Question #194709

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



Subject: Rings and fields


1
Expert's answer
2021-05-25T07:12:54-0400

Solution:

 Consider Z[2]={x+y2x,yZ}\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 Z[2]\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+2y)(a+2b)=0(x+\sqrt{2} y)(a+\sqrt{2} b)=0 gives forxb+ya0x b+y a \neq 0 the equation 2=xa+2ybxb+ya\sqrt{2}=-\frac{x a+2 y b}{x b+y a} , a contradiction to the irrationality of 2\sqrt{2} . If xb+ya=0x b+y a=0 it follows that xa+2yb=0x a+2 y b=0 . Since the determinant of this linear system seen as an equation for x and y is 2b2a22 b^{2}-a^{2} , the only solution is x=y=0x=y=0 if 2b2a202 b^{2}-a^{2} \neq 0 . If 2b2a2=02 b^{2}-a^{2}=0 , the irrationality of 2    a=b=0\sqrt{2} \implies a=b=0 . Thus either x+2y or a+2bx+\sqrt{2} y\ or\ a+\sqrt{2} b is zero.


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