Is there an easier way to prove the Fundamental Theorem of Algebra for polynomials with real coefficients?

Question

My current level of maths does not allow me to understand any of the proofs I was able to find online for the Fundamental Theorem of Algebra. I find it very unsettling having to continue learning about polynomials without being able to grasp such a fundamental property of them that is there exists a complex root for any polynomial with complex coefficients.

My question is, is there a way I can reach the same conclusion of the Fundamental Theorem of Algebra for polynomials with real coefficients instead of complex coefficients. In other words, to prove that at least one complex root exists for a polynomial with real coefficients. Does restricting the coefficients to be real instead of complex make things any easier?

If the answer to my question above is no, then I would appreciate it if you could provide any sources that I could've missed, that explain the Theorem by using the least amount of mathematical notation and advanced concepts as possible.

If I cannot figure this out, I am going to have to accept the existence of a root as an axiom going further, which is something I really don't want to do.

Answer 1

The fundamental theorem of algebra is exactly as difficult for real vs. complex coefficients. The reason is that if $f(x) = f_0 + \dots + f_n x^n$ is any polynomial with complex coefficients, then the polynomial

$$\overline{f(\overline{x})} f(x) = (\overline{f_0} + \dots + \overline{f_n} x^n)(f_0 + \dots + f_n x^n)$$

has real coefficients (because it is invariant under coefficient-wise complex conjugation), and its roots are exactly the roots of $f$ together with their complex conjugates (this is a nice exercise). So any proof of the FTA for real coefficients immediately yields the FTA for complex coefficients.

It speaks well of you that you want to understand this result instead of taking it on faith; unfortunately it is one of the first "genuinely difficult" results one comes across in mathematics. There are no really easy proofs, one has to really understand something. You can find a lovely collection of proofs here on MO but all their prerequisites involve a certain amount of analysis or topology or in one case Galois theory and this is unavoidable. (The Galois theory is used to reduce the amount of analysis necessary to the fact that a real polynomial of odd degree has a real root (edit: and, as Daniel Schepler points out, to the fact that a non-negative real number has a real square root), which follows from the intermediate value theorem.)

FWIW I think this proof shared by Kevin McGerty is the least technical on the list but it still requires some familiarity with analysis.

Answer 2

The fundamental theorem of algebra is, in my opinion, the most striking example of a fact which is comprehensible and relevant very early on mathematically but which really has no simple explanation. What we can do in this situation is try to explain why it is so difficult to prove.

We know that FTA fails for $\mathbb{Q}$: for any $n>1$ the polynomial $$x^n-2$$ is irreducible over $\mathbb{Q}$ despite having rational coefficients. So the key that makes FTA work has to lie in the construction of the reals. There are a few different ways to construct the reals from the rationals - the most common are via Dedekind cuts or (equivalence classes of) Cauchy sequences, but there are others - but in my opinion they are all somehow more about geometry than algebra (this is certainly true for the Dedekind/Cauchy constructions). Basically, we observe that $\mathbb{Q}$ has "holes" and our construction of $\mathbb{R}$ amounts to filling these holes in some mathematically-precise way.

This means that to prove FTA we need to somehow link the geometric ideas that went into the construction of $\mathbb{R}$ with the various algebraic properties that $\mathbb{Q}$ lacks. And this sort of link is going to be rather intricate. This is all very subjective of course, but I think it tells a coherent story of how we should expect FTA to be harder to prove than many similar-sounding statements.

(A further technical subtlety is that we can't focus on polynomials of any fixed degree: for any $n$ there are non-algebraically-closed fields in which every polynomial of degree $<n$ has a root.)

Answer 3

Though the following comments don't directly answer your stated question, perhaps they will be of some help.

A field $F$ is said to be algebraically closed if all the roots of a polynomial with coefficients in $F$ are in $F$. With this terminology, the fundamental theorem of algebra states that $\mathbb{C}$, the field of complex numbers, is algebraically closed.

