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

Question

Bonus question: if it's not, is it a subdomain of some ring of algebraic integers?

This is just something I was thinking about a few weeks ago. I forgot about the concept of algebraic degrees, which I understood at one point when it came up in a discussion about subatomic particles for an article I wrote 20 years ago (my finished article only made passing mention of complex numbers which showed up in several of the scientist's equations and said nothing about algebraic degrees). If I had remembered about algebraic degrees, maybe my confusion would have never had happened in the first place.

Answer 1

All definitions I've seen in my life of $\mathbb{Z}[r]$, where $r\in\mathbb{C}$, is

the least subring of $\mathbb{C}$ including $\mathbb{Z}$ and containing $r$.

It can be easily proved that $\mathbb{Z}[r]$ consists of all numbers of the form $f(r)$, where $f$ is a polynomial with integer coefficients.

So, by definition, $\mathbb{Z}[r]$ is closed under multiplication. Being a subring of $\mathbb{C}$, it is obviously a domain.

Don't be deceived by the fact that $$ \mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\} $$ which is true because $\sqrt{2}$ is algebraic of degree $2$. The number $\sqrt{2}+\sqrt{3}$ is algebraic of degree $4$, so elements in $\mathbb{Z}[\sqrt{2}+\sqrt{3}]$ have a more complicated representation. But the minimum polynomial of $r=\sqrt{2}+\sqrt{3}$ can be easily computed: \begin{gather} r-\sqrt{2}=\sqrt{3}\\ r^2-2r\sqrt{2}+2=3\\ r^2-1=2r\sqrt{2}\\ r^4-2r^2+1=8r^2\\ r^4-10r^2+1=0 \end{gather} Since it's fairly easy to see that $h(X)=X^4-10X^2+1$ is irreducible over the rationals, this is the minimum polynomial of $r$ over $\mathbb{Q}$. Since it is monic, every polynomial $f(X)$ with integer coefficients can be written as $$ f(X)=q(X)h(X)+g(X) $$ where $q$ and $g$ have integer coefficients and $g$ has degree less than $4$. It follows that $$ \mathbb{Z}[r]=\{\,a+br+cr^2+dr^3:a,b,c,d\in\mathbb{Z}\,\}. $$

Answer 2

If by $\mathbb{Z}[\sqrt 2 + \sqrt 3]$ you mean just numbers of the form $a + b (\sqrt 2 + \sqrt 3)$, with $a, b \in \mathbb{Z}$, then the answer is clearly no, it's not closed under multiplication, since for example $(\sqrt 2 + \sqrt 3)^2 = 5 + 2 \sqrt{6}$.

But if instead you mean $\mathcal{O}_{\mathbb{Q}(\sqrt 2 + \sqrt 3)}$, the ring of algebraic integers obtained by adjoining $\sqrt 2 + \sqrt 3$ to $\mathbb{Q}$, then the answer is yes. This is closed by definition, even though it's not always obvious that we have correctly identified all the numbers that belong in the domain.

The number $\sqrt 2 + \sqrt 3$ has $x^4 - 10x^2 + 1$ for a minimal polynomial. This has degree $4$, which means that $\mathcal{O}_{\mathbb{Q}(\sqrt 2 + \sqrt 3)}$ contains algebraic integers of degree less than or equal to $4$, numbers like $5 + 2 \sqrt{6}$ which is obviously of degree $2$, and numbers like $-7$ and $3$.