# Rings, Units and Associates

## Units and Associates

If xy = 1, x is the left inverse of y and y is the right inverse of x.
Clearly x and y are nonzero.
If x has a left inverse w and a right inverse y they are equal.
Write y = 1y = wxy = w(xy) = w1 = w.
In this case x is called "invertible", or a "unit".

The set of units forms a group under multiplication.
If R is the ring, the group of units is denoted R*.
You've seen this before;
**Z**n* is the group of units mod n.

Elements x and y are associates if x divides y and y divides x.
show that "associate" forms an equivalence relation.
Symmetry comes from the definition,
and transitivity (x divides y and y divides z implies x divides z) is straightforward.
Actually we need to be consistent here.
Divides always means left divides, or right divides, as you prefer.
Thus cx = y and ky = z, hence kcx = z.

We sometimes lump an element and all its associates together.
In the integers, 5 and -5 are associates.
When factoring 45,
we don't spend a lot of time worrying about 5 verses -5.
They are essentially the same prime factor;
merely associates of each other.
Of course 5 does not equal -5,
and sometimes the particular associate does matter.

If xy = 0, and x and y are nonzero,
x is a left zero divisor and y is a right zero divisor.

Suppose x has inverse w, and x is a zero divisor.
Write 0 = xy = w(xy) = (wx)y = 1y = y,
which contradicts x being a zero divisor.
Invertible and zero divisor are mutually exclusive.

If the ring is commutative, this theorem and the next are rather pointless.
But I like noncommutative rings.
I think they're cool!
So I will include, from time to time,
some results that are immediate and obvious
when restricted to commutative rings.
If you live in the commutative world you may want to skip these theorems.
Assume every nonzero element is left invertible.
Given v, write uv = 1, which makes u left and right invertible.
Now u is a unit, with inverse v, and that makes v a unit with inverse u.
The ring becomes a division ring.

The expression
1-xy is left invertible iff 1-yx is left invertible.
Start with the latter assumption and write
u*(1-yx) = 1.
We can now construct the left inverse of 1-xy.
Expand (1+xuy)*(1-xy), giving
1-xy+xuy-xuyxy.
Yet uyx = u-1,
so the expression simplifies to 1.
A similar result holds for the right inverse, using similar algebra.
If the right inverse of 1-yx is v,
the right inverse of 1-xy is 1+xvy.
Expand the product and replace yxv with v-1.

## Asymmetric

Let R be the 2×2 matrices of the form [**Z**,**Z**2|0,**Z**2].
Let a = [2,0|0,1] and let b = [0,1|0,0].
Note that a is a left 0 divisor via a*b, but cannot be a right 0 divisor.
Next let R be the ring of endomorphisms on an infinite dimensional K vector space.
Let b take the basis element ei to ei+1, and let a take ei to ei-1,
with e1 mapping to 0.
Now a*b (b followed by a) is the identity map, and a is a left inverse.
However, a cannot be a right inverse, for running a first implies an endomorphism
that is not injective.