Intuition for why finite integral domain is a field So I am not getting the intuition for why every finite integral domain is a field. I mean I saw the proof but still I feel like its somehow not intuitive to me of why finiteness of integral domain implies that it must be a field. For example if we take $\mathbb{Z}$ it is indeed integral domain but not a field since its inverse doesn't exist in $\mathbb{Z}$. If someone could maybe clarify me for a better picture for the intuition that would be great.
 A: The difference between a finite and infinite integral domain (like $\Bbb Z$) is very clear here: multiplication by a fixed element does not make a surjective function of $\Bbb Z$ to $\Bbb Z$ unless the element you picked is $1$ or $-1$. Just consider the map $\Bbb Z\to \Bbb Z$ given by $x\mapsto 2x$.
However, multiplication by a nonzero element is always injective in a domain. Since injectivity and surjectivity are equivalent on finite sets, multiplication by a nonzero element in a finite domain make a surjective map.
So then, for a given nonzero $x$, there always exists a $y$ such that $xy=1$.
A: Rschwieb's answer is likely the best way to think about things here. However, I personally like this approach, which I find very intuitive. Maybe you will find it helpful as well.
Let $R$ be an integral domain and $x$ be a nonzero element of $R$. Then consider the set 
$$S = \{x, x^{2}, x^{3}, \ldots\}$$
It follows that $S$ must be finite, since $R$ is finite. Hence, it follows that $x^{i} = x^{j}$ for some $i > j$ (without loss of generality). But this means that $x^{j}(x^{i-j}-1) = 0$, and since $R$ is an integral domain, this means that $x^{j} = 0$ or $x^{i-j}-1=0$. But $x^{j}\neq 0$ since $x \neq 0$, so we must have $x^{i-j}=1$. Since $i > j, i-j \geq 1$, so $x(x^{i-j-1})=1$, i.e. $x^{-1} = x^{k}$ for some $k \geq 0$. 
Note that this breaks down right at the start for an infinite integral domain - there's no way we can conclude $S$ is finite! You can see this easily with $\mathbb{Z}$. 
