Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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
add comment

2 Answers 2

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.

share|improve this answer
    
$\pm \theta$ 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
add comment

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 $\lvert \Im(a/b) \rvert \le \sqrt{19}/4$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.