If you made it through that last post and thought about it for a while, you might think that I pulled a fast one.

At a few places I commented that if x is any element of an arbitrary ring, then we know that x0 = 0.

That is a certainly a familiar fact about the integers. There we think of x times y as representing the number of objects in a rectangular array with x items along the length and y along the width. If either x or y is 0, then your array will contain nothing at all.

But things are less clear when you are working in an arbitrary ring. In this context “multiplication” is merely an operation that happens to satisfy certain rules. It doesn’t have to look anything like the familiar multiplication of integers.

The element 0, on the other hand, is defined solely in terms of addition. It is the unique ring element y with the property that x+y = x for all ring elements x. No mention of multiplication at all. And addition is itself merely a binary operation that happens to possess certain useful properties.

So why should it be true that an arbitrary ring element, multiplied by the additive identity element, always returns the addtive identity element back again?

The answer lies in the distributive property. In any ring, addition and multiplication are required to interact. With that in mind, watch how the trick is done.

Let x be an arbitrary ring element, and let 0 denote the additive identity element. Then:

x0 = x(0+0) = x0+x0.

But we know that whatever ring element x0 happens to be, it must have an additive inverse. And when we add that additive inverse to both sides of our equation, we get:

x0-x0 = x0+x0-x0,

which implies that x0 = 0, as desired.

So no matter how abstract your rings get, it will always be true that anything times zero is zero.