Answer to Question #194709 in Abstract Algebra for Alia

Question

Prove that (Z(sqrt(2)),+,×) is an integral domain.

Subject: Rings and fields

Answer 1

Solution:

 Consider $\mathbb{Z}[\sqrt{2}]=\{x+y \sqrt{2} \mid x, y \in Z\}$ of the real numbers R.

Since the real numbers R form a ring, the subset $\mathbb{Z}[\sqrt{2}]$ and forms ring if they are closed under the addition and multiplication which is directly checked. It is also integral domain since R as a field is an integral domain.

Alternatively: a nontrivial solution of $(x+\sqrt{2} y)(a+\sqrt{2} b)=0$ gives for$x b+y a \neq 0$ the equation $\sqrt{2}=-\frac{x a+2 y b}{x b+y a}$ , a contradiction to the irrationality of $\sqrt{2}$ . If $x b+y a=0$ it follows that $x a+2 y b=0$ . Since the determinant of this linear system seen as an equation for x and y is $2 b^{2}-a^{2}$ , the only solution is $x=y=0$ if $2 b^{2}-a^{2} \neq 0$ . If $2 b^{2}-a^{2}=0$ , the irrationality of $\sqrt{2} \implies a=b=0$ . Thus either $x+\sqrt{2} y\ or\ a+\sqrt{2} b$ is zero.