# Roots of polynomials in C

• I
Homework Helper
Gold Member
Summary:
Every polynomial in C of degree n has exactly n roots in C.
How do we prove that every polynomial (with coefficients from C) of degree n has exactly n roots in C?

This is not a homework (I wish I was young enough to have homework) I guess this is covered in every typical undergraduate introductory algebra course but for the time being I cant find my algebra book from my undergraduate studies (25 years ago).

anuttarasammyak
Gold Member
Fundamental theorem of algebra, it is called. You will be able to find a good proof on the web.

• WWGD and Delta2
fresh_42
Mentor
It is only true if you count multiplicities. So a complex polynomial has at most ##n## distinct roots. There are quite a lot of very different proofs. Hence the question is: what are we allowed to use?

The shortest answer is: ##\mathbb{C}## is algebraically closed.

martinbn
The shortest answer is: ##\mathbb{C}## is algebraically closed.
That's circular.

• swampwiz, graphking, S.G. Janssens and 1 other person
martinbn
Fundamental theorem of algebra, it is called.
I had an algebra professor who liked to say that it is no longer a fundament theorem in algebra, and it has never been an algebra theorem (every proof requires analysis).

fresh_42
Mentor
That's circular.
No, it is not. You still have a formal induction to perform. It is short.

martinbn
No, it is not. You still have a formal induction to perform. It is short.
How do you prove that ##\mathbb C## is algebraically closed?

fresh_42
Mentor
How do you prove that ##\mathbb C## is algebraically closed?
I haven't said the two statements weren't equivalent.

martinbn
I haven't said the two statements weren't equivalent.
How is that a proof then?

fresh_42
Mentor
How is that a proof then?
It stretches my argument: proof depends on what is acceptable to be used! Admittedly in its extreme form, but anyway.

Here is another one:
All prime ideals in ##\mathbb{C}[x]## are generated by a single linear polynomial. The assertion now follows from Hilbert's Nullstellensatz.

And another one:
Irreducible polynomials over the reals have at most degree ##2## by the mean value theorem. Now all quadratic polynomials can be split over ##\mathbb{C}## by solving Vieta's formulas.

You never start from scratch, pardon, set theory and ZF. So any demand to prove something requires a list of known facts that hasn't been given. Algebraic completeness is simply the shortest fact to deduce the FTA.

martinbn
It stretches my argument: proof depends on what is acceptable to be used! Admittedly in its extreme form, but anyway.

Here is another one:
All prime ideals in ##\mathbb{C}[x]## are generated by a single linear polynomial. The assertion now follows from Hilbert's Nullstellensatz.

You never start from scratch, pardon, set theory and ZF. So any demand to prove something requires a list of known facts that hasn't been given. Algebraic completeness is simply the shortest fact to deduce the FTA.
But Hilbert theorem assumes algebraically closed fields.

Of course you dont have to go to ZF, but showing it is equivalent to something else is not always a poof. They can be both wrong for example. Surely you would agree that showing that the Riemann hypothesis is equivalent to some other statement is not a proof and you dont go to collect you million bucks.

• S.G. Janssens
fresh_42
Mentor
But Hilbert theorem assumes algebraically closed fields.

Of course you dont have to go to ZF, but showing it is equivalent to something else is not always a poof. They can be both wrong for example. Surely you would agree that showing that the Riemann hypothesis is equivalent to some other statement is not a proof and you dont go to collect you million bucks.
And my second solution uses topological completeness of ##\mathbb{R}##. You always use something. The only question is, what this something may be. If you are willing to accept the truth of an equivalent statement, then we're done. Otherwise, you keep going on to ask down the entire road to ZF.

martinbn
The shortest proof I know uses Liouville's theorem, if ##p(z)## has no roots, then ##\frac1{p(z)}## is entire and bounded, thus a constant.

• FactChecker, S.G. Janssens and wrobel
wrobel
I like variational argument. This proof is not shortest one but it is still quite elementary and uses some fundamental ideas that go far outside complex analysis.

Let $$f(z)=\sum_{k=0}^n a_kz^k,\quad a_n\ne 0,\quad n\in\mathbb{N}$$ be a polynomial.

It is easy to see that ##|f(z)|\to\infty## as ##|z|\to\infty##. This property is referred to as coercivity. It implies that the function ##z\mapsto|f(z)|## attains its minimum in ##\mathbb{C}:## $$\exists z':\quad |f(z')|=\inf_{z\in\mathbb{C}}|f(z)|.$$ Then it is not hard to show that this minimum is a root: ##f(z')=0##.

Last edited:
• • swampwiz, S.G. Janssens, martinbn and 1 other person
Homework Helper
Gold Member
No, it is not. You still have a formal induction to perform. It is short.
What exactly do you mean that ##\mathbb{C}## is algebraically closed ( do you mean that every polynomial has at least one root?) and what is this induction you speak about?

fresh_42
Mentor
What exactly do you mean that ##\mathbb{C}## is algebraically closed ( do you mean that every polynomial has at least one root?) and what is this induction you speak about?
You can see from the discussion about it, that it is closely related. This means it depends on the definition of algebraically closed and the phrasing of the fundamental theorem of algebra what is left to prove.

