How do finite fields work?

Question

I know that a field must satisfy a set of axioms, the one that is causing me most discomfort is closure under addition. All the roots of $X^q-X$ are all the elements of a finite field of order $q$. However, I tested this with $X^5-X$, its roots are ${1,-1,0,i,-i}$ however, if I add $i+i=2i$ and $2i$ is not in the field, failing the axiom of closure, how do finite fields work then?

Answer 1

You're not taking the roots in the complex numbers, you're taking the roots in the finite field itself. In the finite field $\mathbb{F}_5$ (the integers $\bmod 5$), there is in fact an element that deserves to be called $i$ (in the sense that it squares to $-1$), namely $2$, and $2 + 2 = 4$ is also in the field (if you like, we have $i + i = -1$; sounds weird, but it all works out $\bmod 5$).

Answer 2

Hint: It depends on the definition of the operations. In $\mathbb{Z}_5$, $2+5=7$ but $7=2$ in $\mathbb{Z}_5$.

Edit: Of course 5 as itself is a representative of the equivalence class $[0]$. $0$ is another such representative. I can't understand why people are going pseudo-pedantic about it.

Answer 3

The main point: the elements of a finite field $Z_p$ ($p$ - prime) are $0, 1, ..., p-1$ (or more precisely, equivalence classes modulo $p$...). Numbers from $\mathbb{C}$ or $\mathbb{R}$ are not in $Z_p$.

For every element of $Z_5$, $x^5-x \equiv 0 \bmod 5$ is true. You can check it yourself. Remember that all field operations in $Z_p$ are modulo $p$, $p=5$ in you case. For instance $2^5 \equiv 2 \bmod 5$.

"Finite Fields" by Rudolf Lidl, Harald Niederreiter is an excellent book on this topic.