The set $R = \{a + b\sqrt{2} + c\sqrt{3}: a \in \Bbb{Z}, c, b \in \Bbb{Q}\}$ is not closed on multiplication, my textbook states. Why is this?
And related to that: why then is $S = \{a + b\sqrt{2} : a, b \in \Bbb{Z}\}$ closed under multiplication?
I figured the following, but I'm totally unsure about the correctness: A multiplication on two numbers of $S$ looks like this: $(\Bbb{Z} + \Bbb{Z}\sqrt{2}) \cdot (\Bbb{Z} + \Bbb{Z}\sqrt{2})$ is of the form $\Bbb{Z}² + 2\Bbb{Z}\Bbb{Z}\sqrt{2} + 2\Bbb{Z}$. Because any $\Bbb{Z}^2$ yields an integer, $2\Bbb{Z}$ yields an integer as well and $2\Bbb{Z}\Bbb{Z}\sqrt{2}$ yields an $\Bbb{Z}\sqrt{2}$, the result can again be expressed as $a + b\sqrt{2} : a, b \in \Bbb{Z}$ and therefore the set is closed under multiplication.
However, a similar argument would also work for $R$ right?
I would appreciate the answer as well as any feedback on the formal correctness of my arguments.