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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?

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    In $\mathbb{F_2}$ note that $-1=1.$ – 2017-02-24
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    Yes, $-1=1$ in $\mathbb F_2$. Recall the construction of $\mathbb F_2 = \mathbb Z/2\mathbb Z = \{\overline{0}, \overline{1}\}$, where $\overline{0} = \{\ldots, -4, -2, 0, 2, 4, 6, \ldots\}$ and $\overline{1} = \{\ldots, -7, -5, -3, -1, 1, 3, 5, 7, \ldots \}$. – 2017-02-24
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    If you just care about the typical integers then just add $p$ until you get a positive number. If there are extensions where they are polynomials in a primitive element, there already mostly formal symbols, so it won't matter. – 2017-02-24
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    @DustanLevenstein, so `-2` equals `1` in $\mathbb{F_3}$? – 2017-02-24
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    @sparkandshine No, $-2=0$. Note that $-2$ and $1$ are not in the same equivalence class in the definitions of $\overline 0$ and $\overline 1$ that I gave above. – 2017-02-24
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    Oh, sorry, I just noticed you were talking about $\mathbb F_3$, not $\mathbb F_2$. In that case, yes, $-2=1$. – 2017-02-24
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    @DustanLevenstein Why not put your first comment as answer? :) – 2017-02-24
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    I've been chewed out for posting answers to questions that are low-hanging fruit before. There's not really any hard and fast rule about it though. – 2017-02-24

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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 $p=3$, $$[0]_3-[1]_3 = [-1]_3 = [2]_3.$$

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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.