Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does anyone know of an example where complex analysis is necessary to prove something in algebra?

I would be particularly interested in results from group theory or Galois theory.

In an ideal answer,

  1. the theorem should be purely algebraic in nature,

  2. the proof should be complex analytic at a crucial step, and

  3. the result should be unprovable (or, at least, unproven) by purely algebraic methods.

To elaborate on these criteria, by (1) I mean that the statement of the theorem should be independent of analysis, i.e. not something like "Let $\alpha=$ (some complex integral). Then $G(\mathbb{Q}(\alpha)/\mathbb{Q})=\ldots$". By (2) I mean that I am looking for something where complex numbers are not merely present, but must be used analytically. So, Maschke's theorem for example would not apply just because it involves vector spaces over $\mathbb{C}$. Lastly, (c) primarily means that I am not looking for alternative proofs of known results, no matter how much simpler they may be (no FTA).

Thank you for reading.

share|improve this question
Surely any statement of complex analysis can ultimately be rewritten, however unpleasantly, as a statement about real functions (the real and imaginary parts of the complex functions the original statement refers to), and all the theorems of complex analysis proven in this context. –  Robert Israel Feb 6 '13 at 2:49
I could say more in an answer if this is what you're looking for, but I don't think it is (also it requires a fair bit more technology than the things you were describing): on a differentiable manifold, one has the de Rham cohomology groups $H^k (X,\mathbb R)$, which extract topological information by, in some sense, organizing differential forms. On a Kahler manifold (a complex Riemannian manifold that is also a symplectic manifold in a compatible way), the complex version of de Rham cohomology admits an orthogonal direct sum decomposition called the Hodge decomposition, which is obtained... –  Tabes Bridges Feb 6 '13 at 3:32
...via analysis of elliptic partial differential equations. Besides the issue of level, this is something of a mishmash of algebra, complex analysis, and real analysis. On the other hand, the basic idea is "group decomposition obtained via analysis," so perhaps this is the sort of thing you want. –  Tabes Bridges Feb 6 '13 at 3:33

4 Answers 4

Take Dirichlet's theorem on arithmetic progressions, or Chebotarev density; I'm sure there are plenty of other analytic results in number theory that will also work.

Sure, theorems proven with L-functions are often density statements, which are analytic in nature, but those can be weakened to existence statements; ie. there exist infinitely many primes of the form $a+nk$ when $(a,n)=1$, and there exist infinitely many primes in an abelian extension with a given Frobenius element.

EDIT: I see on wikipedia that there is an elementary proof known for Dirichlet's theorem. But it came along over 100 years after the $L$-funtion proof.

share|improve this answer

To quote Hartshorne from "Algebraic Geometry":

If $X$ is a nonsingular variety over $\mathbf{C}$, then we can also consider $X$ as a complex manifold. All of the methods of complex analysis and differential geometry can be used to study this complex manifold. ... This is an extremely powerful method, which has produced and is still producing many important results, proved by these so-called "transcendental methods," for which no purely algebraic proofs are known.

He gives a example of using the exact sequence

$$ 0 \to \mathbf{Z} \to \mathbf{C} \xrightarrow{f} \mathbf{C}^* \to 0 $$

where $f(x) = e^{2 \pi i x}$ and studying the cohomology of a complex variety by comparing it to the cohomology of the corresponding complex manifold. Alas, I don't really know enough about the subject to be able to say much about it.

share|improve this answer
This is probably not algebraic enough for the OP's taste, since $\mathbb{C}$ is an inherently analytic object. But it is possible to embed other fields in $\mathbb{C}$ to exploit analytic results originally proven for complex varieties. –  Brett Frankel Feb 6 '13 at 4:48

Sharp bounds on character sums (like Kloosterman sums) often follow from the Riemann hypothesis for suitable $L$-functions of varieties over finite fields, and a piece of the proof of this relies on nonvanishing theorems that ultimately go back to analytic ideas from the proof of the prime number theorem/Dirichlet's theorem.

share|improve this answer

Fermat's Last Theorem. Ultimately, one shows that there are no soutions to a Fermat equation because there are no holomorphic differential forms on $\hat{\mathbb{C}}.$

share|improve this answer
But the cited fact is true for the projective line over any field $k$, so despite the term "holomorphic" it does not in fact require any complex analysis to prove it. –  Pete L. Clark Feb 6 '13 at 5:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.