# Every finite integral domain is a field

## Proof:

An integral domain I is a nonzero commutative ring in which the product of any two nonzero elements is nonzero.

We need to show that every nonzero element has a multiplicative inverse.

Let be a ∈ I and a ≠ 0.

Consider the mapping:

$$!F: I \rightarrow I\\ \qquad x\mapsto ax$$

We have: ax = ay ⇔ a(x-y) = 0

Since I has no zero divisors and a ≠ 0, we have: x = y.

Therefore: F is injective and since I is finite it is also surjective an bijective.

There is an inverse mapping F-1.

Define a-1 = F-1(1), so we have the existence of the inverse.

q.e.d.