Ring of integers

Question

I know that $\mathbb{Z} [\sqrt{3}, \sqrt{7} ]$ is not the ring of integers in $\mathbb{Q} [\sqrt{3}, \sqrt{7} ]$. But, I don't know how to explain. Can someone help in this. Thnx

Answer 1

We have $$\frac{\sqrt7-\sqrt3}2\cdot\frac{\sqrt7+\sqrt3}2=1\,.$$ So, we can expect that both these are also algebraic integer.

For $x=\displaystyle\frac{\sqrt7+\sqrt3}2$, it satisfies $2x-\sqrt3=\sqrt7$, so $4x^2-4\sqrt3 x+3=7$, that is $$x^2-1=\sqrt3x\,.$$ Now square it, that yields to a polynomial of $\Bbb Z[x]$ with leading coefficient $1$ such that $x=\displaystyle\frac{\sqrt7+\sqrt3}2$ is a root of it. (Actually its roots are exactly $\displaystyle\frac{\pm\sqrt7\pm\sqrt3}2$, so it is the minimal polynomial of all of these.)