# Section 3.1 Part 2: Rings

The next part of rings deals with an idea called a Cartesian Product. A simple example would be the ordered pairs used to make two dimensional plots: (x,y). This ordered pair is the cartesian product $\mathbb{R}\times\mathbb{R}$.

##### Definition:

The Cartesian Product of two non empty sets R and S is the set of ordered pairs $R\times S={(r,s):$ $r\in R,s\in S}$

For Example: If $T=\mathbb{Z}_2\times\mathbb{Z}_3$ then $T={(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)}$

In a cartesian product addition and multiplication are defined \begin{align*}(r,s)+(r',s')&=(r+r',s+s')\\ (r,s)(r',s')=(rr',ss')\end{align*}

Note that the cartesian product of two commutative rings has a product that is also commutative, and the product of two fields is not necessarily a field.

#### Theorem 3.1:

Let R and S be rings. The Cartesian product $$R\times S$$ is also a ring with \begin{align*}(r,s)+(r',s')&=(r+r',s+s')\\ (r,s)(r',s')=(rr',ss')\end{align*}

If both R and S have identities $1_R,1_S$ then $R\times S$ has an identity $(1_R,1_S)$.

###### Definition:

A non empty subset S of a ring is a subring of R if S is a ring

Example:

• $E=2\mathbb{Z}$ is a subring of $\mathbb{Z}$
• The odd integers (not closed under addition and does not contain the zero element)
• $\mathbb{Z}$ is a subring of $\mathbb{Q}$ which is a subring of $\mathbb{R}$ which is a subring of $\mathbb{C}$.

### Theorem 3.2:

Suppose R is a ring. A subset S of R is a subring of R if

1. S is closed under addition

2. S is closed under multiplication

3. $0_R\in S$

4. For each $a\in S$, the equation $a+x=0_R$ has a solution x in S

Example:

• $S={0,4}$ of $\mathbb{Z}_8$ is asubring.
• $\mathbb{Q}_5={a+b\sqrt{2}[5]:a,b\in\mathbb{Q}}$ is this a subring of $\mathbb{R}$
• $(a_1+b_1 \sqrt{2}[5]) +(a_2+b_2 \sqrt{2}[5])=(a_1+a_2)+(b_1 +b_2) \sqrt{2}[5]$
• $(a_1+b_1 \sqrt{2}[5]) +(a_2+b_2 \sqrt{2}[5])=a_1 a_2+\sqrt{2}[5] (a_1 b_2 + a_2 b_1)+5 b_1 b_2\in\mathbb{Q}(\sqrt{2}[5])$
• $0_R=0+0\sqrt{2}[5]\in\mathbb{Q} (\sqrt{2}[5])$
• $(a+b\sqrt{2}[5]+x=0_R$ implies $x=-(a+b\sqrt{2}[5]\in\mathbb{Q}(\sqrt{2}[5])$
• This is a commutative ring with identity $1_R=1+0\sqrt{2}[5]$
• Is $\mathbb{Q}$ a field?
• $(a+b\sqrt{2}[5] )x=1$
• In $\mathbb{R}$, $x=\frac{1}{a+b\sqrt{2}[5]}\in\mathbb{Q}(\sqrt{2}[5])$?
• $x=\frac{1}{a+b\sqrt{2}[5]}=\frac{a-b\sqrt{2}[5]}{a^2 - 5 b^2}$