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title: Basic Properties of Rings

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# Basic Properties of Rings 

## :snake: How to show that a ring is a subring?
S is a nonempty subset. Show these two to show that a ring is a subring.
- **Closure under substitution** if $a,b \in S$ then $ab \in S$. 
- **Closure under multiple** if $a,b \in S$ then $a + b \in S$. 

## :hamburger: Units and Zero Divisors
For Ring with Identity, $au = ua = I$, then a is called unit and u is called an inverse of a. 

### Example of units
- In $\mathbb{Z}$, the only units are 1 and -1 because $1*1=1$ and $-1*-1=1$.
- In $\mathbb{Z}_{15}$, 1,2,4,7,8,11,13,14 are units. 

Recall that the definition of field is that all nonzero elements are units. Therefore, $\mathbb{Z}_{15}$ is not a field because not all its nonzero elements is a unit.
### Example of Zero Divisors
- R = $M_n(\mathbb{R})$ has a zero divisor if a matrix A is not a zero matrix and ad-bc = 0. 

## Relationship between Field and Integral Domain 
### All field is an integral domain. 
To show that a ring is an integral domain, we show that if $ab = 0$ then $a= 0$ or $b=0$. Proof is as the following: if $a=0$, then the proof is done. if $a \neq 0$, then in a field, there is a unit and there is an inverse of an unit. Therefore $aa^{-1} = 1$. $ab = 0 \rightarrow a^{-1}ab=0 \rightarrow b =0$. Since $b=0$, we showed that either $a=0$ or $b=0$ if $ab=0$. 
### All finite integral domain is a field. 

To show that a ring forms a field, we show that $aa^{-1} = 1$. Suppose that a set $S = \{a_1, a_2, \cdots, a_n\}$. If we multiply a to all the element of S, then $aS = \{aa_1, aa_2, \cdots, aa_n\}$. Each element is unique because if we say that they are not unique, we get $aa_i = aa_j \rightarrow a(a_i-a_j)=0 \rightarrow a_i = a_j$, which contradicts the initial assumption. In other words, the finite field is an arrangement of the unique elements in an arbitrary order. Now, since an integral domain is a **commutative ring with an identity**, one of the element in the set ($a_1, \cdots, a_n$), must equal to 1. Therefore, for an $a_i$ there but be $aa_1 = 1$. Therefore,  finite integral domain forms a field. 