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I want to know how many finite fields with 10 elements there are. I know for example $\mathbb Z_{11}$ is one of them. How can I count all the possible polynomial fields or even something like $\mathbb Z^{*}_{p}$? Or is $\mathbb Z_{11}$ the only one?

thank you for your help


marked as duplicate by Jyrki Lahtonen abstract-algebra Dec 6 '18 at 10:22

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    $\begingroup$ $\mathbb Z_{11}$ has $11$ elements. $\endgroup$ – lulu Jul 9 '17 at 14:24
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    $\begingroup$ The number of elements of a finite field must be a prime power, so there is no field with $10$ elements $\endgroup$ – Peter Jul 9 '17 at 14:32
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    $\begingroup$ A field $(P,+,*)$ itself is an abelian group for $(P,+)$, so if $P$ is non-trivial, then it has to be a $p$-field for some prime $p$. $\endgroup$ – Ivon Jul 9 '17 at 14:37
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    $\begingroup$ For every prime number $p$ and every positive integer $n$, there exist finite fields of order $p^n$, and all fields of this order are isomorphic. $\endgroup$ – Omnomnomnom Jul 9 '17 at 14:38
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    $\begingroup$ A finite field is also called a galois field. $\endgroup$ – Peter Jul 9 '17 at 14:40

None. A finite field $F$ always has a prime subfield $\mathbf F_p=\mathbf Z/p\mathbf Z$ for some prime $p$ ($p$ is the smallest natural number $n$ such that $n\cdot 1=0$) and it is a finite dimensional vector space over this prime field, hence if its dimension is $r$, it has $p^r$ elements. As a consequence, the cardinality of a finite field is some power of a prime number.

  • $\begingroup$ Yeah you are right, it is $\mathbb Z \to F$. I was preoccupied just now. $\endgroup$ – Ivon Jul 9 '17 at 15:01

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