Return to Homework 12, Glossary, Theorems

**Problem:** (+) Determine if the set {$a + b \sqrt{2} \mid a, b \in \mathbb{Z}$} with the usual addition and multiplication defines a ring. If a ring is not formed, explain why. If a ring is formed, determine whether it has unity and whether it is a field.

**Solution:** In order for this to be a ring, we need to check to see if the three axioms are satisfied.

$\mathscr{R}_1$: We need to see if $\langle R, + \rangle$ is an abelian group. Let $x, y \in R$ such that $x = a_1 + b_1 \sqrt{2}$ and $y = a_2 + b_2 \sqrt{2}$. Then,

(1)
\begin{align} x+y = (a_1 + b_1 \sqrt{2}) + (a_2 + b_2 \sqrt{2}) \end{align}

Since addition is associative, we can rearrange the terms:

(2)
\begin{align} (a_1 + b_1 \sqrt{2}) + (a_2 + b_2 \sqrt{2}) = (a_1 + a_2) + (b_1 + b_2) \sqrt{2} \end{align}

Now,

(3)
\begin{align} y + x = (a_2 + b_2 \sqrt{2}) + (a_1 + b_1 \sqrt{2}) \end{align}

Rearranging again, we get:

(4)
\begin{align} (a_2 + b_2 \sqrt{2}) + (a_1 + b_1 \sqrt{2}) = (a_1 + a_2) + (b_1 + b_2) \sqrt{2} \end{align}

Therefore, $x + y = y + x$ which makes $\langle R, + \rangle$ an abelian group.

$\mathscr{R}_2$: Now we need to see if multiplication is associative. Let $x, y, z \in R$ such that $x = a_1 + b_1 \sqrt{2}, \;y = a_2 + b_2 \sqrt{2}$ and $z = a_3 + b_3 \sqrt{2}$. Then,

(5)
\begin{align} (x*y)*z = [(a_1 + b_1 \sqrt{2})*(a_2 + b_2 \sqrt{2})]*(a_3 + b_3 \sqrt{2}) \end{align}

(6)
\begin{align} = (a_1a_2 + a_1b_2 \sqrt{2} + b_1 \sqrt{2}a_2 + b_1 \sqrt{2}b_2 \sqrt{2})*(a_3 + b_3 \sqrt{2}) \end{align}

(7)
\begin{align} =a_1a_2a_3 + a_1a_2b_3 \sqrt{2} + a_1b_2 \sqrt{2}a_3 + a_1b_2 \sqrt{2}b_3 \sqrt{2} + b_1 \sqrt{2}a_2a_3 + b_1 \sqrt{2}a_2b_3 \sqrt{2} + b_1 \sqrt{2}b_2 \sqrt{2}a_3 + b_1 \sqrt{2}b_2 \sqrt{2}b_3 \sqrt{2} \end{align}

Since we know that multiplication **and** addition are both associative and abelian, as well as that $(\sqrt{a})^2 = a\;\; \forall a \in \mathbb{R}$, we can rearrange the above equation. So,

(8)
\begin{align} (x*y)*z = (a_1a_2a_3 +2a_1b_2b_3 + 2b_1a_2b_3 + 2b_1b_2a_3) + (a_1a_2b_3+a_1b_2a_3 + b_1a_2a_3 +2b_1b_2b_3)\sqrt{2} \end{align}

Now, let's look at x*(y*z):

(9)
\begin{align} x*(y*z) = (a_1 + b_1 \sqrt{2})*[(a_2 + b_2 \sqrt{2})*(a_3 + b_3 \sqrt{2})] \end{align}

(10)
\begin{align} = (a_1 + b_1 \sqrt{2}) * (a_2a_3 + a_2b_3 \sqrt{2} + b_2 \sqrt{2}a_3 + b_2 \sqrt{2}b_3 \sqrt{2}) \end{align}

(11)
\begin{align} = a_1a_2a_3 + a_1a_2b_3 \sqrt{2} + a_1b_2 \sqrt{2}a_3 + a_1b_2 \sqrt{2}b_3 \sqrt{2} + b_1 \sqrt{2}a_2a_3 + b_1 \sqrt{2}a_2b_3 \sqrt{2} + b_1 \sqrt{2}b_2 \sqrt{2}a_3 + b_1 \sqrt{2}b_2 \sqrt{2}b_3 \sqrt{2} \end{align}

Simplifying again leaves us with:

(12)
\begin{align} x*(y*z) = (a_1a_2a_3 +2a_1b_2b_3 + 2b_1a_2b_3 + 2b_1b_2a_3) + (a_1a_2b_3+a_1b_2a_3 + b_1a_2a_3 +2b_1b_2b_3)\sqrt{2} \end{align}

Therefore, $(x*y)*z = x*(y*z)$, hence multiplication is associative.

$\mathscr{R}_3$: Lastly we need to see if the left and right distributive laws hold. Let $x, y,$ and $z$ be defined as in $\mathscr{R}_2$. Now,

(13)
\begin{align} x*(y + z) = (a_1 + b_1\sqrt{2}) * [(a_2 + b_2\sqrt{2}) + (a_3 + b_3\sqrt{2})] \end{align}

(14)
\begin{align} = [(a_1 + b_1\sqrt{2}) (a_2 + b_2\sqrt{2})] + [(a_1 + b_1\sqrt{2}) (a_3 + b_3\sqrt{2})] \end{align}

(15)
\begin{equation} = (x*y)+(x*z) \end{equation}

Hence, the left distributive law holds. Now for the right distributive law:

(16)
\begin{align} (x+y)*z = [(a_1 + b_1\sqrt{2}) + (a_2 + b_2\sqrt{2})] * (a_3 + b_3\sqrt{2}) \end{align}

(17)
\begin{align} = [(a_1 + b_1\sqrt{2}) (a_3 + b_3\sqrt{2})] + [(a_2 + b_2\sqrt{2}) (a_3 + b_3\sqrt{2})] \end{align}

(18)
\begin{equation} = (x*z)+(y*z) \end{equation}

Hence, the right distributive law holds as well.

Since all three axioms are satisfied, we can see that the set {$a + b \sqrt{2} \mid a, b \in \mathbb{Z}$} with the usual addition and multiplication operators is in fact a ring.