# Fundamental theorem of algebra

• ehrenfest
In summary: This is a contradiction, because if p(x) was a non-constant polynomial without a root, then there would be a root of p(x) that isn't a zero of any other polynomial.
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).

## 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)...

anyone?

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

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?

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.

morphism said:
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?

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.

That makes sense.

morphism said:
(why?)

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

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?

ehrenfest said:
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?

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].

ehrenfest said:
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?

NateTG said:
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?

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
$$f_1(x)=p(x)$$
$$g(x)=1$$

NateTG said:
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.

## What is the Fundamental Theorem of Algebra?

The Fundamental Theorem of Algebra states that every non-constant polynomial function with complex coefficients has at least one complex root.

## Why is the Fundamental Theorem of Algebra important?

The theorem is important because it guarantees the existence of solutions to polynomial equations with complex coefficients. It also serves as the foundation for many other theorems and concepts in algebra and complex analysis.

## Who proved the Fundamental Theorem of Algebra?

The theorem was first proved by the mathematician Carl Friedrich Gauss in 1799. However, it was later refined and expanded upon by other mathematicians, including Augustin-Louis Cauchy and George Pólya.

## Can the Fundamental Theorem of Algebra be applied to polynomials with real coefficients?

Yes, the theorem can be applied to polynomials with real coefficients, as the set of complex numbers includes the set of real numbers. This means that the theorem also guarantees the existence of solutions for polynomial equations with real coefficients.

## Is there a practical application for the Fundamental Theorem of Algebra?

Yes, the theorem has many practical applications in fields such as physics, engineering, and computer science. It is used to solve polynomial equations and to analyze and model complex systems.

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