Ring of Integers has no Zero Divisors

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Theorem

The integers have no zero divisors:

$\forall x, y, \in \Z: x \times y = 0 \implies x = 0 \lor y = 0$


This can equivalently be expressed:

$\forall x, y, \in \Z: x \ne 0 \land y \ne 0 \implies x \times y \ne 0$


Proof

Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the formal definition of integers.

That is, $\eqclass {\tuple {a, b} } \boxminus$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxminus$.

$\boxminus$ is the congruence relation defined on $\N \times \N$ by $\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$.


In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, as suggested.


From the method of construction, $\eqclass {c, c} {}$, where $c$ is any element of the natural numbers $\N$, is the identity of $\struct {\Z, +}$.

To ease the algebra, we will take $\eqclass {0, 0} {}$ as a canonical instance of this equivalence class.


We need to show that:

$\forall a, b, c, d \in \N: \eqclass {a, b} {} \times \eqclass {c, d} {} = \eqclass {0, 0} {} \implies \eqclass {a, b} {} = \eqclass {0, 0} {} \lor \eqclass {c, d} {} = \eqclass {0, 0} {}$


From Natural Numbers form Commutative Semiring, we can take it for granted that:

addition and multiplication are commutative and associative on the natural numbers $\N$
natural number multiplication is distributive over natural number addition.


So:

\(\ds \eqclass {a, b} {} \times \eqclass {c, d} {}\) \(=\) \(\ds \eqclass {0, 0} {}\)
\(\ds \leadsto \ \ \) \(\ds \eqclass {a c + b d, a d + b c} {}\) \(=\) \(\ds \eqclass {0, 0} {}\)
\(\ds \leadsto \ \ \) \(\ds a c + b d + 0\) \(=\) \(\ds a d + b c + 0\)
\(\ds \leadsto \ \ \) \(\ds a c + b d\) \(=\) \(\ds a d + b c\)


We have to be careful here, and bear in mind that $a, b, c, d$ are natural numbers, and we have not defined (and, at this stage, will not define) subtraction on such entities.


Without loss of generality, suppose that $\eqclass {c, d} {} \ne \eqclass {0, 0} {}$.

Then $c \ne d$.

Without loss of generality, suppose also that $c > d$.


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From Ordering in terms of Addition, $\exists p \in \N: d + p = c$ where $p > 0$.


Then:

\(\ds a c + b d\) \(=\) \(\ds a d + b c\)
\(\ds \leadsto \ \ \) \(\ds a \paren {d + p} + b d\) \(=\) \(\ds a d + b \paren {d + p}\)
\(\ds \leadsto \ \ \) \(\ds a d + a p + b d\) \(=\) \(\ds a d + b d + b p\)
\(\ds \leadsto \ \ \) \(\ds a p\) \(=\) \(\ds b p\)
\(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds b\)
\(\ds \leadsto \ \ \) \(\ds \eqclass {a, b} {}\) \(=\) \(\ds \eqclass {0, 0} {}\) Construction of Inverse Completion: Equivalence Class of Equal Elements


Similarly for when $c < d$.

Thus:

$\eqclass {c, d} {} \ne \eqclass {0, 0} {} \implies \eqclass {a, b} {} = \eqclass {0, 0} {}$

A similar argument shows that:

$\eqclass {a, b} {} \ne \eqclass {0, 0} {} \implies \eqclass {c, d} {} = \eqclass {0, 0} {}$


The equivalence between the two forms of the statement of this theorem follows from De Morgan's Laws: Conjunction of Negations and the Rule of Transposition.

$\blacksquare$


Sources

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