For a Hecke L-function, if all of the local eigenvalues are roots of unity, is it an Artin L-function?
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4$\begingroup$ Wikipedia tells you: "In mathematics, a Hecke L-function may refer to: (i) an L-function of a modular form, (ii) an L-function of a Hecke character." Which one are you talking about? $\endgroup$– JoëlCommented Feb 2, 2017 at 15:31
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Whatever your answer to my question in comment, the answer to the title question is yes. Take the case of the $L$-function attached to a modular form $f$, which we assume an eigenform for almost all Hecke operators (since otherwise there is no decomposition of the L-function into Euler product and the phrase "local eigenvalues" has no meaning.)
By Deligne, we know then that the local eigenvalues at $p$ for almost every $p$ are Weil numbers, more precisely they are algebraic numbers whose every embedding in $\mathbb C$ has absolute values $p^{(k-1)/2}$ where $k$ is the weight of your modular form (cf. the last corollary of this link for instance). If just one of those eigenvalue is a root of unity, this already forces $k=1$, in which case we know by Deligne-Serre that the Galois representation attached to $f$ has finite image, that is that $L(f,s)$ is an Artin $L$-function.
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$\begingroup$ An Hecke L-function might also be associated to a Hecke-Maass form of weight $0$, in which case I don't think we know how to answer the question. $\endgroup$ Commented Feb 2, 2017 at 17:52
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1$\begingroup$ @Denis, yes. The notion of Hecke $L$-function is hopelessly ambiguous. $\endgroup$– JoëlCommented Feb 2, 2017 at 18:05
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2$\begingroup$ Indeed! (Not to mention the perverse possibility of the L-function of a holomorphic form, normalized so that the critical line is at 1/2...) $\endgroup$ Commented Feb 2, 2017 at 18:10