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I know there are finite fields like $\mathbb F_2$, $\mathbb F_4$ or the $\mathbb Z/n\mathbb Z$ for prime $n$ with modulo operations. For other special $n$, I've seen fields $\mathbb F_n$ with $n$ elements being constructed. And of course there are the usual infinite fields (take $\mathbb Q$ and so on).

So I wonder: Let $M$ be an arbitrary set that contains at least two elements. Can you always find operations $+ : M\times M \to M$ and $\cdot : M \times M \to M$, such that $(M, +, \cdot)$ is a field?

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    The cardinality of finite field always must be a power of prime number. Thus, for example, you can not build a field of 20 elements.2012-01-12
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    To complete Norbert's comment : You can see from this Wikipedia page : http://en.wikipedia.org/wiki/Finite_field that a finite fields exists if and only its cardinal is a power of a prime number.2012-01-12
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    Related: http://math.stackexchange.com/questions/8683992016-08-22

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