Math and science::Algebra::Aluffi

Rings. Zero-divisors.

Zero-divisor

An element $a$ in a ring $(R, +, \cdot)$ is a left zero-divisor iff there exists an element $b \neq 0$ in $R$ such that $a \cdot b = 0$ .

An element $a$ in a ring $(R, +, \cdot)$ is a right zero-divisor iff there exists an element $b \neq 0$ in $R$ such that $b \cdot a = 0$ .

Lambda perspective

The left case describes the 2-input function $\cdot_{R} : R \times R \to R$ having the first parameter fixed at $a$ . This forms a function $m_{a} : R \to R$ . Lambda calculus would call this partial application.

Injectivity and surjectivity

Not a left zero-divisor iff left-multiplication is injective

The partial application of $\cdot_{R} : R \times R \to R$ by $a$ as the first argument is an injective function iff $a$ is not a left zero-divisor.

Can you recall the proof?

When a ring has finite elements, all of which are not zero-divisors, it is a field.

Proof

The proposition is easily grasped in the contrapose: if $a$ composes on the left with distinct elements $b$ and $c$ to produce the same element $d$ (i.e. not injective), then $a$ is a left zero-divisor. The reverse implication is also true.

Proof. Forward case. Assume $a b = a c$ . Then $a \cdot b - a \cdot c = 0$ and $a \cdot (b - c) = 0$ . As $b$ and $c$ are distinct elements, $c$ 's inverse does not bring $b$ to zero, and so we have found a non-zero element $(b - c)$ which $a$ composes to produce zero.
Reverse case. If $a$ is a left zero-divisor, then there is a non-zero element $x$ such that $a \cdot x = 0$ . And so both $a \cdot 0$ and $a \cdot x$ map to the same element (not injective).

Intuition

Below, the function $\cdot_{R}$ with lambda type $R \to R \to R$ is partially applied, $\cdot_{R} (a)$ , and the resulting function of type $R \to R$ has its mapping drawn out:

What is interesting to note is that the non-injectivity imposed by the double mapping to $d$ ( $a \cdot b = a \cdot c = d$ ) creates two "holes" in the co-domain. The second hole arises arises as there are two elements $e_{+}$ and $(b - c)$ which map to $e_{+}$ . So the co-domain holes come in pairs. This is the essence of the proposition: for every co-domain position (non-zero) that the mapping doubles up on, there will be another doubling up at $e_{+}$ .

Injectivity implies inverse exist

If $R$ has finite elements, then the injectivity of the partial $\cdot_{R} (a)$ implies that $a$ has a right-inverse.

Rings introduce more jargon to describe this idea:

Not a left zero-divisor iff a two-sided unit (if $R$ is finite)

Proof.Let $a$ be one of the finite elements of $R$ . If $a$ is not a left zero-divisor then the partial $\cdot_{R} (a)$ is injective by the proposition above. Injectivity implies surjectivity when $R$ is finite. $a$ has a left-inverse, which is the element that it maps to $e_{\cdot}$ . The partial function $\cdot_{R} (a)$ is both injective and surjective so is a bijection and thus invertible. TODO: how to know that the inverting function maps to an element of $R$ ? With both inverses, $a$ is a two-sided unit.

When $R$ is not assumed to be finite, the discussion is more subtle. See p123 in Aluffi for more details.

Source

Aluffi p122