Answer to Question #133618 in Abstract Algebra for PRATHIBHA ROSE C S

Question

Prove that Z[√2]={a+b√2;a,b element of Z} is a ring under the ordinary addition and multiplication

Answer 1

$\mathbb{Z}[\sqrt{2}]$ is a ring under the ordinary addition and multiplication because:

1) $\mathbb{Z}[\sqrt{2}]$ is closed under addition

If $x,y\in \mathbb{Z}[\sqrt{2}]$ , then $x=a+b\sqrt{2}$ and $y=c+d\sqrt{2}$ , where $a,b,c,d\in \mathbb{Z}$ .

$x+y=(a+b\sqrt{2})+(c+d\sqrt{2})=(a+c)+(b+d)\sqrt{2}\in \mathbb{Z}[\sqrt{2}]$

2) addition is associative ( $\mathbb{Z}[\sqrt{2}]$ is a subset of $\mathbb{R}$ )

3) there exists $0=0+0\cdot \sqrt{2}\in \mathbb{Z}[\sqrt{2}]$

4) $\forall x\in \mathbb{Z}[\sqrt{2}],\ x=a+b\sqrt{2}, \ \ \exists (-x)\in \mathbb{Z}[\sqrt{2}], \ -x=(-a)+(-b)\sqrt{2}$

5) addition is commutative ($\mathbb{Z}[\sqrt{2}]$ is a subset of $\mathbb{R}$ )

6) $\mathbb{Z}[\sqrt{2}]$ is closed under multiplication

If $x,y\in \mathbb{Z}[\sqrt{2}]$ , then $x=a+b\sqrt{2}$ and $y=c+d\sqrt{2}$ , where $a,b,c,d\in \mathbb{Z}$ .

$x\cdot y=(a+b\sqrt{2})(c+d\sqrt{2})=(ac+2bd)+(ad+bc)\sqrt{2}\in \mathbb{Z}[\sqrt{2}]$

7) multiplication is associative ($\mathbb{Z}[\sqrt{2}]$ is a subset of $\mathbb{R}$ )

8) distributivity ($\mathbb{Z}[\sqrt{2}]$ is a subset of $\mathbb{R}$ )