# [SOLVED] Algebra help

• Apr 27th 2005, 01:37 PM
Devilutza
[SOLVED] Algebra help
Hello everybody,
I have some algebra problems that I cannot solve and I'm hoping for some help.

1. Let C be a countable set. Prove that any linear well-ordered on C with the property that whatever c in C there are only finitely elements c` with c`<c, is unduced from the canonical order on N via a bijection N-> C. (N - natural no)

2. Prove that all algebraic numbers (all roots of polynomials with rational coefficients) form a countable subfield of C (complex).

3. Find a representation of the ring C[[x]] as an inverse limit of rings which have finite dimension over C.(complex).

4. Prove that the field Qp is uncountable. (Qp=field of fractions of Zp)

Well... there are some more but these are enough for now. Hope to get good complete answers as soon as possible. Thank you all.

Bye Bye
• Apr 27th 2005, 10:41 PM
Shmuel
Number 4 is straight forward. You just need a surjection onto the reals (well in this case the non-negative reals which are also uncountable).

Hint. Given a number p any real number can be written in the form
SUM from j=n to infinity(ajp^-j) with 0<=aj<p for some n E Z.
Do you know any representations of Qp that is similar?
The surjection doesn't not need any homomorphic properties.

Easier than a diagonal argument. Though there is also a diagonal argument.
• May 3rd 2005, 11:33 AM
hpe
Quote:

Originally Posted by Devilutza
2. Prove that all algebraic numbers (all roots of polynomials with rational coefficients) form a countable subfield of C (complex).

To show that this is a countable set, it is sufficient to show that the set of all polynomials with integer coefficients is countable (rational coefficients can be made integer by using common denominators). This set is the union of the sets of all such polynomials of degree n, n >=0. Each such set is countable (equivalent to the cartesian product of finitely many countable sets), and their countable union therefore is also countable.

To show it's a field, show that it is closed under addition, multiplication, negation, and taking the reciprocal. E.g. if p(z) = 0, then q(-z) = 0 for q(x)=p(-x), and r(z^(-1)) = 0 where r(x) = x^np(x^(-1)), n = degree of p.