abstract algebra - Numbers of elements in field

The question is list all those integers $n$ such that $1 \leq n \leq 10$ and there exists a field with $n$ elements.

My approach:

Of course $n$ can take values $2,3,5,7$ since they are prime and we know $ \mathbb Z_p$ is a field iff $p$ is a prime.

Next I will use the concept that if $p(x)$ is a irreducible polynomial then the ideal $$ is maximal, and if $I$ is a maximal ideal of a ring $R$ then $R/I$ is a field.

Now I consider the ideal $$ over $\mathbb F_2(x)$, since $x^2+x+1$ is irreducible in $\mathbb F_2(x)$ then $\mathbb F_2(x)/$ is a field. Any element of this field will have a form $ax+b$, where $(a,b) \in \mathbb F_2$, this implies this field has $4$ elements.

By the same type of argument, I have found the fields $\mathbb F_2(x)/$ and $\mathbb F_3(x)/$ have $8$ and $9$ elements respectively.

However I cannot find any field containing $6$ or $10$ elements. How to confirm whether there really are fields containing $6$ or $10$ elements?

Now I found that there are fields with $2,3,4,5,7,8,9$ elements. My method is a bit tedious. Are there any simple and direct computation to find so?

Answer

A finite field has prime characteristic.

If $R$ is any unital ring then there is a unique unital ring homomorphism
$\phi:\Bbb Z\to R$. Either $\phi$ is injective ($R$ has characteristic zero) or $\ker R=n\Bbb Z$ ($n\in\Bbb N$); in this case $R$ has
characteristic $n$. Then $\Bbb Z/n\Bbb Z$ is a subring of $R$.

If $R$ has composite characteristic $n=ab$ (a proper factorisation)
then $n1_R=0$ in $R$ but $n1_R=(a1_R)(b1_R)$ and $a1_R\ne0$
and $b1_R\ne0$. Therefore $R$ has zero-divisors and is not a field.

Therefore every finite field has prime characteristic. It
has a subfield $\Bbb Z/p\Bbb Z$, and as it is a vector space

over this subfield, it has $p^d$ elements for some $d\in\Bbb N$.