Fundamental theorem of algebra

[ehrenfest]

[SOLVED] fundamental theorem of algebra

Homework Statement

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).

Homework Equations

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)...

 

[ehrenfest]

anyone?

 

[morphism]

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

 

[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?

 

[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.

 

[ehrenfest]

However, the FTA tells us that p(x) splits into linear factors -- and each of these linear factors occurs in g(x).

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?

 

[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 - a_n) for some k in C.

 

[ehrenfest]

That makes sense.

(why?)

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

 

[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?

 

[NateTG]

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?

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

 

[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].

 

[NateTG]

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].

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

 

[ehrenfest]

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

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?

 

[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

 

[ehrenfest]

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

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.