Ok, as far as I understand a field has to look like $\mathbb{F}_{p^n}$. But why? What is the explanation?

  • $\begingroup$ A field can't have non-zero zero divisors, but $3\times4\equiv0\bmod12$. $\endgroup$ – J. W. Tanner May 22 at 23:29
  • $\begingroup$ Are you asking why the cardinality (size) of a field only takes certain values, e.g. why there is no field with $12$ elements? $\endgroup$ – Gone May 22 at 23:33

I assume you mean why you can't have a field with $12$ elements. The Cliff's notes version is

1) If $F$ is a finite field then you must have some $n$ with $1+ 1 + \cdots + 1 = 0$, with $n$ ones in the sum. The smallest such $n$ is called the characteristic. If $n$ is not prime, say $n = ab$, then $0 = a(1 + \cdots + 1)$ there are $b$ 1's in the sum; but fields don't have zero divisors. So $n$ is a prime, $p$.

2) Show that the $p$ elements $0, 1, 1+1, 1+1+1, \ldots$ must form a subfield of $F$, called the prime subfield, say $K$.

3) Linear algebra: Most of the linear algebra you know over $\mathbb{R}$ can also be done over an arbitrary field. Show that $F$ can now be viewed as a vector space over $K$. Vector spaces have a fixed dimension, so $F$ must have a dimension $d$ over $K$. So as a $K$-vector space, $F$ is isomorphic to $K^d$, just as any real vector space of dimension $d$ is isomorphic to $\mathbb{R}^d$. Then note $|F| = |K^d| = p^d$.

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