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I'm having difficulties with a homework problem from Algebra by Hungerford.

Let $R$ be the following subring of the complex numbers: $R = \{a + b(1 + \sqrt{19}i)/2 \mid a, b \in \mathbb{Z}\}$. Then $R$ is a principal ideal domain that is not a Euclidean domain.

I've shown that it is not a Euclidean domain, but I can't seem to show that it is a principal ideal domain. There's only a few details that are bothering me.

Here is a proof outline that I've been given:

We will use the function $N(x) = x\bar{x}$, and notation $\theta = (1 + \sqrt{19}i)/2$. Let $I$ be an ideal in $R$, and choose $b \in I \backslash \{0\}$ which minimizes $N$ (i.e. $N(b) \le N(a)$ for all $a \in I \backslash \{0\}$).

  1. If $I \backslash Rb$ is non-empty, then there is $a \in I \backslash Rb$ with $\lvert Im(a/b) \rvert \le \sqrt{19}/4$ (of course $Im(x)$ is the imaginary part of $x$).
  2. If $\lvert Im(a/b) \rvert < \sqrt{3}/2$ then there is $m \in \mathbb{Z}$ with $N(a/b - m) < 1$.
  3. If $\sqrt{3}/2 \le \lvert Im(a/b) \rvert \le \sqrt{19}/4$ then $0 \le \lvert Im(2a/b-\theta) \rvert \le \sqrt{3} - \sqrt{19}/2$ [I think this should be $\sqrt{19}/2 - \sqrt{3}$]. Therefore there is an $m \in \mathbb{Z}$ with $N(2a/b - \theta - m) < 1$.
  4. Conclude that $I = Rb$.

The only part I'm having trouble with is part 1. Everything else is clear to me.

How can I find such an $a$?

Thanks for any help! And thanks to Gerry Myerson for clearing up part 3.

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See my answer here for many further details. – Bill Dubuque Nov 29 '12 at 3:31
I have trouble understanding why Step (3) implies Step (4). My concern is why can't $2a/b-\theta-m=0$, which would not contradict the minimality of $N(b)$? – Conan Wong Feb 6 '13 at 0:12
up vote 1 down vote accepted

For first sentence of part 3, you have $${\sqrt3\over2}\le\Im(a/b)\le{\sqrt{19}\over4}{\rm\ or\ }-{\sqrt3\over2}\ge\Im(a/b)\ge-{\sqrt{19}\over4}$$ so $$\sqrt3\le\Im(2a/b)\le{\sqrt{19}\over2}{\rm\ or\ }-\sqrt3\ge\Im(2a/b)\ge-{\sqrt{19}\over2}$$ The imaginary part of $\theta$ is $\sqrt{19}/2$, so $$\sqrt3-{\sqrt{19}\over2}\le\Im((2a/b)-\theta)\le0{\rm\ or\ }-\sqrt3+{\sqrt{19}\over2}\ge\Im((2a/b)+\theta)\ge0$$ which is not exactly what you claim, but if you are willing to put up with $\pm\theta$ in place of $\theta$ it is good enough to get what you need.

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$\pm \theta$ is certainly good enough: as you probably saw, the point of saying $N(2a/b - \theta - m) < 1$ is simply that $N(b)N(2a/b - \theta - m) = N(2a - \theta b - mb) < N(b)$ contradicts $b$'s minimality. – smackcrane Nov 29 '12 at 3:42

It turns out, as usual, that part 1 is pretty simple.

Note that if $x \in I \backslash Rb$, then for any $y \in R$, $x - yb \in I \backslash Rb$ because certainly it is in I, and if it is in $Rb$ then $x \in Rb$ is a contradiction.

Now $$\Im((x - yb)/b) = \Im(x/b - y) = \Im(x/b) - \Im(y).$$ Let $y = n\theta$, so that $\Im(y) = n \sqrt{19}/2$, and $\Im((x - yb)/b) = \Im(x/b) - n \sqrt{19}/2$. There is some $n$ such that $\Im(x/b) - n \sqrt{19}/2 > 0$ and $\Im(x/b) - (n + 1) \sqrt{19}/2 < 0$.

Therefore either $\lvert \Im((x - n\theta b)/b) \rvert = \lvert \Im(x/b) - n \sqrt{19}/2 \rvert \le \sqrt{19}/4$ or $\lvert \Im((x - (n + 1)\theta b)/b) \rvert = \lvert \Im(x/b) - (n + 1) \sqrt{19}/2 \rvert \le \sqrt{19}/4$, and so we have some $a \in I \backslash Rb$ (either $a = x - n\theta$ or $a = x - (n + 1)\theta$) with $\lvert \Im(a/b) \rvert \le \sqrt{19}/4$.

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