# Fundamental theorem of algebra

1. Feb 8, 2008

### ehrenfest

[SOLVED] fundamental theorem of algebra

1. The problem statement, all variables and given/known data
Theorem: If F is a field, then every ideal of F[x] is principal.
Use the above theorem to prove the equivalence between these two theorems:

Fundamental Theorem of Algebra: Every nonconstant polynomial in C[x] has a zero in C.

Nullstellansatz for C[x]: Let f_1(x), ...,f_r(x) in C[x] and suppose the every alpha in C that is a zero of all r of these polynomials is also a zero of a polynomial g(x) in C[x]. Then some power of g(x) is in the smallest ideal of C[x] that contains the r polynomials f_1(x),..., f_r(x).

2. Relevant equations

3. The attempt at a solution
Lets prove the FTA implies that Nullstellansatz first. Say p(x) generates the ideal. I think its pretty clear that all the alpha's will be zeroes of p(x), right? So, what does that mean about the relationship between p(x) and powers of g(x). All the alphas are zeroes of both...but why does that necessarily mean p(x)q(x)=g(x)^n for some q(x)...

2. Feb 10, 2008

### ehrenfest

anyone?

3. Feb 10, 2008

### morphism

Since fi(x) = ai(x)p(x) for each i, then every root of p(x) is a root of each fi(x).

4. Feb 10, 2008

### ehrenfest

So, what you are saying is that the roots of p(x) and g(x) are identical? Does that imply that g(x)^n will have p(x) as a factor for some n?

5. Feb 10, 2008

### morphism

The roots of p(x) and g(x) don't have to be identical -- for instance, g(x) can have more roots. However, the FTA tells us that p(x) splits into linear factors -- and each of these linear factors occurs in g(x). Consequently, g(x)/p(x) is in C[x] and g(x) = (g(x)/p(x)) * p(x) is in I.

6. Feb 10, 2008

### ehrenfest

That's what the FTA usually means to me: that any polynomial over the reals factors into linear and irreducible quadratic factors, and the quadratic formula tells us that the irreducible quadratic factors reduce to linear complex factors.

Is the statement that "every nonconstant polynomial in C[x] has a zero in C" equivalent to that in some obvious way that I am missing?

7. Feb 10, 2008

### morphism

Yeah. Say p(x) has a root a1. Then (x - a1) divides p(x), and p(x)/(x - a1) is either constant or has a root a2. If it's the latter then (x - a2) divides p(x)/(x - a1), and p(x)/((x - a1)(x - a2)) is either constant or has a root a3. And so on. This process will eventually terminate (why?) leaving us with the fact that p(x) = k (x - a1)(x - a2)...(x - an) for some k in C.

8. Feb 10, 2008

### ehrenfest

That makes sense.

The process eventually terminates because the degree of the polynomial is finite and you reduce it at each iteration.

9. Feb 11, 2008

### ehrenfest

OK. Now the other direction. So, we need to take an arbitrary polynomial f(x) in C[x] and somehow plug it into the Nullstellansatz to get a zero. So, we probably want to look at the ideal generated by f(x). Then I am kind of stuck. How do I choose the r polynomials and how do I know that <f(x)> is the smallest ideal that contains them?

10. Feb 11, 2008

### NateTG

Can you relate the roots of a polynomial to its factors?

11. Feb 11, 2008

### ehrenfest

Umm...yes, what are you getting at? How do you know that f(x) has any nonconstant factors. It is just an arbitrary polynomial in C[x].

12. Feb 11, 2008

### NateTG

Well, the greatest common factor of the f's is almost a factor of g, right?

13. Feb 12, 2008

### ehrenfest

You are speaking of "the f's" and "g", but all we have right now is one polynomial f(x) in C[x] and we need to show that it has a zero. Will you please tell what the "f's" are and what g is?

14. Feb 12, 2008

### NateTG

Sorry, I was still thinking in the other direction...

Assume, by contradiction, that $p(x)$ is a non-constant polynomial without a root.
Then let
$$f_1(x)=p(x)$$
$$g(x)=1$$

15. Feb 12, 2008

### ehrenfest

Yes! That does it. You can multiply 1 by itself as many times as you want and it will never fall into the ideal <p(x)> when deg(p(x)) > 0.