Andrew,
Thank you for helping me to have a flash of insight.  For any prime
power p^n there is a unique Galois field of order p^n.  For p^1 the field
is isomorphic to the integers mod p, as we discussed after class.  For
p^n, GF(p^n) is isomorphic to the direct product (n times) of the
integers mod p.  I think that is all of the finite fields but I will
check on the case of composite numbers.  Probably I should have thought
of this sooner because finite fields were the subject in the projective
geometry seminar today.
        As for a=a and b=b => a+b=a+b, I think you can't say anything
about it really because Rudin leaves '=' undefined.
        I would prefer not to have you do the book proofs in lecture.
Those are available in the book.
        Something I got from your statement of your teaching philosophy
has been helpful to me in 1350 lecture.  Every day I start by reminding
them about what we talked about last time and previewing the subject for
today.  I don't know if it does them any good but it certainly helps me
feel more organized.  This is my first teaching experience so I can use the
help.

I noticed something interesting about the finite field of 5 elements.
There is a sense in which we can think of this field as
{0, 1, -1, i, -i}.
Why?  Because it is isomorphic to the integers mod 5 and so
2^2 = 3^2 = 4 = -1,
and  1^2 = 4^2 = 1.

See you Tuesday.
Cary

Not quite. All xi must be positive. So,
your counterexample cannot be applied.

Let me repeat the problem:
Let x1, x2, ..., xn be positive elements of R. 
Suppose x1x2...xn = 1. 
Prove that x1 + x2 + ... + xn >= n.

This is a particular case of the famous
inequality that the geometric mean
is not larger then the arithmetic mean.

See problems 7-8 from Shilov.