I have a few questions in ring and fields theory.
First of all, I was trying to show that the field of quotients of $\frac{\mathbb{Z}_{12}}{\langle 4 \rangle}$ is exactly itself, once it is a field. Is easy once $\langle 4 \rangle$ is a maximal ideal and $\mathbb{Z}_{12}$ has unity and is commutative, then this quotient is a field. Is this right?
The second question is, why cannot exist a integral domain with 10 elements?
On
The prime ideals of $\mathbf Z/12\mathbf Z$ are the images by the canonical map $\mathbf Z\to\mathbf Z/12\mathbf Z$ of the prime ideals of $\mathbf Z$ which contain $12$, i.e. they are $\;2\mathbf Z/12\mathbf Z$ and $\;3\mathbf Z/12\mathbf Z$.
An elementary proof that a finite domain $A$ has a number of elements that is a prime power:
Consider the ring homomorphism $\;c\colon\mathbf Z\to A$, $\;n\mapsto n\cdot 1_A$. As $A$ is finite, $\ker c=r\mathbf Z, r\neq 0$. By the first isomorphism theorem, $\;\mathbf Z/r\mathbf Z$ is isomorphic to a subring of $A$, hence it is an integral domain, and $r$ is prime.
Now $\mathbf Z/r\mathbf Z$ is a finite field with $r$ elements, and $A$ is a finite dimensional vector space over that field, say it has dimension $n$. This implies $A$ has $r^n$ elements.
Note: it is very easy to prove a finite integral domain is a field: to show each $a\ne 0$ has an inverse in $A$, consider the multiplication-by-$a$ endomorphism of $A$. As $A$ is an integral domain, this homomorphism is injective (this means $a$ is not a zero-divisor on $A$). But as $A$ is finite, for any map from $A$ to $A$, injective $\iff$ surjective. Hence $1$ is attained by this multiplication, i.e. $a$ has an inverse.
I almost feel this must be wrong, since it seems so obvious and nobody else has pointed it out, but:
Actually $\langle 4\rangle$ is not a maximal ideal. And in fact the quotient $\Bbb Z_{12}/\langle 4\rangle$ is not even an integral domain, since $2(2)=0$.