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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.

Bill Dubuque
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David R.
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    To ask a rhetorical question, what is the definition of $\mathbb{Z}[x]$? – Joshua Mundinger Jun 26 '14 at 20:40
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    Please do not assume that readers have seen the title of your question; that is only for the main page. The body of your question should have all the information necessary to answer the question. The other information that is needed is context: where did you encounter the problem? What have you tried already? – Carl Mummert Jun 26 '14 at 20:51
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    imho, the bonus question is the more intersting question. my instintc is that it is a subdomain of some ring, but I have no idea what taht ring would be. –  Jun 26 '14 at 20:51
  • To answer the question in the title, "yes", because if $T$ is a subring of a ring $R$, and $z \in R$, then $T[z]$ is the smallest subring of $R$ containing $z$. In particular $\mathbb{Z}[x]$ is a ring for every $x \in \mathbb{C}$. If you have another definition of $\mathbb{Z}[x]$ you really need to state it explicitly. – Carl Mummert Jun 26 '14 at 20:52
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    It appears that you may not be familiar with the meaning of the ring adjunction notation $,R[\alpha].,$ See for example here and here. – Bill Dubuque Jun 26 '14 at 20:58
  • @CarlMummert, you write: "Please do not assume that readers have seen the title of your question; that is only for the main page." That's a common position, but why? The title of the question is printed above the body of the question. – goblin GONE Jun 26 '14 at 21:02
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    @user18921: it is just like in email, where one can't assume that the recipient will read the subject line when reading the message. The title above the question is really only for locating the question on the main page; many readers will begin reading just beside the up arrow, not at the title. So the body needs to include enough information to make the question clear. When I read this question, the first thing I saw was "bonus question". – Carl Mummert Jun 26 '14 at 21:06
  • Would you add some details to your question so that it can be reopened? – egreg Jun 27 '14 at 20:48
  • @egreg I can try, I don't know if this helps any. You still get the points either way, right? – David R. Jul 03 '14 at 21:39
  • @DavidR. It's not a question of points; the better the question, the more helpful is for future readers. – egreg Jul 03 '14 at 21:41

2 Answers2

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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 $$ \mathbb{Z}[r]=\{\,a+br+cr^2+dr^3:a,b,c,d\in\mathbb{Z}\,\}. $$

egreg
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  • I will just note that in some texts (Hungerford), [ ] is used for polynomial rings and polynomial evaluation rings, whereas ( ) is used for generators. – Forever Mozart Jun 26 '14 at 21:24
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    Thank you very much, egreg for your correct and well-explained answer. Thank you also to those who posted wrong answers. Those who rained hate on the wrong answerers should be ashamed of themselves. – David R. Jun 27 '14 at 20:23
  • @ForeverMozart Isn't that the same thing? $R[a]$ is the image of $R[X]$ under the substitution morfism $X \mapsto a$, but also the smallest ring containing $R$ and $a$ – Jens Renders Jul 31 '17 at 14:31
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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 $3$.

Bob Happ
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