The Ring of Q(√2)

Recall from the Rings page that if $+$ and $*$ are binary operations on the set $R$, then $R$ is called a ring under $+$ and $*$ denoted $(R, +, *)$ when the following are satisfied:

• 1. For all $a, b \in R$ we have that $(a + b \in R)$ (Closure under $+$).

• 2. For all $a, b, c \in R$, $a + (b + c) = (a + b) + c$ (Associativity of elements in $R$ under $+$).

• 3. There exists an $0 \in R$ such that for all $a \in R$ we have that $a + 0 = a$ and $0 + a = a$ (The existence of an identity element $0$ of $R$ under $+$).

• 4. For all $a \in R$ there exists a $-a \in R$ such that $a + (-a) = 0$ and $(-a) + a = 0$ (The existence of inverses for each element in $R$ under $+$).

• 5. For all $a, b \in R$ we have that $a + b = b + a$ (Commutativity of elements in $R$ under $+$).

• 6. For all $a, b \in R$ we have that $a * b = b * a$ (Closure under $*$).

• 7. For all $a, b, c \in R$, $a * (b * c) = (a * b) * c$ (Associativity of elements in $R$ under $*$).

• 8. There exists a $1 \in R$ such that for all $a \in R$ we have that $a * 1 = a$ and $1 * a = a$ (The existence of an identity element $1$ of $R$ under $*$).

• 9. For all $a, b, c \in R$ we have that $a * (b + c) = (a * b) + (b * c)$ and $(a + b) * c = (a * c) + (b * c)$ (Distributivity of $*$ over $+$).

We will now look at the ring of $\mathbb{Q}(\sqrt{2})$.

Let $\mathbb{Q}(\sqrt{2})$ be the set of real numbers of the form $a + b\sqrt{2}$ where $a, b \in \mathbb{Q}$. Let $+$ denote standard addition and $*$ denote standard multiplication. We will show that then $\mathbb{Q}(\sqrt{2})$ forms a ring with these operations by verifying each of the axioms above.

Let $x, y, z \in \mathbb{Q} (\sqrt{2})$ where $x = a + b\sqrt{2}$, $y = c + d \sqrt{2}$ and $z = e + f \sqrt{2}$ for $a, b, c, d, e, f \in \mathbb{Q}$. Clearly the $\mathbb{Q}(\sqrt{2})$ is closed under $+$ since:

(1)

\begin{align} \quad x + y = (a + b \sqrt{2}) + (c + d \sqrt{2}) = (a + c) + (b + d)\sqrt{2} \in \mathbb{Q} (\sqrt{2}) \end{align}

The operation $+$ is also associative since:

(2)

\begin{align} \quad x + (y + z) = (a + b \sqrt{2}) + [(c + d\sqrt{2}) + (e + f \sqrt{2})] = (a + b\sqrt{2}) + [(c + e) + (d + f)\sqrt{2}] \\ = [(a + c) + (b + d)\sqrt{2}] + (e + f\sqrt{2}) = [(a + b\sqrt{2}) + (c + d\sqrt{2})] + (e + f\sqrt{2}) = (x + y) + z \end{align}

The identity element of $+$ is $0 = 0 + 0 \sqrt{2}$ since:

(3)

\begin{align} \quad x + 0 = (a + b\sqrt{2}) + (0 + 0\sqrt{2}) = (a + 0) + (b + 0)\sqrt{2} = a + b\sqrt{2} = x \end{align}

(4)

\begin{align} \quad 0 + x = (0 + 0\sqrt{2}) + (a + b\sqrt{2}) = (0 + a) + (0 + b)\sqrt{2} = a + b\sqrt{2} = x \end{align}

For each element $x$, the additive inverse of $x$ is $-x = -a -b\sqrt{2}$ since:

(5)

\begin{align} \quad x + (-x) = (a + b\sqrt{2}) + (-a - b\sqrt{2}) = (a - a) + (b - b)\sqrt{2} = 0 + 0\sqrt{2} = 0 \end{align}

(6)

\begin{align} \quad (-x) + x = (-a - b \sqrt{2}) + (a + b\sqrt{2}) = (-a + a) + (-b + b)\sqrt{2} = 0 + 0\sqrt{2} = 0 \end{align}

The operation $+$ is commutative since:

(7)

\begin{align} \quad x + y = (a + b\sqrt{2}) + (c + d\sqrt{2}) = (a + c) + (b + d)\sqrt{2} = (c + a) + (d + b)\sqrt{2} = (c + d\sqrt{2}) + (a + b\sqrt{2}) = y + x \end{align}

The set $\mathbb{Q}(\sqrt{2})$ is closed under $*$ since:

(8)

\begin{align} \quad x * y = (a + b\sqrt{2}) * (c + d\sqrt{2}) = ac + ad\sqrt{2} + bc\sqrt{2} + 2bd = (ac + 2bd) + (ad + bc)\sqrt{2} \in \mathbb{Q}(\sqrt{2}) \end{align}

The operation $*$ is associative since:

(9)

\begin{align} \quad x *(y * z) = (a + b\sqrt{2}) * [(c + d\sqrt{2}) * (e + f\sqrt{2})] = (a + b\sqrt{2}) * [(ce + 2df) + (cf + de)\sqrt{2}] \\ = (ace + 2adf) + (acf + ade)\sqrt{2} + (bce + 2bdf)\sqrt{2} + 2(bcf +bde) \\ \end{align}

And:

(10)

\begin{align} \quad (x * y) * z = [(a + b\sqrt{2}) * (c + d\sqrt{2})] * (e + f\sqrt{2}) = [(ac + 2bd) + (ad + bc)\sqrt{2}] * (e + f\sqrt{2}) \\ = (ace + 2bde) + (acf + 2bdf)\sqrt{2} + (ade + bce)\sqrt{2} + 2(adf + bcf) \end{align}

Comparing the two equations above we see that they are indeed equal.

The identity element for $*$ is $1 + 0\sqrt{2} \in \mathbb{Q} (\sqrt{2})$.

Left distributivity also holds for $*$ since:

(11)

\begin{align} \quad x * (y + z) = (a + b\sqrt{2}) * [(c + d\sqrt{2}) + (e + f\sqrt{2})] \\ = (a + b\sqrt{2}) * [(c + e) + (d + f)\sqrt{2}] = (ac + ae) + (ad + af)\sqrt{2} + (bc + be)\sqrt{2} + 2(bd + bf) \end{align}

And:

(12)

\begin{align} \quad x * y + x * z = (a + b \sqrt{2}) * (c + d \sqrt{2}) + (a + b\sqrt{2}) * (e + f\sqrt{2}) = [(ac +2bd) + (ad + bc)\sqrt{2}] + [(ae + 2bf) + (af + be)\sqrt{2}] \end{align}

Comparing the two equations above and we see that they are equal. We can similarly show that right distributivity holds for $*$, which is left to the reader.

Therefore the set $\mathbb{Q}(\sqrt{2})$ satisfies all of the ring axioms with respect to the operations of $+$ and $*$ and so $(\mathbb{Q}(\sqrt{2}), +, *)$ is a ring.