Answer to Question #135827 in Abstract Algebra for YASHU

Question

Show that the set of all real numbers of the form a+b√2 where 'a' and 'b' are integrals forms an integral domain

Answer 1

Denote by $K$ the set of all real numbers of the form $a+b\sqrt{2}$ where $a$ and $b$ are integers, that is

$K=\{ a+b\sqrt{2}\ |\ a,b\in\mathbb Z\}.$

To prove that $(K,+,\cdot)$ is a ring it is sufficient to prove that $(K,+,\cdot)$ is a subring of the ring $(\mathbb R,+,\cdot)$ of real numbers.

Let $a+b\sqrt{2}, c+d\sqrt{2}\in K$, where $a,b,c,d\in\mathbb Z.$ Then $(a+b\sqrt{2})+ (c+d\sqrt{2})=(a+c)+(b+d)\sqrt{2}\in K$, since $a+c,b+d\in\mathbb Z,$

$(a+b\sqrt{2})- (c+d\sqrt{2})=(a-c)+(b-d)\sqrt{2}\in K$, since $a-c,b-d\in\mathbb Z,$

$(a+b\sqrt{2})\cdot (c+d\sqrt{2})=(ac+2bd)+(ad+bc)\sqrt{2}\in K$, since $ac+2bd, ad+bc\in\mathbb Z.$

Therefore, $(K,+,\cdot)$ is a subring of the ring $(\mathbb R,+,\cdot)$, that is $(K,+,\cdot)$ a ring.

Since $(\mathbb R,+,\cdot)$ is a commutative ring, $(K,+,\cdot)$ is a commutative ring as well. The number $1=1+0\sqrt{2}$ is the identity of the ring $(K,+,\cdot)$. It follows that $(K,+,\cdot)$ is a commutative ring with identity. Since for any real numbers the equality $\alpha\cdot\beta=0$ implies that $\alpha=0$ or $\beta=0,$ in particular, for $a+b\sqrt{2}, c+d\sqrt{2}\in K$, the equality $(a+b\sqrt{2})(c+d\sqrt{2})=0$ implies that $a+b\sqrt{2}=0$ or $c+d\sqrt{2}=0.$

Therefore, $(K,+,\cdot)$ is an integral domain.