (2,1,2)^t$ and $v_3 = (3,1,1)^t$
First, is this correct ?
Second, I don't see what to do now. In the exercise before this one, we verified that $\mathfrak{B}$ is a basis of $\mathbb{R^3}$
It seems to me that $p$ could be any prime number, because $\mathbb{F}_p^3$ is in three dimensions, and $\mathfrak{B}$ spans all of $\mathbb{R^3}$. Why would $\mathfrak{B}$ only be a basis for some prime numbers and not others, if we know that $\mathfrak{B}$ spans all of $\mathbb{R^3}$ ?
Thanks for your help
 A: If you have $n$ column vectors, each with $n$ coordinates, you can form a matrix from them and then the determinant is $0$ if and only if the columns are linearly dependent over the field you're working in. So your next step will be to compute the determinant of a $3\times3$ matrix.
What can go wrong in characteristic $p$? Well, consider a simpler example:
$$ \mathcal{B}=\left\{\begin{pmatrix}1\\0\end{pmatrix},\begin{pmatrix}1\\n\end{pmatrix}\right\}. $$
This will be a basis for $\mathbb{F}^2$ over any field $\mathbb{F}$ where $n\ne0$, but will be linearly dependent if $n=0$. Thus, it will be linearly dependent over any field of characteristic $p$, where $p$ is a prime factor of $n$. (Note it's not about how "large" $n$ or $p$ are relative to each other.)
The determinant in this case is $7$. That means, for instance, over the integers they are linearly independent - there is no "$\mathbb{Z}$-linear combination" which yields the zero vector. However, in the integers mod $7$ for instance, we're basically asking about an integer combination yielding not just the zero vector, but potentially any vector with multiples of $7$ in it (which would be the zero vector mod $7$). So for instance,
$$ 1\begin{pmatrix}1\\2\\0\end{pmatrix} +
2\begin{pmatrix}2\\1\\2\end{pmatrix} + 
3\begin{pmatrix}3\\1\\1\end{pmatrix} =
\begin{pmatrix}14\\7\\7 \end{pmatrix} $$
does not tell us our vectors are linearly dependent over $\mathbb{Z}$, but it does tell us they are linearly dependent mod $7$. (They will be independent in any other characteristic, though.)
