(related to Proposition: Algebraic Structure of Integers Together with Addition and Multiplication)

We have to show that \((\mathbb Z, +, \cdot)\), i.e. the set of integers \(\mathbb Z\), together with the operations addition "\(+\)" and multiplication "\(\cdot\)" forms an integral domain. We will do so by demonstrating, that \((\mathbb Z, +, \cdot)\) is a commutative ring, which is not the zero ring and in which the integer \(0\) is the only zero divisor.

In order to show that \((\mathbb Z, +,\cdot)\) is a commutative ring, we check the following properties:

- \((\mathbb Z, + )\) is a commutative group. This has been shown in a corresponding proposition.
- \((\mathbb Z, \cdot )\) is a commutative monoid. This requires the following (sub)properties:
- We have already shown that the multiplication of integers is commutative, thus \((\mathbb Z,\cdot)\) is commutative.
- We have already shown that the multiplication of integers is associative, thus \((\mathbb Z,\cdot)\) is associative.
- \((\mathbb Z,\cdot)\) includes the identity \(1_{\in\mathbb Z}:=[1_{\in\mathbb N},0_{\in\mathbb N}]\), which have been shown in the proposition integers one is neutral with respect to the multiplication of integers.

- \((\mathbb Z, +, \cdot)\) is distributive, which has been shown in the corresponding proposition. We have just shown that \((\mathbb Z, + ,\cdot)\) is a commutative ring with \(1:=[1,0]\) as the identity of the commutative monoid \((\mathbb Z,\cdot)\). Obviously, \((\mathbb Z, + ,\cdot)\) is not the zero ring, because \(1\in\mathbb Z\). It remains to be shown that \(0_{\in\mathbb Z}:=[0_{\in\mathbb N},0_{\in\mathbb N}]\) is the only zero divisor.

Suppose, there would be other (but \(0\)) zero divisors, say \(x,y\in\mathbb Z\) with \(x\neq 0\), \(y\neq 0\), and still \(x\cdot y=0\). By definition of integers, suppose \(x=[a,b],y=[c,d]\) for some natural numbers \(a,b,c,d\in\mathbb N\). By definition of multiplying integers, we also have that \[\begin{array}{ccl} x\cdot y=[a,b] \cdot [c,d] := [ac + bd, ad + bc]. \end{array} \]

Because both \(x\) and \(y\) are not equal \(0\), we have that \([a,b]\neq[0,0]\) and \([c,d]\neq[0,0]\), which means that \(a\) and \(b\) are not simultaneously equal \(0\), and that \(c\) and \(d\) are not simultaneously equal \(0\). There are \(7\) possibilities for this and we have to check, in which cases the natural numbers \(ac + bd\) and \(ad + bc\) both equal zero:

\(a\) | \(b\) | \(c\) | \(d\) | \(ac\) | \(bd\) | \(ad\) | \(bc\) | \(ac+bd\) | \(ad+bc\) |
---|---|---|---|---|---|---|---|---|---|

\(\neq 0\) | \(\neq 0\) | \(\neq 0\) | \(\neq 0\) | \(\neq 0\) | \(\neq 0\) | \(\neq 0\) | \(\neq 0\) | \(\neq 0\) | \(\neq 0\) |

\(= 0\) | \(\neq 0\) | \(\neq 0\) | \(\neq 0\) | \(=0\) | \(\neq 0\) | \(= 0\) | \(\neq 0\) | \(\neq 0\) | \(\neq 0\) |

\(\neq 0\) | \(=0\) | \(\neq 0\) | \(\neq 0\) | \(\neq 0\) | \(=0\) | \(\neq 0\) | \(= 0\) | \(\neq 0\) | \(\neq 0\) |

\(= 0\) | \(\neq 0\) | \(= 0\) | \(\neq 0\) | \(=0\) | \(\neq 0\) | \(= 0\) | \(=0\) | \(\neq 0\) | \(= 0\) |

\(= 0\) | \(\neq 0\) | \(\neq 0\) | \(= 0\) | \(=0\) | \(= 0\) | \(= 0\) | \(\neq 0\) | \(= 0\) | \(\neq 0\) |

\(\neq 0\) | \(= 0\) | \(= 0\) | \(\neq 0\) | \(=0\) | \(=0\) | \(\neq 0\) | \(=0\) | \(= 0\) | \(\neq 0\) |

\(\neq 0\) | \(= 0\) | \(\neq 0\) | \(= 0\) | \(\neq 0\) | \(=0\) | \(=0\) | \(=0\) | \(\neq 0\) | \(= 0\) |

We have seen that in all \(7\) cases \[\begin{array}{ccl} [ac + bd, ad + bc]\neq[0,0], \end{array} \] which demonstrates that the product of integers \(x\cdot y\) is never equal (integer) zero. This is a contradiction to our hypothesis \(x\cdot y=0\). This completes the proof.

∎

**Kramer Jürg, von Pippich, Anna-Maria**: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013