Integers do not form Field

Corollary of Invertible Integers under Multiplication
The integers $\left({\Z, +, \times}\right)$ do not form a field.

Proof
For $\left({\Z, +, \times}\right)$ to be a field, it would require that all elements of $\Z$ have an inverse.

However, from Invertible Integers under Multiplication, only $1$ and $-1$ have inverses (each other).

Example
Take $2$, for example.

In the field of rational numbers, we have that $2 \times \dfrac 1 2 = 1$ and so $2$ has an inverse in $\Q$.

But that inverse is not in $\Z$.