## 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.*