The equivalent of multiplying by $-1$ in $GF(4)$?

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In $\mathbb{R}$, we have the notion of multiplying an element by $-1$. For example:

$12 \in \mathbb{R}$, then $12 \times (-1) = -12 \in \mathbb{R}$,

$7 \in \mathbb{R}$, then $7 \times (-1) = -7 \in \mathbb{R}$.

Now, assume we are in $GF(4)$ and its definition is a standard definition with elements $\{0, 1, \omega, \bar{\omega}\}$. We will label them as $\{0, 1, 2, 3\}$, respectively:

$$ \begin{array}{|c|cccc|}\hline + & 0 & 1 & 2 & 3\\\hline 0 & 0 & 1 & 2 & 3\\ 1 & 1 & 0 & 3 & 2\\ 2 & 2 & 3 & 0 & 1\\ 3 & 3 & 2 & 1 & 0\\\hline \end{array}\qquad \begin{array}{|c|cccc|}\hline - & 0 & 1 & 2 & 3\\\hline 0 & 0 & 1 & 2 & 3\\ 1 & 1 & 0 & 3 & 2\\ 2 & 2 & 3 & 0 & 1\\ 3 & 3 & 2 & 1 & 0\\\hline \end{array}\qquad \begin{array}{|c|cccc|}\hline \times & 0 & 1 & 2 & 3\\\hline 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 2 & 3\\ 2 & 0 & 2 & 3 & 1\\ 3 & 0 & 3 & 1 & 2\\\hline \end{array}\qquad \begin{array}{|c|cccc|}\hline \div& 0 & 1 & 2 & 3\\\hline 0 & - & 0 & 0 & 0\\ 1 & - & 1 & 3 & 2\\ 2 & - & 2 & 1 & 3\\ 3 & - & 3 & 2 & 1\\\hline \end{array}\qquad $$

It is usually understood that $-x$ is the additive inverse so that for any element $x \in GF(4)\,\,$, $x + (-x) = 0$.

I am interested in multiplying by $-1$ in $GF(4)$ defined as the same mathematical meaning as ordinary multiplying by $-1$ in $\mathbb{R}$. How would I go about doing something like this?

In the case where you would like more context, this deals with finding determinant of a matrix with elements in $GF(4)$ and multiplying by $-1$ is a necessary condition.

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$${GF(4)= \{0,1,w,w^2\}}$$ Now, ${w+1=w^2}$, so ${w^3=w \cdot w^2=w \cdot(1+w)=w+w^2=w + (1+w)=1}$, so ${w^2}$ is multiplicative inverse of ${w}$.

  • Now if you create the same field with ${Z_2[x]}$ with irreducible polynomial ${X^2+X+1}$, then you will have the set ${\{0,1,X,X+1\}}$
  • in ${GF(2^n)}$ we consider ${1=-1}$ , Hence ${X-1}$ is same as ${X+1}$ as mentioned by @Stinking Bishop
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Well, $GF(4)$ has $GF(2) = \{0,1\}$ as a subfield. In this subfield, $-1=1$ as you work modulo 2.