Let $\mathbb{F}_p$ be a finite field of characteristic $p > 2$, for a fixed $p$. I will consider only prime fields, not $GF(p^n)$. Represent the $p$ elements of the field as integers $\{0,1,\ldots p-1\}$. Now lets define the "positive elements" as $\{0,2,4\ldots p-1\}$ and "negative elements" as $\{1,3,5\ldots p-2\}$.
(In other words the sign bit is the least significant bit of the integers's binary representation).
I was wondering if there is a way to determine the sign of $\frac{u}{v}$ given $u,v \in \mathbb{F}_p$ without computing the multiplicative inverse? Or in other words, is there a faster way than computing $uv^{p-2}$ to determine the sign bit of $\frac{u}{v}$?
(Besides using Extended Euclidean algorithm for computing the inversion, which is faster than $v^{p-2}$).
No, this is not consistent; you cannot think of them as being positive and negative in this way and get something meaningful. For example, in $\mathbb{Z}_{3}$, you have $1$ as negative and $2$ as positive. So $2^{-1} = 2$ is positive, and $1^{-1} = 1$ is negative. In $\mathbb{Z}_{5}$, $3$ is negative and $3^{-1} = 2$ is positive.
The real problem is that if your finite field does not have prime order, you cannot even use this definition of positive and negative for elements not in the base field.
A more common thing is to consider the elements that are squares in the field as being "positive" and the nonsquares as "negative". Then you have that the inverse of a nonsquare is a nonsquare, inverse of a square is a square.