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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?

GuPe
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  • @5xum Yea, I spoke too quickly. – Kitegi Nov 03 '15 at 21:32
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    When $q$ is a power of a prime $p$, the field of $q$ elements is isomorphic to the spitting field of $X^{q}-X \in \mathbb{F}{p}[X]$, where $\mathbb{F}{p}$ is the field $\mathbb{Z}/p\mathbb{Z}$. It is not the splitting field of $X^{q}-X \in \mathbb{Q}[X]$. – Geoff Robinson Nov 03 '15 at 21:33

3 Answers3

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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 $\bmod 5$).

Qiaochu Yuan
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    Obligatory (recent) SMBC: http://www.smbc-comics.com/index.php?id=3913 – Eric Towers Nov 03 '15 at 23:26
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    Incidentally, one way to make sense of this equation $i + i = -1$ is that although the finite field $\mathbb{F}_5$ can be described as $\mathbb{Z}/(5)$, it can also be described as $\mathbb{Z}[i]/(2i + 1)$, where $\mathbb{Z}[i]$ denotes the Gaussian integers (https://en.wikipedia.org/wiki/Gaussian_integer). – Qiaochu Yuan Nov 04 '15 at 07:27
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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.

Jantar88
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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.

Neeraj Kumar
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