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I've been learning about Dedekind zeta functions and some basic L-functions in my introductory algebraic number theory class, and I've been wondering why some functions are called L-functions and others are called zeta functions. I know that the zeta function is a product of L-functions, so it seems like an L-function is somehow a component of a zeta function (at least in the case of Artin L-functions, they correspond to specific representations). Is this the idea behind the distinction between "zeta function" and "L-function"? How do things generalize to other kinds of zeta- and L-functions?

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  • $\begingroup$ Currently, I feel like there is no consensus among the answers, with Hansen's and Buzzard's answers reflecting one view, and engelbrekt and Will Sawin's reflecting another. $\endgroup$ Commented Nov 14, 2012 at 18:08
  • $\begingroup$ It is likely that "all are true", I think... as in my add-on answer below. $\endgroup$ Commented Nov 14, 2012 at 19:00
  • $\begingroup$ There is a reason that Prof. Nick Katz's course this semester is on equidistribution of L-functions over finite fields at not equidistribution of zeta functions over finite fields. $\endgroup$
    – Will Sawin
    Commented Nov 14, 2012 at 19:44
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    $\begingroup$ Dear Davidac, As Paul Garrett says, "all are true". On the one hand, $\zeta$-functions are traditionally associated to the trivial Galois representation, or an entire scheme of finite type over $\mathbb Z$ (or its cohomology with trivial coefficients, if one wants to think sheaf-theoretically), while $L$-functions are what happens when you allow ``twists'' in what would otherwise be $\zeta$-functions (think of Dirichlet $L$-functions compared to the Riemann $\zeta$-function), such as non-trivial Galois reps., or non-trivial sheaves, etc. But there is no hard and fast rule. ... $\endgroup$
    – Emerton
    Commented Nov 15, 2012 at 3:04
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    $\begingroup$ ... and you shouldn't be very concerned about possible differences in usage. Regards, Matthew $\endgroup$
    – Emerton
    Commented Nov 15, 2012 at 3:10

5 Answers 5

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Let me say first that a Dedekind zeta function is always a product of Artin L-functions. It is the structure of the Galois closure which is relevant here. Let me give a nice example which is indicative of the general case. Let $p(x) \in \mathbb{Z}[x]$ be an irreducible cubic, and let $\alpha$ be a root of $p$. Then $K=\mathbb{Q}(\alpha)$ has trivial automorphism group, and its Galois closure (say $L/\mathbb{Q}$) is an S3-extension. The group S3 has three irreducible representations: the trivial representation, the "sign representation" $\chi$ which is also one-dimensional, and an irreducible two-dimensional representation which we will call $\rho$. Then we have the relations $\zeta_K(s)=\zeta_{\mathbb{Q}}(s)L(s,\rho)$ and $\zeta_L(s)=\zeta_{\mathbb{Q}}(s)L(s,\chi)L(s,\rho)^2$. The proofs of these facts are part of the formalism of Artin L-functions.

Generally, the distinction is really a matter of history. Certain objects were named zeta functions - Hasse-Weil, Dedekind - while Dirichlet chose the letter "L" for the functions he made out of characters. However, one feature is that "zeta" functions tend to have poles, and they often "factor" into L-functions. These vagaries are made more precise in various places, for example Iwaniec-Kowalski Ch. 5 and some survey articles on the "Selberg class" of Dirichlet series.

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    $\begingroup$ It sounds like you are saying that every irreducible cubic has a splitting field which has Galois group $S_3$, which is false. Similarly, it is false that Q(\alpha) has trivial automorphism group. Or maybe I'm misreading? $\endgroup$ Commented Jun 9, 2010 at 16:52
  • $\begingroup$ @Cam, you are right, I should have said "irreducible cubic with $S3$ splitting field". $\endgroup$ Commented Jun 9, 2010 at 17:03
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Although no-one else seems to have suggested this, my personal take on this is that there's no difference whatsoever. Sure, if people start talking about "Dedekind zeta functions" or "Artin $L$-functions" then there starts to be relations amongst these specific choices. But for me, all of these are special cases of automorphic $L$-functions, which, in a parallel universe, could just have easily have been called automorphic zeta-functions. History tells us that varieties have zeta functions, Dirichlet characters have $L$-functions, number fields have zeta functions, elliptic curves have $L$-functions and so on. But they're all just instances of the same thing really (at least conjecturally)---they're all just (simple combinations of) automorphic $L$-functions, which, as I say, could easily have been called automorphic zeta-functions.

