# Why is it “easier” to work with function fields than with algebraic number fields?

I just bought a copy of Jürgen Neukirch's book Algebraic Number Theory. While browsing through it I found a section titled § 14. Function Fields in chapter I. In it the author describes some aspects of an analogy between function fields and algebraic number fields.

This led me to google for a while and I ended up reading the Wikipedia entry for Global Field. And this is where my question comes from. In the last sentence of that entry there's the following passage, which I find really interesting:

It is usually easier to work in the function field case and then try to develop parallel techniques on the number field side. The development of Arakelov theory and its exploitation by Gerd Faltings in his proof of the Mordell conjecture is a dramatic example.

Unfortunately, being as dramatic as it is, the example mentioned does not tell me anything because not even the Wikipedia entry on Arakelov Theory is somehow close to give even a small hint as to what it is about.

So I would like to ask for some insight and/or examples that illustrate why it is said to be easier to work with function fields than with algebraic number fields and then try to develop parallel techniques for the number field case.

Thank you very much for any help.

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I'd recommend taking a look at Rosen's "Number Theory on Function Fields". It really is interesting to see a lot of theorems proved much more easily in the function field setting. – John M Jul 21 '11 at 5:32
@John M Thanks, that books seems really nice. – Adrián Barquero Jul 21 '11 at 17:41
– Dylan Moreland Jul 22 '11 at 3:51

## 3 Answers

One answer is that we can take formal derivatives. For example, Fermat's last theorem is rather difficult but the function field version is a straightforward consequence of the Mason-Stothers theorem, whose elementary proof crucially relies on the ability to take formal derivatives of polynomials.

There is no obvious way to extend this construction to integers in a way that preserves its good properties. If there were, then the abc conjecture (of which Mason-Stothers is the function field version) would be trivial, which it's not. There is a thing called the arithmetic derivative, but it is of course not linear, and it doesn't seem to me to be very easy to prove anything with it.

The problem is that if we want to think of $\mathbb{Z}$ as being analogous to a function field, then the "field" that it's a function field over is the field with one element, so if a reasonable notion of formal derivative exists here it needs not to be $\mathbb{Z}$-linear, but to be $\mathbb{F}_1$-linear, whatever that means... if we understood what that meant, perhaps we could construct the "correct" version of the arithmetic derivative and presumably prove the abc conjecture.

Arakelov theory addresses another difference between function fields and number fields, which is the existence of Archimedean places. Over a function field all places are non-Archimedean and I understand this makes various things easier, but I don't know much about this so someone else should chime in here.

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Incidentally, there's nothing formal about the derivative of a polynomial.  Given a polynomial map, the derivative is the canonical map from the tangent bundle of the source to the pullback of the tangent bundle of the target. – Scott Carnahan Jul 21 '11 at 5:26

The primary reason that function field arithmetic is simpler than number field arithmetic is due to the existence of nontrivial derivations. With the availability of derivatives many things simplify.

E.g. for polynomials derivatives yield easy algorithms for squarefree testing, squarefree part, etc. Contrast this to the integer case. No feasible (polynomial time) algorithm is currently known for recognizing squarefree integers or for computing the squarefree part of an integer. In fact it may be the case that this problem is no easier than the general problem of integer factorization. This problem is important because one of the main tasks of computational algebraic number theory reduces to it (in deterministic polynomial time). Namely the problem of computing the ring of integers of an algebraic number field depends upon the square-free decomposition of the polynomial discriminant when computing an integral basis.

From derivatives also come Wronskians and associated measures of independence. For example, this is what is at the heart of Mason's trivial high-school level proof of the ABC theorem for polynomials - which is a difficult important open problem for numbers. From Mason's theorem follows immediately a trivial two-line proof of FLT for polynomials. If there existed some sort of analogous "derivative for integers" that yielded the corresponding ABC theorem, then it would yield an analogous trivial proof of FLT for integers (more precisely it would yield asymptotic FLT, i.e. FLT for all sufficiently large exponents).

Such observations have motivated searches for "arithmetic analogues of derivations". For example, see Buium's paper by that name in Jnl. Algebra, 198, 1997, 290-99, and see his book Arithmetic differential equations.

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Let's consider an example to see why function fields are easier:

Let $q$ be a prime, and consider the global function field, $\mathbb F_q(T)$. An ideal $\mathfrak a$ of $\mathbb F_q[T]$ is just the principal ideal $\mathfrak a=(f)=(T^d+a_{d-1}T^{d-1}+\dots+a_0)$. The norm $N\mathfrak a=q^d$, and you can see that there are exactly $q^d$ ideals of norm $q^d$.

Then the zeta function over this field is $$\zeta_{\mathbb F_q[T]}(s)=\sum_{\mathfrak a \neq 0}N\mathfrak a^{-s}=\sum_{d=0}^\infty q^d(q^d)^{-s}=1/(1-q^{1-s})$$

That's a very simple expression for the zeta function. Note that it has no zeros, so it trivially satisfies the Riemann Hypothesis.

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– John M Jul 23 '11 at 6:16
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