# What is the explanation for why a field cannot have certain values like e.g. 12? [duplicate]

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

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