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  • $\begingroup$ I feel like this is a good strategy when speaking of an ideal world in which Langlands Correspondence is known or assumed. In more formal contexts, where we only assume the current state of knowledge, it might be useful to make a distinction. $\endgroup$
    – S. Carnahan
    Commented Nov 28, 2009 at 0:41
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    $\begingroup$ The point of my post is that I cannot see any distinction between the two names and it seems to me that whenever we have chosen one name over the other it is just a random historical coincidence. I don't understand your comment. Are you saying that, before it was known that all elliptic curves were modular, there was a clear argument for calling the L-function of an elliptic curve an L-function rather than a zeta function?? Bewildered, $\endgroup$ Commented Nov 28, 2009 at 7:52
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    $\begingroup$ I think that historically zeta functions were generating functions of sequences of counting numbers, and that to a large extent the phrase still retains that meaning outside number theory. $\endgroup$
    – engelbrekt
    Commented Nov 29, 2009 at 23:39
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    $\begingroup$ Kevin, can you think of a counterexample to Will Sawin's stipulation? $\endgroup$ Commented Nov 14, 2012 at 18:09
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    $\begingroup$ Sorry, I'm not sure what I was thinking when I wrote that comment. $\endgroup$
    – S. Carnahan
    Commented Nov 15, 2012 at 4:09
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L-functions depend on characters (or representations), zeta functions do not (or correspond to a trivial character). For example, $L(s,\chi) = \sum_{n = 1}^{\infty}\chi(n)n^{-s}$ where $\chi$ is a Dirichlet character. Supposing that $\chi_0$ is the trivial character modulo $q$, we get $L(s,\chi_0) = \zeta(s)\prod_{p|q}(1 - p^{-s})$ where $\zeta(s)$ is the Riemann zeta function.

So the Dirichlet L-functions generalize the Riemann zeta function. The Dedekind zeta function also generalizes in the same way, to L-functions with Hecke Grössencharacters.

(Removed a false statement at the end; as David Hansen points out, one can get a factorization into Artin L-functions (belonging to the Galois closure of the extension) even when the extension is not Galois by factorizing the Dedekind zeta function of the Galois closure and taking away some factors)

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  • $\begingroup$ Is the latter true if the number field is Galois? I at least know it's true for cyclic extensions. $\endgroup$ Commented Nov 26, 2009 at 18:09
  • $\begingroup$ Yes, it is true if the extension is Galois. $\endgroup$
    – engelbrekt
    Commented Nov 26, 2009 at 23:01
  • $\begingroup$ I downvoted this due to the falsity of the second-to-last sentence. If edited I will upvote again, and edit my post below accordingly. $\endgroup$ Commented Nov 27, 2009 at 4:52
  • $\begingroup$ In light of David Hansen's remarks, a more precise version of my answer above is that if the extension is Galois, the Dedekind zeta function factorizes into Artin L-functions belonging to that extension. $\endgroup$
    – engelbrekt
    Commented Nov 27, 2009 at 6:07
  • $\begingroup$ This one here seems to be clearly the right answer among all. That by Will Sawin of course makes essentially the same point. $\endgroup$ Commented Aug 28, 2014 at 12:39
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I do like the other answers, too, but it seemed silly to append comments to all... :

To my perception, first, I tend to not feel a difference between "zeta function attached to ..." and "L-function attached to..." if only because usage is variable.

