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I know the very basic basics of $\mathbb{Z}[\sqrt{14}]$. Numbers in it are of the form $a + b \sqrt{14}$, with $a, b \in \mathbb{Z}$. Numbers like $-3 + \sqrt{14}$ and $7 - 8 \sqrt{14}$. The norm function is $N(a + b \sqrt{14}) = a^2 - 14b^2$, which I'm told is not an Euclidean function even after the absolute value adjustment.

Proposition $4.11$ in this paper by Franz Lemmermeyer

http://www.rzuser.uni-heidelberg.de/~hb3/publ/survey.pdf

mentions $\mathbb{Z}\left[\sqrt{14}, \frac{1}{2}\right]$. I don't think I've ever read about a domain like that before, except perhaps in a very general way that the specifics eluded me.

I'm guessing $\mathbb{Z}\left[\sqrt{14}, \frac{1}{2}\right]$ contains all the same numbers of $\mathbb{Z}[\sqrt{14}]$ as well as some other numbers. What's the form of those other numbers? What are some concrete examples of those other numbers?

P.S. Proposition $4.11$ is on page $14$ of $56$ of the PDF. Looks like a brilliant survey. I need to print it out and sit down to read it beginning to end.

Mr. Brooks
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    $\frac12+\sqrt{14}$ would be one such number. – Clayton Jan 03 '17 at 22:52
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    So in $\frac{a}{2} + \frac{b \sqrt{14}}{2}$ the parity of $a$ and $b$ need not match? – Mr. Brooks Jan 03 '17 at 22:53
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    Answer to the question in the title: $$\frac12$$ – Did Jan 03 '17 at 23:04
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    @Mr.Brooks To answer your comments-question, no, definitely not; $\mathbb{Z}[\frac{1+\sqrt{14}}2]$ is a (pretty small) subset of this ring. – Steven Stadnicki Jan 03 '17 at 23:10
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    fyi: the notation $,\Bbb Z[\sqrt{14},1/2],$ denotes the ring adjunction of $,1/2,$ to $,\Bbb Z[\sqrt{14}],,$ i.e. the smallest subring of $,\Bbb Q(\sqrt{14}),$ containing $, \Bbb Z[\sqrt{14}],,$ and $,1/2\ \ $ – Bill Dubuque Jan 03 '17 at 23:16
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    @Steven: in the usual notation, ${\mathbb Z}[a]$ denotes the smallest ring containing $\mathbb Z$ and $a$. Since $z = \frac{1+\sqrt{14}}2$ satisfies the equation $z^2-z-3 = 1/4$, the two rings in question actually do coincide. The pretty small subset should refer to the additive subgroup ${\mathbb Z} \oplus z {\mathbb Z}$. –  Jan 04 '17 at 11:42
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    @franzlemmermeyer Oh, good point, mea culpa - I'm so used to $\mathbb{Z}[\frac12(1+\sqrt{D})]$ for $D$ odd where the 2-adic valuation never gets beyond -1. – Steven Stadnicki Jan 04 '17 at 16:44

2 Answers2

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An element of this ring can be considered a polynomial in $\frac12$, with coefficients in $\mathbf Z[\sqrt{14}]$. Reducing all terms to the same denominator, an element can ultimately be written as $$\frac{a+b\sqrt{14}}{2^n} \quad (a,b\in\mathbf Z).$$

For readers aware of localisations, it also may be described as the ring of fractions of $\mathbf Z[\sqrt{14}]$ w.r.t. the multiplicative subset of powers of $\frac12$.

Bernard
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  • This is true, but it deserves much further explanation given the level of the question (the OP is probably not familiar with localizations, so that language may be of little help to understand such a simple adjunction). – Bill Dubuque Jan 03 '17 at 23:19
  • @Bill Dubuque: I've added some details. Hope it's clear enough… – Bernard Jan 03 '17 at 23:24
  • Indeed I am not. I've seen the term in chapter headings in at least two different books, but not Alaca & Williams, which is what I have checked out currently. – Mr. Brooks Jan 05 '17 at 22:03
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If $a,b \in \mathbb R$, then $\mathbb{Z}[a,b]$ is the smallest ring that contains $a,b$; it is the same as the set of all polynomial expressions in $a,b$ with coefficients in $\mathbb{Z}$.

When $a=\sqrt{14}$ and $b=1/2$, all powers of $a$ reduce to an integer or to an integer times $a$. There is no reduction for powers of $b$, but all fractions can be reduced to the same denominator.

Therefore, a typical element of $\mathbb{Z}\left[\sqrt{14}, \frac{1}{2}\right]$ is of the form $\dfrac{u+v\sqrt{14}}{2^n}$, for $u,v \in \mathbb Z$ and $n \in \mathbb N$.

lhf
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