It's worth pointing out that constructing some algebraically closed field is not such a difficult thing to do. Let's say we start with $\mathbb{Q}$, the field of rational numbers, and we want to expand $\mathbb{Q}$ to an algebraically closed field. Conceptually, the simplest thing we could try is to begin by taking all the roots of all polynomials with coefficients in $\mathbb{Q}$ (this is the set of all algebraic numbers, denoted by $\overline{\mathbb{Q}}$), and adjoining them to $\mathbb{Q}$. Any algebraically closed field containing $\mathbb{Q}$ must certainly, at a minimum, contain all the algebraic numbers. Next, it might seem that we have to iterate, and adjoin all roots of polynomials with coefficients in $\overline{\mathbb{Q}}$. However, there are two very nice facts:

Fact 1. $\overline{\mathbb{Q}}$ is a field.

Fact 2. The root of a polynomial with coefficients in $\overline{\mathbb{Q}}$ is already in $\overline{\mathbb{Q}}$.

So we don't need to iterate after all; we get our algebraically closed field after just one step of "manually adding all roots."

It turns out to be much easier to prove Fact 1 and Fact 2 than to prove the Fundamental Theorem of Algebra; see this question for example. So if you're looking for a more easily proved theorem as a "stepping stone" on the way to the fundamental theorem of algebra, you might start by trying to understand the proofs of Fact 1 and Fact 2.

As a next step, it turns out that the arguments used to prove Fact 1 and Fact 2 can (if you know a bit of set theory and use the axiom of choice) be fairly straightforwardly adapted to expand any field $F$ to a larger field $\overline{F}$ that is algebraically closed. In particular, we can take $F = \mathbb{R}$, the field of real numbers, and expand $\mathbb{R}$ to a larger field $\overline{\mathbb{R}}$ that is algebraically closed. This result gets us kind of close to the actual fundamental theorem of algebra, and we have used only fairly straightforward arguments.

However, "kind of close" is not all the way there. What is remarkable about the fundamental theorem of algebra is that to get $\overline{\mathbb{R}}$ (which is secretly equal to $\mathbb{C}$), all you need to adjoin to $\mathbb{R}$ are the roots of the equation $x^2+1=0$, and you're done. You need not "manually add" all roots of all polynomials with real coefficients; this one equation does the job. This remarkable fact is not true of just any old field $F$; it is a special property of $\mathbb{R}$ that is shared by very few other fields (most importantly, so-called real closed fields). This explains why the proof of the fundamental theorem of algebra is subtle and tricky; it must somehow use special properties of $\mathbb{R}$ that are not enjoyed by just any old field such as $\mathbb{Q}$.

ADDENDUM: As Will Jagy alluded to in a comment, it is possible to pinpoint exactly what is special about $\mathbb{R}$ in this context. It turns out—though this is not so easy to prove—that if $F$ is a field in which every polynomial whose degree is a prime number has a root, then $F$ is algebraically closed (Shipman, Improving the Fundamental Theorem of Algebra). In the case of $\mathbb{R}$, it is easy to prove that every (monic) polynomial $f$ of odd degree has a root—if $x$ is a sufficiently large positive number then $f(x)>0$ and $f(-x)<0$, so the intermediate value theorem (which, by the way, is a large part of the motivation for the definition of the real numbers—we want to "fill in all the gaps" between the rational numbers) implies that $f$ must have at least one root somewhere between $-x$ and $x$. So the only "missing" prime is 2. This is a partial explanation (though not a complete proof) of why all you have to do to turn $\mathbb{R}$ into an algebraically closed field is to deal with the fact that it doesn't contain all its square roots.

Let me also recommend to you the book by Fine and Rosenberger, The Fundamental Theorem of Algebra, which contains many different proofs and auxiliary material. There is sure to be something in there which will help you understand this amazing and difficult theorem.

Answer 4

Suppose $f$ is a polynomial function with $f(0) \neq0$. Let $z = re^{i\theta}$, and let $C_r$ be the curved traced by the values of $f(z)$ for a fixed $r$ as $\theta$ ranges from $0$ to $2\pi$. For small enough $r$, $0$ will be outside $C_r$. For large enough $r$, $0$ will be inside $C_r$. Therefore, there is some $r$ in between such that $0$ lies on $C_r$, which means that there is a number $z$ such that $|z|=r$ and $f(z)=0$. I.e., $f$ has a root.

This isn't an entirely rigorous proof (for one thing, the question "What if $C_r$ is not a simple curve" has to be dealt with), but it strikes me as both simple enough to be understandable, and rigorous enough to be satisfying.

Answer 5

For the source part of the question: A proof is outlined in Multivariable Calculus With Applications by P. Lax and M. Terrell, Springer 2017.