How fundamental is the fundamental theorem of algebra?

Question

Despite its name, its often claimed that the fundamental theorem of algebra (which shows that the Complex numbers are algebraically closed - this is not to be confused with the claim that a polynomial of degree n has at most n roots) is not considered fundamental by algebraists as it's not needed for the development of modern algebra. My question is - what are the major uses of the theorem, and to which extent can they justify the claim that the theorem is fundamental for something?

An example I think of is the Jordan canonical form for matrices, but I don't think it suffices.

Answer 1

I can immediately think of four important areas for which the FTA is actually quite fundamental. Maybe other people may come up with more contributions.

1. Algebraic Geometry (as already touched in Agusti Roig's answer). In particular, we should couple the FTA with Lefschetz's Principle which basically says that the complex projective space is sort of "universal" environment for algebraic geometry in characteristic 0.

2. Classification of algebraic structures over local or global fields, where one uses constantly that $\Bbb C$ has no non trivial finite extensions (a trivial consequence of the FTA)

3. The theory of representations of groups, where the TFA plays a big role making the theory of complex representations simpler.

4. Galois theory and algebraic number theory, where the FTA allows to realize $\Bbb C$ as a good environment for studying finite extensions of $\Bbb Q$ (such in the basic resut that a number field of degree $n$ always admits $n$ independent embedings into $\Bbb C$) and allowing the use of the "geometry of numbers" (Minkowski's theorem) to prove important arithmetic results such as the finiteness of the class number and the structure of the units (Dirichlet's theorem)

Answer 2

One cannot over-emphasize the point that one of the primary benefits of working over an algebraically closed field such as $\mathbb C$ is the immense simplification gained by linearizing what would otherwise be much more complicated nonlinear phenomena. The ability to factor polynomials completely into linear factors over $\mathbb C$ enables widespread linearization simplifications of diverse problems. An example familiar to any calculus student is the fact that integration of rational functions is much simpler over $\mathbb C$ (vs. $\mathbb R$) since partial fraction decompositions involve at most linear (vs. quadratic) polynomials in the denominator. Analogously, one may reduce higher-order constant coefficient differential and difference equations (i.e. recurrences) to linear (first-order) equations by factoring them as linear operators over $\mathbb C$ (i.e. "operator algebra").

More generally, such simplification by linearization was at the heart of the development of abstract algebra. Namely, Dedekind, by abstracting out the essential linear structures (ideals and modules) in number theory, greatly simplified the prior nonlinear Gaussian theory (based on quadratic forms). This enabled him to exploit to the hilt the power of linear algebra. Examples abound of the revolutionary breakthroughs that this brought to number theory and algebra - e.g. it provided the methods needed to generalize the law of quadratic reciprocity to higher-order reciprocity laws - a longstanding problem that motivated much of the early work in number theory and algebra.

Answer 3

It's easy to be jaded about FTA after mathematical history has run for another two centuries. Sure, it is not the number one most important result any more, or the center of any research program (though understanding the algebraic closure of Q could be considered as half of number theory). But consider the situation around 1800. In addition to solution of algebraic equations one had new methods of constructing numbers, using power series, integrals and other limits. Algebra and number theory dealt with the first situation, to a limited extent, and analysis showed that the second type of construction could be iterated but still stay within the same realm of numbers. There was still the possibility that solving equations with $\pi$ and $e$ as coefficients could require an entirely new type of super-transcendental analysis. Fundamental Theorem of Algebra is self-defeating in this sense: it shows that nothing more was needed than complex numbers. But this is not clear in a world where you don't know that FTA is true.

To get an idea what algebraic geometry looked like without complex numbers, look up Newton's classification of degree 3 algebraic curves in the plane, $P_3(x,y)=0$, using real coordinates. The reason this work is obscure today is that there are many dozens of cases compared to the complex projective version. As in Lie groups and topology, looking from the universal covering (C) downward, modulo some Galois-theoretic details (R), is usually easier than working from the bottom up.

Suppose you want to evaluate the integral, from $- \infty$ to $+\infty$, of a rational function (one with integer coefficients would illustrate the point). The answer will involve $\pi$ and the usual method for finding it will use the specific location of the roots in the complex plane, so is more specific to complex numbers than the existence of roots in an algebraic closure. There are some non-usual methods that stay entirely within the real numbers, but they are nonstandard because they are more complicated, and harder to understand and adapt to other problems.

Algebraic geometry in general has a transcendental part -- periods, Hodge theory, uniformization, etc -- that, in the present state of knowledge, cannot be fully substituted by algebraic methods over fields of characteristic 0 (or p). Sometimes Lefschetz principle or reduction to positive characteristic can be used, sometimes not, or the theory is unknown.

Answer 4

Well, there is something called "Algebraic Geometry". You may check the first pages of most of the books on that subject and you'll see something like: "Let $k$ be an algebraically closed field..." [for instance, $\mathbb{C}$]. Indeed, this is the beginning of Hartshorne's book.

The reason? It's much harder to do geometry over the real numbers, where all your theorems about the set of solutions of some system of polynomial equations (the primary object of Algebraic Geometry) can miserably fail, just because there are equations like $x^2 + 1 = 0$ with no solution at all.

For instance, the existence of polynomial equations like $x^2 + 1 = 0$ with no solution over the reals, ruins the basic bijection between prime ideals in $\mathbb{R}[x_1, \dots , x_n]$ and irreducible algebraic sets in the affine space $\mathbb{R}^n$, which is central for Algebraic Geometry.

This doesn't mean you cannot do Algebraic Geometry over the reals, only it's another business.

Answer 5

Perhaps it is worth expanding my answer. (Forgive me if I get any details wrong.) If you are interested in Lie groups, you are interested in their classification. A connected Lie group is the quotient of its universal cover by a discrete subgroup of its center, so to classify connected Lie groups it suffices to classify simply connected Lie groups, then compute their centers. This classification is essentially equivalent to the classification of finite-dimensional real Lie algebras. The Levi decomposition says that to classify such Lie algebras it is more or less enough to classify solvable and semisimple Lie algebras. The former classification problem, as I understand it, is wild; however the latter is tractable. The basic idea is first to study complex semisimple Lie algebras and then to study their real forms.

A big reason this is possible is that the complex numbers are algebraically closed. This means that eigenvalues always exist, and from there the entire classification is possible.