Z[√2] as a ring

Detailed Solution

For $a+b\sqrt2,\ c+d\sqrt2\in S$, the product $(a+b\sqrt2)(c+d\sqrt2)=(ac+2bd)+(ad+bc)\sqrt2$ has integer coefficients, so $S$ is closed under multiplication, ruling out (A). $S$ is a commutative ring with unity $1=1+0\sqrt2$, inheriting these properties from $\mathbb R$. However, $S$ is not a field: for a generic nonzero element such as $2=2+0\sqrt2$, its inverse in $\mathbb R$ is $1/2$, which is not of the form $a+b\sqrt2$ with integer $a,b$, so $2$ has no inverse within $S$, ruling out (C). Since $S$ is a subring of the field $\mathbb R$, it inherits the "no zero divisors" property (a product of two nonzero real numbers is never zero), ruling out (D). Thus the correct option is $\boxed{\mathrm{B}}$.