Is $\mathbb{Z}[\sqrt{2} + \sqrt{3}]$ integrally closed?

Question

Is $\mathbb{Z}[\sqrt{2} + \sqrt{3}]$ integrally closed? (Or could it have a relation to another domain like $\mathbb{Z}[\sqrt{-3}]$ does with $\mathbb{Z}[\omega]$?) Also, is it UFD? What are its units?

I have never read about this domain in any book, though i did read try to read one book with very general techniques that could theoretically be applied here.

reason I even known about this domain is because their was a question on this website last month asking if it was closed under multiplication. A lot of people fell into the trap of thinking that its quadratic but its not, myself included. Their was a whole bunch of snarky comments from people who know its quartic (is that the right term? not quadratic but only one person explained the difference -- he did a great job answering the question and explaining the misconception, and maybe from his answer one can derive the answers to the questions I'm asking today, but I'm not sure how if that's the case.

Answer 1

Interestingly enough this is not integrally closed. (spelling not corrected)

We determine the rings of integers of $\mathbb{Z}[\sqrt{2}+\sqrt{3}]$ and $\mathbb{Q}[\sqrt{2},\sqrt{3}]$

We prove first that

$$a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \in \mathbb{Z}[\sqrt{2}+\sqrt{3}]$$ if and only if $b$ and $c$ have the same parity and $d$ is even.

An additive basis is given by, $\{1,\sqrt{2}+\sqrt{3}, 2\sqrt{2}, 2\sqrt{6}\}$. We show this as follows,

$(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}$ so $2\sqrt{6} \in\mathbb{Z}[\sqrt{2}+\sqrt{3}]$ this gives on multiplying $2\sqrt{6}$ by the original expression we have

$$6\sqrt{2} +4\sqrt{3}\in\mathbb{Z}[\sqrt{2}+\sqrt{3}]$$ from which we can easily get $2\sqrt{2}, 2\sqrt{3} \in \mathbb{Z}[\sqrt{2}+\sqrt{3}]$.

Next,

$$(\sqrt{2}+\sqrt{3})^3=11\sqrt{2}+9\sqrt{3}.$$

Now noting that $2\sqrt{3}=2(\sqrt{2}+\sqrt{3})-2\sqrt{2}$ we see that $1,(\sqrt{2}+\sqrt{3}),(\sqrt{2}+\sqrt{3})^2,(\sqrt{2}+\sqrt{3})^3$ are all expressed in terms of these our basis.

From which we easily deduce the claim.

Now let us determine the ring of integers of $\mathbb{Q}(\sqrt{2},\sqrt{3})$

Since we have automorphisms such as $\sqrt{2},\sqrt{3} \mapsto -\sqrt{2},\sqrt{3}$ and $\sqrt{2},\sqrt{3} \mapsto \sqrt{2},-\sqrt{3}$ and $\sqrt{2},\sqrt{3} \mapsto -\sqrt{2},-\sqrt{3}$ It is easily seen that any algebraic integer has the form

$$\frac{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}}{2}.$$

If we write this as $\frac{\alpha+\beta\sqrt{3}}{2}$ and take its norm over $\mathbb{Q}(\sqrt{2})$ we get an algebraic integer in $\mathbb{Q}(\sqrt{2})$ and we know the integers here are of the form $n+m\sqrt{2}$, $n,m \in \mathbb{Z}$. Skipping the details of the calculation we find $4|a^2+c^2+2(b^2+d^2)$ and $2|ab+cd$ from which a little simple number theory gives that $a$ and $c$ are even and $b$ and $d$ have the same parity.

Thus we see that the integers in $\mathbb{Q}(\sqrt{2},\sqrt{3})$ have the form

$$a+b\sqrt{2}+c\sqrt{3}+d\frac{\sqrt{2}+\sqrt{6}}{2}.$$
It remains to check that this last is in fact algebraic integer. So let $$\alpha=\frac{\sqrt{2}+\sqrt{6}}{2}\in \mathbb{Q}[\sqrt{2},\sqrt{3}]$$ then

$$\alpha^2=2+\sqrt{3}$$ therefore

$$(\alpha-2)^2=3$$ so

$$\alpha^4-4\alpha^2+1=0$$

so $\alpha$ is an algebraic integer.

Note that this shows that $\mathbb{Z}[\sqrt{2},\sqrt{3}]$ is not integrally closed.

Answer 2

The full set of $\mathbb Q$-conjugates of $\alpha=\sqrt2+\sqrt3$ is the set $\{\pm\sqrt2\pm\sqrt3\}$, and so you see that the minimal $\mathbb Q$-polynomial for $\alpha$ is $X^4-10X^2+1$. Continuing, you easily see that $1/\alpha=-\sqrt2+\sqrt3$, while the minimal polynomial for $\alpha$ also shows that $1/\alpha=10\alpha-\alpha^3$. Since $\sqrt3=\frac12(\alpha+1/\alpha)$, you see that $\sqrt3=\frac92\alpha-\frac12\alpha^3$, so that $\sqrt3$ is not in $\mathbb Z[\alpha]$.

Answer 3

Hint $\,\ \alpha = \sqrt 2 + \sqrt 3\ \Rightarrow\ \dfrac{\alpha^2-1}2\, =\, 2+\sqrt 6$