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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.

jacob78
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    To know the sources you might've missed we'd have to know where you already looked. – Randall Sep 11 '23 at 16:24
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    I can't imagine that it is possible to prove FTA with less machinery than this https://www.ime.usp.br/~oliveira/FTAAUTHOR.pdf Personally I don't think restriction to the real case is going to help. – ancient mathematician Sep 11 '23 at 16:41
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    I think it’s worth challenging the general premise – not that it necessarily applies to the OP’s case, but to point out its limits. “I’m temporarily accepting a theorem as valid because the proof is too advanced for my current self or uses concepts I’m currently unfamiliar with” is not bad. At a high level, it even becomes necessary, because in a lot of math, problem statements are simpler and shorter than the answers. – Aphelli Sep 11 '23 at 16:49
  • Check this: https://math.stackexchange.com/q/3195875/42969 – Martin R Sep 11 '23 at 17:26
  • there are proofs based on this: real coefficient, odd degree, there is a real root. I won't claim the jump to even degree is easy. – Will Jagy Sep 11 '23 at 17:29
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    @WillJagy I think that there are fields over which every odd-degree polynomial has a root but there are irreducible polynomials of arbitrarily large even degree. So - unless my memory is wrong - the jump to even degree is truly subtle. – Noah Schweber Sep 11 '23 at 17:32
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    Take a look at this Numberphile video with David Eisenbud which provides, not quite a proof, but a convincing argument that only requires minimal knowledge of complex numbers. https://youtu.be/shEk8sz1oOw. – Euro Micelli Sep 12 '23 at 04:40
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    @NoahSchweber Theorems 3 and 4 of Improving the Fundamental Theorem of Algebra by Joe Shipman imply your claim, at least if "irreducible" is replaced by "rootless." Shipman shows that much more is true. For example, you can pick any prime $p$, and there are fields over which every degree-$\ell$ polynomial has a root (where $\ell$ is any prime different from $p$) but for every $k$ there is a polynomial of degree $kp$ that is rootless. In this sense, 2 is not any different from any other prime. – Timothy Chow Sep 12 '23 at 12:20
  • Hi you would be surprise with iterated method like continued fraction or Newton's method .In one word try with https://math.stackexchange.com/questions/4738155/got-a-factored-version-of-the-taylors-series Ps: there is hidden fractal in polynomials and the golden ratio – Miss and Mister cassoulet char Sep 12 '23 at 11:20
  • I presented two well known proofs in my blog post https://paramanands.blogspot.com/2013/06/fundamental-theorem-of-algebra-two-proofs.html?m=0. Both proofs do need some bit of analysis. I doubt if there are easier versions available. – Paramanand Singh Oct 08 '23 at 16:15

5 Answers5

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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.

Qiaochu Yuan
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    When I first read the question my immediate instinct was exactly your first sentence, but I couldn't put my finger on why (it was just gut). This explains it very nicely. – Randall Sep 11 '23 at 18:03
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    Technically, the Galois theory proof also uses the fact that complex numbers have square roots, which follows from the fact that nonnegative real numbers have square roots (again a consequence of the intermediate value theorem). – Daniel Schepler Sep 12 '23 at 17:29
  • one thing that might be interesting is how simple a "proof-like-object" could be made. specifically, Kevin McGerty's proof can be simplified a lot (losing most of the rigor along the way but retaining most of the spirit) – Oscar Smith Sep 13 '23 at 05:43
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    specifically, the simplified version of the proof is something like: let p be an n degree polynomial who's first non constant term is ax^k. |p| must have a global minimum at some finite point x, because for large x, |x^n| dominates and goes to infinity. assume for contradiction p(x)!=0. for points b near x, p(b) looks like p(x)+ax^k. since |ax^k| goes below 0, x is not the minimum point of |p|. – Oscar Smith Sep 13 '23 at 05:46
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    @OscarSmith |ax^k| goes below 0 ... looks like you lost the analogy. :) I'd use p(b+x) and b small? Or maybe a(b-x)^k? – Yakk Sep 14 '23 at 21:33
  • that's what I get for doing this on a phone. – Oscar Smith Sep 15 '23 at 00:05
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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.)

Noah Schweber
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  • Actually what you say (constructing $\mathbb R$ by completion of $\mathbb Q$) is the easiest part, because more intuitive and relying on Cauchy sequences which appear early in undergraduate studies. More difficult to grasp is, I believe, the proof that there is always one root in $\mathbb C$. – Jean-Armand Moroni Sep 11 '23 at 17:28
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    @Jean-ArmandMoroni I agree. The point of my answer is that because the construction of the real numbers is of fundamentally different "flavor" than the result we're trying to extract, we should expect the extraction of that result to be hard. – Noah Schweber Sep 11 '23 at 17:29
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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.

Timothy Chow
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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.

Acccumulation
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    Of course you mean a polynomial of degree $\ge 1$. The interesting part of this, I think, is showing that $0$ is inside $C_r$ for large enough $r$, which you can do using the fact that (if the degree is $d$ and the coefficients are $a_j$) $|f(z) - a_d z^d| < |a_d z^d|$ when $|z|$ is large enough. – Robert Israel Sep 13 '23 at 01:17
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    ... so basically it's using Rouché's theorem. But if you're going to bring complex analysis into it, you might as well prove it using Liouville's theorem. – Robert Israel Sep 13 '23 at 01:29
  • @RobertIsrael Phrasing it as $f(x) \neq 0$ rather than degree $\geq 1$, it's (hopefully) more obvious why that assumption can be made; if $f(0)=0$, then clearly there's the root $z=0$. – Acccumulation Sep 13 '23 at 03:49
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    The point is that if you just say $f(0) \ne 0$, $f$ could be a nonzero constant, which of course has no roots. – Robert Israel Sep 13 '23 at 04:08
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For the source part of the question: A proof is outlined in Multivariable Calculus With Applications by P. Lax and M. Terrell, Springer 2017.

Bob Terrell
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