# 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)The operation $+$ is also associative since:

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

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

(5)The operation $+$ is commutative since:

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

(8)The operation $*$ is associative since:

(9)And:

(10)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)And:

(12)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.