Proofs in Linear Algebra: Countable Sets, Algebraic Numbers, and Fields

[blaster]

Linear Algebra proof

I would appreciate any help with any of the foolowing:

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)

 

[snoble]

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_{j=n}^\infty a_j p^{-j} with 0 \le a_j &lt; p for some n\in \mathbb{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.

 

[M98Ranger]

1. To prove that any linear well-ordered on C with the given property is induced from the canonical order on N via a bijection, we need to show that there exists a bijection f: N -> C such that for any two elements n, m in N, n < m if and only if f(n) < f(m) in C.

First, we know that C is countable, which means that there exists a bijection g: N -> C. Let's define a new function h: N -> C such that h(n) = g(n+1). This function essentially shifts the elements of C by one, so that the first element of C is now mapped to the second element, the second element to the third, and so on.

Next, let's define a new function f: N -> C such that f(n) = g(1) if n = 0, and f(n) = h(n-1) if n > 0. This function essentially maps the first element of N to the first element of C, and then maps the remaining elements of N to the remaining elements of C in the same order.

Now, let's consider any two elements n, m in N. If n < m, then we know that f(n) = h(n-1) < h(m-1) = f(m) in C, since h is a bijection. And if f(n) < f(m) in C, then we can easily see that n < m, since f(n) = g(1) if n = 0, and f(n) = h(n-1) if n > 0.

Therefore, we have shown that there exists a bijection f: N -> C such that for any two elements n, m in N, n < m if and only if f(n) < f(m) in C. This proves that the linear well-order on C is induced from the canonical order on N via a bijection, as desired.

2. To prove that all algebraic numbers form a countable subfield of C, we first need to show that the set of all polynomials with rational coefficients is countable. This is because each polynomial can be represented as a finite sequence of rational coefficients, and the set of all finite sequences of rational numbers is countable.

Next, we know that the set of all roots of a polynomial with rational coefficients is finite. This