Division in a finite field, $1/2$ in $\mathbb{Z}_5$ is $3$ I was watching my math lecture for a course and we were doing inverse matrices in finite fields in $\mathbb{Z}_5$. The determinant was $1 / (6-4)$ so $1/2$ which apparently means it is 3. I have no idea where 3 comes from, is $1$ over $2$ the same as division or does that mean something else?
I am entirely new to fields and have very limited knowledge on them, but my searches never revealed anything about division or what this would be for the determinant.
 A: Note that $2\cdot 3=6=5+1\equiv 1\pmod{5}$. This means that in the ring $\mathbb{Z}_5$, the element $2$ is invertible  and the inverse of $2$ (what you call $1/2$), is $3$.
A: Division is defined the same way in modular arithmetic as it is on the regular number line. In other words, $\frac12$ is the unique number such that $2\cdot \frac12 = 1$. However, since modular arithmetic works differently from regular arithmetic, the interpretation of the above definition is very different: We have $2\cdot 3 = 1$, which means that $3$ actually fulfills the defining property of $\frac12$. We can also easily check that there are no other such numbers, so it is unique. That is why $\frac12 = 3$.
Note for later: If you ever have to solve quadratic equations in modular arithmetic (at least as long as it's modulo an odd prime), the formula $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ still works, except once again, the definition of $\sqrt{{}\cdot{}}$ is the same (sortof; there isn't anything like positive / negative to distinguish the two square roots of a number), but the interpretation is different, with, for instance, $\sqrt{-1} = \pm 2$ modulo $5$.
