Domain

A crone \( R \) is a domain if the product of any two nonzero elements in \( R \) is itself non-zero

Note that in other places they may call this an integral domain

Product Zero Implies Factor Zero Domain Equivalence

Suppose that \( R \) is a crone. It is a domain if and only if for every \( a, b \in D \), \( ab = 0 \) implies \( a = 0 \) or that \( b = 0 \)

Since \( R \) is a domain if and only if for any \( x \neq 0 \) and \( y \neq 0 \) it implies that \( x y \neq 0 \), this statement is equivalent to it's contrapositive: \( xy = 0 \) implies that \( x = 0 \) or \( y = 0 \) as needed.

Domain iff Cancellation

Let \( R \) be a crone. \( R \) is a domain if and only if given \( r, a, b \in R \) we have \( ra = rb \land r \neq 0 \implies a = b\)

For the rightward implication we can use properties of a ring to see the following \[ \begin{gather} ra &= rb \iff \\ ra - rb &= 0 \iff \\ r \left( a - b \right) &= 0 \iff \\ \end{gather} \]

Therefore we conclude that either \( a - b = 0 \) or \( r = 0 \), the latter of which is impossible through our assumption that \( r \neq 0 \), therefore \( a = b \).

We move on to the leftward implication where our goal is to show that \( R \) is a domain, we do so by chasing it's corollary. So let \( a, b \in R \) and assume that \( ab = 0 \), note that at this point if \( a = 0 \), our proof is over, therefore we assume that \( a \neq 0 \), note that we can rewrite \( ab = 0 \) to \( ab = a0 \), then by our assumption we can conclude that \( b = 0 \), which proves it.

Since \( \mathbb{ Z } _ 4 \) forms a ring then we can then notice the following: \[ \begin{align} \left( [ 2 ] x + [ 1 ] \right) ^ 2 &= [ 2 ] [ 2 ] x ^ 2 + [ 2 ] [ 1 ] x + [ 1 ] [ 2 ] x + [ 1 ] [ 1 ] \\ &= [ 4 ] x ^ 2 + [ 4 ] x + [ 1 ] \\ &= [ 0 ] x ^ 2 + [ 0 ] x + [ 1 ] \\ &= [ 1 ] \end{align} \] Where we're noticing that the degree of the product has decreased, this is occuring because in \( \mathbb{ Z } _ 4 \) there are non-zero elements which have a product of \( 0 \).

Unit

Suppose that \( R \) is a ring, then an element \( u \in R \) is said to be a **unit** if there exists some element \( v \in R \) such that
\[
uv = vu = 1 _ R
\]

Zero Divisor

A non-zero element \( x \) in a ring is said to be a zero divisor if there exists some other non-zero \( y \) such that \( x y = 0 \)

Nilpotent

Let \( R \) be a crone, then an element \( x \in R \) is said to be **nilpotent** if there is some \( n \in \mathbb{ N } _ 1 \) such that \( x ^ n = 0 _ R \)

Nilpotent Implies Zero or Zero Divisor

If \( x = 0 _ R \) then our proof is done, so instead let's assume that \( x \neq 0 _ R \), we know that it's nilpotent, considering \( \left\{ a \in \mathbb{ N } _ 1: x ^ a = 0 _ R \right\} \subseteq \mathbb{ N } _ 0 \) it has a least element (as it's non-empty) say \( l \) such that \( x ^ l = 0 _ R \), note that \( l \gt 1 \) because \( x ^ 1 \neq 0 _ R \)

In that case we can see \( x ^ l = x ^ { l - 1 } x ^ 1 \), since \( l \) was the smallest natural number such that \( x ^ l = 0 _ R \) then we must have \( x ^ { l - 1 } \neq 0 _ R \) since \( l - 1 \in \mathbb{ N } _ 0 \). Thus we can conclude that \( x x ^ { l - 1 } = 0 _ R \) where \( x, x ^ { l - 1 } \neq 0 _ R \) so that by definition \( x \) is a zero divisor.

