Find all the intermediates rings between $D=\{(x,x)\mid x\in \mathbb{Q}\}$ and $\mathbb{Q}[\sqrt{2}]\times\mathbb{Q}[\sqrt{3}]$. Let $D\subseteq E\subseteq \mathbb{Q}(\sqrt{2})\times \mathbb{Q}[\sqrt{3}]$. 
 If $E=A\times B$ for some $A\subseteq \mathbb{Q}[\sqrt{2}]$ and $B\subseteq \mathbb{Q}\sqrt{3}$. Then $A$ and $B$ must be fields because any intermediate ring between algebraic field extensions is a field (here $\mathbb{Q} \subseteq \mathbb{Q}[\sqrt{2}]$ is algebraic extension of degree 2 so there is no proper ring between $\mathbb{Q}$ and $\mathbb{Q}[\sqrt{2}]$ and similarly there is no proper field between $\mathbb{Q}$ and $\mathbb{Q}[\sqrt{3}]$). So in this case the intermediate rings are $\mathbb{Q}\times\mathbb{Q}$, $\mathbb{Q}\times\mathbb{Q}[\sqrt{3}]$, $\mathbb{Q}[\sqrt{2}]\times\mathbb{Q}$.
I don't know whether there are any other intermediate rings or not.
Thank you.
 A: If $z=(a+b\sqrt 2,c+d\sqrt 3)\in E$, let $p(x)$ be the minimal polynomial for $a+b\sqrt{2}$. If $b\neq 0$, this is of degree $2$, and no number of the form $c+d\sqrt{3}$ is a root of it. So $p(z)=(0,c_1+d_1\sqrt{3})\in E,$ with $c_1+d_1\sqrt{3}\neq 0$.
Then also $(2c_1,2c_1)-(0,c_1+d_1\sqrt{3})=(2c_1,c_1-d_1\sqrt{3})\in E$. Multiplying, we get $(0,c_1^2-3d_1^2)\in E$, and hence $(0,1)\in E$ and $(1,0)=(1,1)-(0,1)\in E$.
So if $b\neq 0$ then all of $\mathbb Q\times\mathbb Q\subseteq E$. Likewise if $d\neq 0$.
Now, if $z=(a,c)\in E$ with $a\neq c$ then $(1,0)=\left(\frac1{a-c},\frac1{a-c}\right)\left((a,c)-(c,c)\right)\in E$, So again, $\mathbb Q\times\mathbb Q\subseteq E$.
If $\mathbb Q\times\mathbb Q\subseteq E$, then if $(a+b\sqrt 2,c+d\sqrt{3})\in E$ we have $(b\sqrt{2},d\sqrt{3})\in E$. And then $(1,0)(b\sqrt{2},d\sqrt{3})=(b\sqrt{2},0)\in E$ and likewise $(0,d\sqrt{3})\in E$. So if $b\neq 0$ for any element of $E$ then $(\sqrt{2},0)\in E$, and if $d\neq 0$ for any element of $E$ then $(0,\sqrt{3})\in E$. 
From here, it is easy to classify all possible $E$.
