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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.

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    $\frac12+\sqrt{14}$ would be one such number. – 2017-01-03
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    So in $\frac{a}{2} + \frac{b \sqrt{14}}{2}$ the parity of $a$ and $b$ need not match? – 2017-01-03
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    Answer to the question in the title: $$\frac12$$ – 2017-01-03
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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. – 2017-01-03
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    fyi: the notation $\,\Bbb Z[\sqrt{14},1/2]\,$ denotes the [ring adjunction](http://math.stackexchange.com/a/15463/242) 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\ \ $ – 2017-01-03
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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 ${\mathbb Z} \oplus z {\mathbb Z}$. – 2017-01-04
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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. – 2017-01-04

2 Answers 2

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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 $n \in \mathbb N$.

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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 $\frac12$.

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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). – 2017-01-03
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    @Bill Dubuque: I've added some details. Hope it's clear enough… – 2017-01-03
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    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. – 2017-01-05