Constant times Nilpotent is Nilpotent

Suppose \( R \) is a crone and that \( x \) is nilpotent, then \( r x \) is nilpotent for any \( r \in R \)

Since \( x \) is nilpotent we have some \( n \in \mathbb{ N } _ 1 \) such that \( x ^ n = 0 _ R \), since \( R \) is commutative, then it can be shown by induction that an equation of the form \( \left( ab \right) ^ n = a ^ n b ^ n \), in other words \( \left( r x \right) ^ n = r ^ n x ^ n = r ^ n 0 _ R = 0 _ R \) which proves that \( rx \) is nilpotent

One plus Nilpotent is Nilpotent

Suppose that \( R \) is a crone, and \( x \) nilpotent then \( 1 _ R + x \) is a unit

If \( x ^ 1 = 0 _ R \), it clearly holds so lets assume that \( x ^ n = 0 _ R \) for some \( n \gt 1 \).

We claim that \[ v := 1 - x + x ^ 2 - x ^ 3 + ... + \left( -1 \right) ^ { n - 1 } x ^ n - 1 = \sum _ { i = 1 } ^ { n - 1 } \left( -1 \right) ^ i x ^ i \] is \( 1 + x \)'s multiplicative inverse.

To see this true, we use distributivity, \[ \left( 1 + x \right) \left( \sum _ { i = 1 } ^ { n - 1 } \left( -1 \right) ^ i x ^ i \right) = \sum _ { i = 1 } ^ { n - 1 } \left( -1 \right) ^ i x ^ i + \sum _ { i = 1 } ^ { n - 1 } \left( -1 \right) ^ i x ^ { i + 1 } \] Define the following \[ S _ 1 \left( k \right) := \sum _ { i = 1 } ^ { k - 1 } \left( -1 \right) ^ i x ^ i = 1 - x + x ^ 2 - x ^ 3 + ... + \left( -1 \right) ^ { k - 1 } x ^ { k - 1 } \] and that \[ S _ 2 \left( k \right) := \sum _ { i = 1 } ^ { k - 1 } \left( -1 \right) ^ i x ^ { i + 1 } = x - x ^ 2 + x ^ 3 - x ^ 4 + \ldots \left( -1 \right) ^ { k - 2 } x ^ { k - 1 } + \left( -1 \right) ^ { k - 1 } x ^ { k - 1 } \]

Note that \( S _ 2 \left( n \right) = S _ 2 \left( n - 1 \right) \) because \( S _ 2 \left( n \right) \)'s last term is \( \left( -1 \right) ^ { n - 1 } x ^ { n } = 0 _ R \) since \( x ^ n = 0 _ R \) .

Through finite induction, we can show that for any \( j \in [ 2 \ldots n ], S _ 1 \left( j \right) + S _ 2 \left( j - 1 \right) = 1 \) which allows us to conclude that \( S _ 1 \left( n \right) + S _ 2 \left( n - 1 \right) = 1 \) and then \( S _ 1 \left( n \right) + S _ 2 \left( n \right) = 1 \) from the previous paragraph.

Finally this is good because now we continue the chain of equalities we mentioned at distributivity \[ \begin{align} \left( 1 + x \right) \left( \sum _ { i = 1 } ^ { n - 1 } \left( -1 \right) ^ i x ^ i \right) &= \sum _ { i = 1 } ^ { n - 1 } \left( -1 \right) ^ i x ^ i + \sum _ { i = 1 } ^ { n - 1 } \left( -1 \right) ^ i x ^ { i + 1 }\\ &= S _ 1 \left( n \right) + S _ 2 \left( n \right) \\ &= 1 \end{align} \] Therefore \( 1 + x \) is a unit, as needed.

The Sum of a Unit and a Nilpotent Element is a Unit

Suppose \( R \) is a crone and let \( u \) be a unit and \( x \) a nilpotent element in \( R \) , then \( u + x \) is a unit.

Since \( u \) is a unit there exists some \( v \) such that \( u v = v u = 1 \), if that's the case then \( \left( u + x \right) v = 1 + vx \), we know that \( vx \) is nilpotent and thus \( 1 + vx \) is a unit, so we have some \( a \) such that \( \left( 1 + vx \right) a = 1 \).

But \( \left( u + x \right) v = 1 + vx \) so in total we're saying \( \left( \left( u + x \right) v \right) a = 1 \) by associativity we have \( \left( u + x \right) \left( v a \right) = 1 \) and closure of multiplication shows us indeed \( u + x \) has a multiplicative inverse so that \( u + x \) is a unit.

Field is a Ring with Units

A field is a ring \( R \) in which every non-zero element \( r \in R \) is a unit