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If $\mathbb{F}=\{ 0,1,x,y\}$ is a field with four elements, why is the characteristic 2?

In a field with four elements, the characteristic is 2 or 3(since it has to be a prime). Is there any way to determine precisely the characteristic of this field, and in general any finite field?

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4 Answers

up vote 4 down vote accepted

If a field $\mathbb{F}$ contains a field $\mathbb{K}$, then $\mathbb{F}$ is a vector space over $\mathbb{K}$.

A field of $4$ elements cannot contain a field of $3$ elements, because then it would be a vector space over $\mathbb{F}_3$, and so would necessarily have cardinality $3^k$ for some $k$. But $4\neq 3^k$.

(Alternatively: the additive subgroup generated by $1$ must have order dividing the order of the field, by Lagrange's Theorem. So the subfield generated by $1$ must have order $2$; it cannot have order $3$.)

A finite field must be a vector space over the field generated by $1$; hence its order will be $p^k$ for some prime $p$ and some positive integer $k$, and the characteristic will then be $p$.

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Forget the multiplication. Since $(\Bbb F,+)$ is a group, we must have $1+1+1+1=4=0$.

Now put back the multiplication in the picture. We have a field where $$ 4=2\cdot2=0. $$ Thus $2=0$, i.e. the characteristic is $2$.

Note that this works as well for any field with $p^n$ elements where $p$ is prime (which includes all finite fields).

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Hint $\ $ A finite field, having no zero-divisors, must have prime characterisic $\rm\:p.\:$ Thus it is a vector space of finite dimension $\rm\:n\:$ over $\rm\:\mathbb F_p,\:$ so has cardinality $\rm\:p^n.$

Thus, given the cardinality, or any integer known to be a prime power $\rm\:p^n,\:$ one can compute $\rm\:p\:$ very quickly (in fact in polynomial time).

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If it had characteristic $3$ then $-1$ is distinct from $1.$ WLOG let $x=-1$ so $\mathbb{F}= \{ 0, 1 , -1, y \}.$ But none of those $4$ elements can be $-y.$

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