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} .


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) .


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


  • 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


  • 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}
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