FTA = every polynomial has a zero
Closure = every polynomial splits into linear factors

Closure ##\Longrightarrow ## FTA: obvious
FTA ##\Longrightarrow ## Closure:
##\deg p =1:\quad p=x-z_1## Induction basis.
##\deg p=n:\quad \exists z_1\, : \,p(z_1)=0 \Longrightarrow p(x)=(x-z_1)p_1(x) \wedge \deg p_1=n-1##
Induction hypothesis: ##p_1(x)=(x-z_2)\cdot\ldots\cdot (x-z_{n}) ##
Induction step: ##p(x)=(x-z_1)p_1(x)=(x-z_1)(x-z_2)\cdot\ldots\cdot (x-z_{n}) ##

This is an extreme example, as the two statements are equivalent or even the same if you define and phrase them accordingly. It only should have demonstrated that it is important to describe where to start, i.e. what can be taken for granted.

You can also see from the discussion above, that there are dozens of very different proofs. Each one uses a different technique from a different mathematical field. This is why it is so important to describe the starting point of a possible proof, in this case more than in any other.

• Delta2
lavinia
Gold Member
• Delta2
WWGD
Gold Member
I like variational argument. This proof is not shortest one but it is still quite elementary and uses some fundamental ideas that go far outside complex analysis.

Let $$f(z)=\sum_{k=0}^n a_kz^k,\quad a_n\ne 0,\quad n\in\mathbb{N}$$ be a polynomial.

It is easy to see that ##|f(z)|\to\infty## as ##|z|\to\infty##. This property is referred to as coercivity. It implies that the function ##z\mapsto|f(z)|## attains its minimum in ##\mathbb{C}:## $$\exists z':\quad |f(z')|=\inf_{z\in\mathbb{C}}|f(z)|.$$ Then it is not hard to show that this minimum is a root: ##f(z')=0##.
What if we restricted to a ball about the origin and then we can use that Continuous functions on compact sets achieve extrema and show minimum is 0? Edit: Maybe we can use Analyticity and then f(z) has finitely many lacunary values and go from there?

fresh_42
Mentor
What if we restricted to a ball about the origin and then we can use that Continuous functions on compact sets achieve extrema and show minimum is 0? Edit: Maybe we can use Analyticity and then f(z) has finitely many lacunary values and go from there?
The shortest proof I know uses Liouville's theorem, if ##p(z)## has no roots, then ##\frac1{p(z)}## is entire and bounded, thus a constant.

wrobel
What if we restricted to a ball about the origin and then we can use that Continuous functions on compact sets achieve extrema and show minimum is 0? Edit: Maybe we can use Analyticity and then f(z) has finitely many lacunary values and go from there?
Let's hold an elementary argument. Let ##z_k## be a minimizing sequence:
$$|f(z_k)|\to\inf_{z\in\mathbb{C}}|f(z)|,\quad |f(z_1)|\ge |f(z_2)|\ge\ldots$$
Such a sequence exists by definition of inf.

By coercivity this sequence is bounded ##\sup_k|z_k|<\infty.## Thus it contains a convergent subsequence: ##z_{k_j}\to z'##.

Then assume that ##f(z')\ne 0## and consider an expansion
$$f(z)=f(z')+c(z-z')^p+(z-z')^{p+1}Q(z-z'),\quad c\ne 0.$$

Last edited:
• WWGD and S.G. Janssens
mathwonk
Homework Helper
2020 Award
a polynomial map extends to a holomorphic map from the riemann sphere to itself, sending infinity to infinity. this map sends every closed, hence compact, set to a compact, hence closed, set. Moreover, if the polynomial is non constant, it is locally equivalent to a map of form x-->x^n, n≥ 1, (the proof uses the inverse function theorem), and is thus an open map. Thus the image of the riemann sphere is a subset of the riemann sphere which is both open and closed, hence either empty or the entire sphere. Since it is not empty, it is all of the riemann sphere. hence the image of the complex plane is the entire complex plane. in particular, zero is in the image, so every non constant polynomial has a zero.

Last edited:
wrobel
It seems that historically it was the first pure existence theorem. One does not solve an equation explicitly and even does not provide any constructive procedure to obtain a solution but just asserts that the solution exists

• Delta2
wrobel
I read somewhere that when David Hilbert explained to students what pure existence theorems are he said "there exist a person in this auditorium who has the least number of hair on his\her body.

lavinia
Gold Member
I have spoken to mathematicians who talk about irreducible complexity in mathematics. The idea is that theorems all require a certain amount of inescapable work. I am not sure if this has ever been rigorously measured or not.

A short proof, as @fresh_42 has emphasized, does not mean that is requires less work. It just takes a lot of work as a given - e.g. the use of Louiville's Theorem to prove that every complex polynomial has a root depends on foundational work in complex analysis. As he points out the shortest proof is " therefore all polynomials have a root". This proof assumes all of the necessary arguments as given.

All of the proofs of the FTA seem to depend on the topology of the real line and of the complex plane and this is a lot of foundational work. The paper that claimed a purely algebraic proof uses the hyperreals so I wonder if this is just a long end run around the usual approach. I haven 't read the paper.

For myself, the fascinating question about the Fundamental Theorem of Algebra is how it depends upon notions of calculus in the plane and becomes straight forward when one recognizes that polynomials are analytic functions. One wonders about the unity of mathematics when algebraic questions require analysis to be solved. There are many questions that seem to become tractable through analysis - e.g. questions in number theory. Why is this?

A related question is how much structure is ultimately needed to prove a theorem. Complex polynomials need not be viewed as entire functions but merely as smooth functions. In this case one needs to count singularities and show that a polynomial can be extended to a smooth function of the 2 sphere into itself - as @mathwonk explained in post #21. This is also the idea underlying the "coercivity" proof that @wrobel indicates in post #14.

• fresh_42
martinbn
• 