Markers: Give 1 point for any reasonably correct solution
(note some questions are is not assessed). Give the full point for any
solution to number 3 which is even vaguely reasonable.


1. If Z+Z\alpha is a subring then \alpha^2 is certainly in the set.
Conversely if \alpha^2 is in Z+Z\alpha then any power of \alpha is
so is any linear combination of powers of \alpha. But this set is
a ring (closed under +, times and additive inverses). In each of the
sample numbers just calculate the square and see if it is in the Z span
of 1 and the number. For instance

        \alpha = (1/2) + (\sqrt 5)i/2

gives

        \alpha^2 = (1/4)(-4 +2i \sqrt 5 ) = -(1/2) + \alpha

If this were a lin comb of 1 and alpha so would 1/2 be and  
we would have integers m and n so

        1/2  =  m + n(1/2 + i\sqrt 5/2) = (n/2 + m) +  n i \sqrt 5)

so n=0 and m is a whole number contradiction.  [ 1 point for mostly correct]

2. The darkened points include 2+ root(3)i  and  -3 + 2 root(3)i. These
are perpendicular and all the others are an integer times 2 + root(3)i
plus an integer times -3 + 2 root(3) i.  Every element of the form
m + n root(3)i can be obtained by adding one such element to
one of the following elements: 0, a+root(3)i , b+root(3)i  where
a is -1,0, or 1  and b is -2, =1, or 0. So in every element of
the ring is congruent to exactly one of these seven elements. 
Other similar solutions are possible.


3. The complex plane is 'tiled' by rectangles  of
area say A whose vertices are the elements of O. The
ideal generated by a+bi  is generated by rectangles
of area  (a^2+b^2)A. Roughly, there are a^2+b^2 little
rectangles filling up each big one. 

4. The condition on the function e:  a+bi |-> a^2+b^2 is that
for every  pair of elements p and q of the ring O with q nonzero there 
are s and r so

          p = qs+r   where       r=0   or   e(r) < e(q)

This is the same as saying for any pair p,q the absolute value of
 
     p/q - s

is less than one for some s in O. Any complex number is arbitrarily
close to one of the form p/q (even taking q to be an ordinary integer)
so this is the same as saying any complex number is at most distance
one from an element of O.    [1 mark for mostly correct]

5. The solution is basically given in the question. If this ideal
I were principal then the number of cosets [O:I] would be m^2=3n^2
where m+ni is a generator. But 2 is not of this form so I is not principal.
 [ markers be careful to check studens say something at least slightly
different from statement of question to get one mark]

6. Again the solution is in the hints. [Markers check they have added
at least some comments of their own for one point]


7. Let's ask, what are the invertible elements in C(R,R)? These are
functions f(t) such that there exists a g(t) with f(t)g(t)=1. These
are the nowhere vanishing functions (which take either strictly
positive or strictly negative values). 

If f(t) is a divisor of the sin function then f(t)g(t)=sin(t) for
all t, where g(t) is a continuous function. It follows that the
zeroes of f(t) are all integer multiples of pi.

Now, if f(t) takes more than one zero, consider two adjacent zeroes a,b.
We then have an interval [a,b]  where a and b are integer multiples
of pi,  such that  f(a)=f(b)=0 but f(t)  is not zero  if a < t < b.
Choose any point c  between a and b. I have f(c) is not zero and
consider two functions  defined

g(t)= f(c)        if t less than or equal to c
g(t) = f(t)/f(c)   if  t greater than c


h(t) = f(t)/f(c)  if t less than or equal to c
h(t) = f(c)        if  t greater than c. 

We have h(t)g(t) = f(t) obviously and h(t) and g(t) aren't invertible
so f(t) is not irreducible. This proves irreducible divisors of sin
take at most one zero and so the sin function can't be a product of finitely
many.  Of course then it does not have a unique factorization as a product
of irreducibles, it has no such factorization.
[markers: be generous! this is hard! 1 point for anything reasonable
but note in class I made the statement that t divides into sin(t)]

8. This challenge question is not assessed. Let r be a nonzero
element of R.  Let's show it is contained in just finitely
many ideals. Suppose r is contained in an ideal I. 
Then because ideals are principal we have an element s with I=Rs.
Then r is an element of Rs  so r is a multiple of s. We know
R has unique factorization by a theorem in the course. So
if the prime factorization of r is  p_1p_2...p_n with the p_i
possibly including repetitions, then  all divisors of r
are obtained by multiplying one of the 2^n  subsequences of (p_1,...,p_n)
and multiplying by an invertible element. The invertible element
does not affect the ideal so there are at most 2^n ideals containing r.
Note there could be fewer because some p_i may be an invertible element
times another. 

The next part of 8 is a challenge question and I say "see me if you solve
it."  The question is, what extra properties does a principal ideal domain
have to have in order to be Euclidean.