Lemma 4.2 in M. Ram Murty, Selberg conjectures and Artin L-functions(1994), states that under Ramanujan conjecture, an irreducible cuspidal automorphic representation of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$ gives rise to a primitive L-function.

Is the converse true? That is, assuming Ramanujan conjecture, if a degree $n$ primitive L-function comes from an automorphic representation of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$, is this representation necessarily irreducible and cuspidal?


Yes, and this is true without the Ramanujan conjecture (but see also Peter Humphries' comment below).

If $\pi$ is not irreducible, say $\pi=\pi_1\oplus\pi_2$, then $L(s,\pi)=L(s,\pi_1)L(s,\pi_2)$.

If $\pi$ is not cuspidal, then by Langlands' theory of Eisenstein series, there is a nontrivial partition $n=\sum n_j$ and cuspidal irreducible representations $\pi_j$ of $\mathrm{GL}_{n_j}(\mathbb{A}_\mathbb{Q})$ such that $\pi$ is parabolically induced from $\prod\pi_j$ on $\prod\mathrm{GL}_{n_j}(\mathbb{A}_\mathbb{Q})$ as a Levi subgroup of $\mathrm{GL}_n(\mathbb{A}_\mathbb{Q})$. Then, $L(s,\pi)=\prod L(s,\pi_j)$.

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    $\begingroup$ It's not true without the Ramanujan conjecture if by "primitive $L$-function" you mean "primitive element of the Selberg class whose Dirichlet series coefficients satisfy $a(n) \ll_{\varepsilon} n^{\varepsilon}$", which is what I imagine the OP meant. $\endgroup$ – Peter Humphries Mar 21 at 1:25
  • $\begingroup$ @PeterHumphries: I guess I thought of primitivity within automorphic $L$-functions or some extended Selberg class. But you are absolutely right. $\endgroup$ – GH from MO Mar 21 at 3:32
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    $\begingroup$ Many thanks to the both of you. $\endgroup$ – Sylvain JULIEN Mar 21 at 9:43

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