Are finite fields subfields of real numbers field? I am trying to prove that if a set of $n-$dimensional vectors, $v_1, \dots, v_m$, are linearly independent over a finite field $GF(p)$, then they are linearly independent over the real field.
I came across this link, Linear independence of vectors over larger fields, which provides a proof for a similar concept but for a subfield $\mathbb{F}$ of a larger field $\mathbb{G}$. I hope I can extend this argument if I can show that a finite field $GF(p)$ is a subfield of the real field.
Thanks
 A: A subfield has the same characteristic as the ambient field, in particular, any subfield of $\mathbb R$ has characteristic $0$, thus cannot be finite.
A: A finite field can't be a subfield of the real field since in the finite field $1 + 1 + \cdots + 1 = 0$ (the number of summands is the characteristic of the field).
A: Finite fields are not subfields of $\mathbb{R}$ because of,for example, torsion. 
To focus on the question in the body of your text -- how are you comparing the vectors, whose entries are a priori elements of $F_q$, with vectors whose entries are real? 
If your $F_q$ is such that $q$ is a prime, then you can try to pick integer representatives - i.e. above $[a] \in F_q$, thought of as a residue class, you pick the integer $a$. The determinant of this matrix of integers will necessarily be nonzero if the vectors are independent over $F_q$, because $mod q$ the determinant will be nonzero, and this will prove that the matrix of integers, thought of now as real vectors, consists of independent vectors. (Mod q is a ring homomorphism, so you can apply it before or after the computation of the determinant.)
In general you could try to look for a surjective ring homomorphism $R \to F_q$, and an inclusion $R \to \mathbb{\mathbb{C}}$. I'm not sure in general what is true here, but I would guess that you could try to adjoin various complex numbers to $\mathbb{Z}$, and produce maps down to $F_q$ by the universal property for the polynomial ring. (For example, consider $R = \mathbb{Z} [\sqrt{2}] = \mathbb{Z} [x] / (x^2 - 2)$. Maps from this ring to any other ring $A$ are determined by sending $x$ to a square root of $2$ in $A$. So if you took $F_3$, and adjoined a square root of $2$ to it, i.e. you worked with $A = F_3[x]/(x^2 - 2)$, then you get a map from $R$ to $A$, that takes a polynomial (mod $x^2 - 2$) and reduces the coefficients modulo 3. $R \subset \mathbb{R}$, so you can play this game. However, I'm not sure to what extent this procedure is ad-hoc, or if there are general statements one can make.)
