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\centerline{ \bigbf Notes on III.3, Problem 8, September 6, 1989 }
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\centerline { \bigbf Math 401 -- Second Course in Abstract Algebra, I}
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\centerline { \bigbf Prof. Daniel Grayson }
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Problem 8, III.3, was the following.  We consider the subring $R$ of $\C$,
where $R = \Z + \Z \omega$ and $\omega = \displaystyle { 1 + \sqrt{-19} \over 2}$.  We are to
prove that $R$ is a principal ideal domain.  

Here is an approach that seems to work, and was devised jointly with several
students who came to my office hours yesterday.  Suppose $I$ is an ideal of
$R$, and consider its elements as vectors in the complex plane.  Pick the
shortest nonzero vector $v$ in $I$, i.e., the one which has $|v|$ as small as
possible.  Now try to prove that $I = vR$, thus showing that $I$ is principle.

Here are some tools which make this proof possible to carry out.

The ideal $I$ is a subroup of $\C$ isomorphic to $\Z \oplus \Z$.
Moreover, $I$ is a discrete subgroup of $\C$, which means that each element of
$I$ is contained in an open disk which contains no other element of $I$.
The quotient $\C/I$ is an abelian group, and can also be thought of as a torus,
because $(\R \oplus \R) / ( \Z \oplus \Z ) = \R/\Z \oplus \R/\Z = S^1 \oplus
S^1 $.  The quotient $\C/I$ also has a Riemannian metric which it inherits from
$\C$, and since it is compact, its total area is a finite real number, which we
will call $\Area(\C/I)$.

In fact, one can introduce this notion of area without referring to Riemannian
metrics at all, as follows.  Choose a basis over $\Z$ for $I$, so that $I = \Z
w_1 \oplus \Z w_2$, say.  Then define $\Area(\C/I)$ to be the area of the
parallelogram spanned by $w_1$ and $w_2$ in the complex plane.  One must check that
the number obtained this way is independent of the choice of basis $\{w_1,w_2\}$,
and this can be done by making use of the fact that the change of basis matrix
connecting two bases of $I$ is a two-by-two matrix with integer entries with
determinant equal to $1$ or $-1$.

Now one proves the following formulas.
$$
	|v|^2 { \sqrt 3 \over 2 } \le \Area(\C/I)		\tag 1
$$
$$
	\Area(\C/R) = {\sqrt{19} \over 2}			\tag 2
$$
$$
	{ \Area(\C/I) \over \Area(\C/R) } = \#(R/I)		\tag 3
$$
$$
	\#(R/vR) = |v|^2					\tag 4
$$
Formula (1) is the hardest to prove, and involves the idea that $\Area(\C/I)/|v|^2$
will be smallest when $I$ corresponds to a tiling of the complex plane by
equilateral triangles.

It is possible to combine the formulas above to prove that 
$$
	\#(I/vR) \le \sqrt{19/3} < 3
$$
Finally, one can prove that no $R$-module can have two elements, by examining
what multiplication by $\omega$ could do to the nonzero element, in light of
the equation $\omega^2 - \omega + 5 = 0$.  It then follows that $I = vR$, and
thus that $I$ is principal.

I hope that someone can come up with a simpler approach to this problem before
the semester is out.

We came up with no ideas for proving that $R$ is not a Euclidean domain.


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