How to do subtraction over Galois field or finite field?

Question

I know a field exists additive inverses. What is the additive inverse of $a$ in finite field? Is it $-a$? If yes, however, for instance, $-1$ is not in $\mathbb{F_2}=\{0, 1\}$.

What is the result for $(0-1)$ in $\mathbb{F_2}$? I guess it is equivalent to $1$. If yes, why?

Answer 1

Recall the construction $\mathbb F_p = \mathbb Z/p\mathbb Z$ by equivalence classes of integers $$[a]_p := \{n \in \mathbb Z \mid n \equiv a \mod p\},$$ where $n \equiv a \mod p$ if and only if $n-a$ is a multiple of $p$. The definition of subtraction can be passed to integers via $[a]_p-[b]_p = [a-b]_p$, and it can be verified that this is well-defined. In the case $p=2$, $$[0]_2-[1]_2 = [-1]_2 = [1]_2,$$ and for $p=3$, $$[0]_3-[1]_3 = [-1]_3 = [2]_3.$$

Answer 2

Every finite field has some prime P as its characteristic - which means, for each element a, Pa = (a + a + ... [P instances of a] ... + a + a) is zero. It naturally follows that for any element a, -a is the element (P-1)a.