Second, in many settings (analytic/automorphic or geometric/motivic or...) a zeta function is an L-function with relatively trivial "further data", whatever that means in context. So, Dedekind zeta functions are Hecke L-functions with trivial data, for example. Analogously for schemes without or with non-trivial sheaf, in the other world. This general rule is certainly not strict... depending on usage.

Third, there are the systematic, partly proven, partly conjectural, miracles that "larger" zetas factor into "smaller" L-functions. Classfield theory and such. This does highlight the ambiguity in "usage", namely, that some "base" is necessary to understand "triviality", etc.

Edit: again significantly contingent on "usage"... If we say that an "L-function" (or "zeta function") "has an analytic continuation (provable by us)", then this would accidentally disallow Hasse-Weil zeta/L-functions of general varieties/schemes/whatever, because in this year we know "few" cases wherein we can prove this, although conjecturally it is mostly-always so (meaning that poles, if any, are finite and describable). Similarly, factorization into Artin L-functions is in one way completely fine (for decades), but, in another, unsatisfactory since we do not know their holomorphy, ... so might decide that they're not yet (in 2012) fully-legitimate "L-functions"? And/or that the "factorization" of Dedekind zetas into such things is not entirely satisfactory (as in a comment). I would not be surprised that such "technicalities" persist in things I know less about. :)

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  • $\begingroup$ What conjectures are you referring to? My understanding was that all observed factorization can be explained by decomposition of Galois representations into irreducible Galois representations. Are you referring to conjectural factors smaller than that? Or something else? $\endgroup$
    – Will Sawin
    Commented Nov 14, 2012 at 19:37
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    $\begingroup$ Dear Will, For example, it is not known in general that Artin L-functions are holomorphic (rather than just merely meromorphic) in the $s$-plane. If they have poles, then the "factorization" into Artin $L$-functions is not such a good factorization. Regards, Matthew $\endgroup$
    – Emerton
    Commented Nov 14, 2012 at 21:04
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Zeta functions arise from schemes. $L$-functions arise from schemes + a sheaf on that scheme.

Obviously, the zeta function of a number field $K$ is the zeta function of $\operatorname{Spec} K$.

The $L$-function of an elliptic curve is the $L$-function of its Tate module. Its zeta function is $\frac{\zeta(s)\zeta(s-1)}{L(s,E)}$, where $\zeta(s)$ is the Riemann zeta function.

The factorization of a zeta function into $L$-functions is the factorization of the etale cohomology into irreducible Galois representations. The poles arise from the factors that are Tate twists of trivial representations, in particular from $H^0$ and $H^{2d}$.

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  • $\begingroup$ I don't understand the 2nd sentence of the 3rd paragraph. $\endgroup$
    – temp
    Commented Nov 14, 2012 at 19:07
  • $\begingroup$ Dear Will, In the elliptic curve case, you seem to have written down the zeta-function of $E$ (as a scheme over $\mathbb Z$). Regards, Matthew $\endgroup$
    – Emerton
    Commented Nov 14, 2012 at 19:26
  • $\begingroup$ @Emerton: I forgot a few words. Thanks! @temp: I am boldly trying to subsume every other answer into mine. The "poles" comment is from David Hansen's answer. Although I guess you actually have more sources of poles. You can see that the zeta function of an elliptic curve has a pole arising from each $\zeta$ term, which themselves arise from the etale cohomology groups $H^0$ and $H^2$, and a bunch of poles from the inverted $L$ function term. $\endgroup$
    – Will Sawin
    Commented Nov 14, 2012 at 19:31
  • $\begingroup$ Oh, thanks for the edit. From what I understand, L-functions are from representations (of the Galois group or so) and zeta functions are ... err, I don't know where they come from, but when they appear they have their own flavor. $\endgroup$
    – temp
    Commented Nov 14, 2012 at 19:40
  • $\begingroup$ Zeta functions come from all the Galois representations that come from the etale cohomology of a single scheme with coefficients in the constant sheaf, put together. $\endgroup$
    – Will Sawin
    Commented Nov 15, 2012 at 2:03

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