Why is $\{a + b\sqrt2 + c\sqrt3 : a\in\Bbb{Z}, b, c \in\Bbb{Q}\}$ not closed under multiplication? 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.
 A: The underlying concept is the degree of an extension - which is the same as the dimension of the extension as a vector space over $\mathbb Q$.
Extending by either $\sqrt 2$ or $\sqrt 3$ produces an extension of degree $2$. When we extend by both the degrees multiply so we get an extension of degree $4$. The formulation you give has purported degree 3 with basis $\{1, \sqrt 2, \sqrt 3\}$ - so when you know about degrees of extensions you will know that you are looking for another basis element (as the excellent accepted answer notes $\sqrt 6$ is the obvious choice - and this illustrates why $4$ is the natural degree).
There is much to learn about how this works, and the conditions under which the degrees multiply (which is intimately linked with the factorisation of polynomials). There are extensions of degree $3$, but they cannot be quadratic extensions, or contain quadratic elements, because a quadratic element always leaves a factor $2$ behind.
A: I think you want to show $\sqrt{2}\sqrt{3} = \sqrt{6}\notin R$.
A: Suppose $a+b\sqrt2+c\sqrt3+d\sqrt6=0$ where $\gcd(a,b,c,d)=1$. Then
$$
a^2+2ab\sqrt2+2b^2=3c^2+6cd\sqrt2+6d^2\tag{1}
$$
Since $\sqrt2\not\in\mathbb{Q}$, $(1)$ implies
$$
ab=3cd\tag{2}
$$
and
$$
a^2+2b^2=3(c^2+2d^2)\tag{3}
$$
$(2)$ implies that
$$
3\mid a\text{ or }3\mid b\tag{4}
$$
and consequently, $(3)$ and $(4)$ imply that
$$
3\mid a\text{ and }3\mid b\tag{5}
$$
$(2)$ and $(5)$ imply that
$$
3\mid c\text{ or }3\mid d\tag{6}
$$
$(3)$, $(5)$, and $(6)$ imply that
$$
3\mid c\text{ and }3\mid d\tag{7}
$$
Thus, $3\mid\gcd(a,b,c,d)$. $\quad\rightarrow\leftarrow\quad\square$

$\{a+b\sqrt2:a,b\in\mathbb{Z}\}$ is closed under multiplication since
$$
(a+b\sqrt2)(c+d\sqrt2)=(ac+2bd)+(ad+bc)\sqrt2
$$
A: Suppose $\sqrt 6$ belongs to $R$, i.e., $\sqrt 6 = a + b \sqrt 2 + c \sqrt 3$ for $a\in \mathbb Z$, and $b, c \in  \mathbb Q$. Squaring both sides, we see $6 = a^2 + 2b^2 + 3c^2 + 2 ab \sqrt 2 + 2 ac \sqrt 3 + 2 bc \sqrt 6$. Plugging in our expression for $\sqrt 6$ into this second equation, we have $6 = a^2 + 2b^2 + 3c^2 + 2 ab \sqrt 2 + 2 ac \sqrt 3 + 2 bca + 2b^2c \sqrt 2 + 2bc^2\sqrt 3$. It is easy to see $\sqrt 2 \notin \mathbb Q(\sqrt 3)$ (an easier application of same proof by contradiction I'm making here). This tells us the coefficients of $\sqrt 2$ and $\sqrt 3$ sum to 0, i.e., $ab + b^2c = 0$ and $ac + bc^2 = 0$. If $b=0$, either $a$ or $c$ equal 0, and we get a contradiction (can you see why?). We encounter a similar contradiction if $c=0$. If $b\neq 0$ and $c \neq 0$, $a = -cb$. Then $6 = 2b^2 + 3c^2 + 3b^2c^2$. Since you have that $a = -bc\in \mathbb Z$, you see that $(bc)^2 = 1$, else the $3 b^2c^2$ term is at least 6, so the sum on the right exceeds 6. But if $bc = 1$ or $-1$, either $b$ or $c$ will exceed (or equal) 1. Show $2(1/c^2) + 3 (c^2)$ exceeds 3 (find global minimum and see that this is greater than 3).
