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Finite fields are split up into two parts. Prime fields, arithmetic is simply mod p.A prime fields takes the form $GF(p)$, where $p$ is prime. Why for extension fields, eg, of the form $GF(p^m),m>1$, are the elements expressed as polynomials, and not numbers.

Thanks

Colbi
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Sam Gregg
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    Polynomials? The fields $GF(p^m)$ are often constructed as quotient rings of the polynomial ring $GF(p)[x]$ with the ideal generated by an irreducible polynomial moded out, and the elements are then cosets of polynomials. But that's just one way of thinking about them. The same you could say that there are no complex numbers, the number $a+bi$ is just the polynomial $a+bx$ that is identified with all the other polynomials that differ from it by a multiple of $x^2+1$. But I usually think of them as numbers with a specified zero of that polynomial in place (like $i$ in place of $x$). – Jyrki Lahtonen Jun 08 '16 at 15:20
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    I guess the question is : "what is a number ?" then. How would you like to "express" elements of a finite field ? – Captain Lama Jun 08 '16 at 15:22
  • Ok, it does take some time getting used to. That learning process is not unlike the one when you learn to think of elements of $\Bbb{Z}_n$ as cosets as opposed to integers. Only after that process can you write, e.g. $2=14$ in $\Bbb{Z}_3$ instead of $2\equiv 14\pmod3$, or $\overline{2}=\overline{14}$. – Jyrki Lahtonen Jun 08 '16 at 15:25
  • But (trying to be more helpful), we can think of bigger finite fields much like the prime fields. For example, if $p\equiv-1\pmod4$, we can think of $GF(p^2)$ as $\Bbb{Z}[i]/\langle p\rangle$. That is, as Gaussian integers with both real and imaginary parts viewed as integers modulo $p$. But, that is just one way. And won't work with $p=5$. We can similarly view $GF(25)$ as $\Bbb{Z}[\omega]/\langle 5\rangle$, where $\omega=(-1+i\sqrt3)/2$ is a complex cubic root of unity. – Jyrki Lahtonen Jun 08 '16 at 15:29
  • Modulo $p=17$ both ways of looking at it work. We get $GF(17^2)$ as $\Bbb{Z}[i]/\langle 17\rangle$ as well as $\Bbb{Z}[\omega]/\langle 17 \rangle$. And this is why we won't define $GF(17^2)$ one way or the other. Both work equally well, and it depends on the problem you study which (if either) works better for you. We like to keep this liberty of defining the field $GF(p^m)$ the way that suits us. If there were a universal recipe using algebraic integers that way would have a lot of merit. But as we just saw, the simplest way of getting $GF(p^2)$ depends on the choice $p$. – Jyrki Lahtonen Jun 08 '16 at 15:35
  • So an answer to your question might read: There is no way of getting all the fields $GF(p^m)$ as quotient rings of the same number ring (=a subring of complex numbers). That's why we use quotients of polynomial rings. OTOH we can get each and every one of the fields $GF(p^m)$ as a quotient ring of some ring $R$ of numbers. It's just that the choice of $R$ necessarily depends on both $p$ and $m$, and that's why it looks a bit more complicated than with the prime fields. – Jyrki Lahtonen Jun 08 '16 at 15:39
  • And (before the smart alecs get to me), because we can embedd $\Bbb{Z}[x]$ as a subring of $\Bbb{R}$, we actually could do this e.g. with the choice $R=\Bbb{Z}[\pi]$, but that's not useful at all :-) – Jyrki Lahtonen Jun 08 '16 at 15:41

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