0
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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 $p$.

$\endgroup$
4
$\begingroup$

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).

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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 $-y.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.