Given a newform $f(z)$ (of weight $k$) and a prime $p,$ consider the classical Galois representation $$\rho_{f,p}: \text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \to \text{GL}_2(\mathbb{Z}/p\mathbb{Z}).$$ In the case of elliptic curves, due to Weil pairing, we know that the primitive $p^{\text{th}}$ root of unity $\zeta_p$ is in the field corresponding to the kernel of the representation above. In the general case of modular forms (let's say $k>2$), is it known whether some power of $\zeta_p$ is in the field determined by the kernel? For power, I guess $(p-1,k-1)$ could be a candidate.
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$\begingroup$ $\det\rho_{f,p}$ cuts out the maximal abelian extension inside the field fixed by the kernel of $\rho_{f,p}$. You are asking if this contains $\zeta_p$, which is asking if the determinant is a power of the cyclotomic character. If you twist the form by a Dirichlet character of order >2, the determinant changes that gives you examples when it is not in already for $k=2$. $\endgroup$– Chris WuthrichCommented Dec 17, 2022 at 11:21
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$\begingroup$ Could you please elaborate on what you mean by the determinant cuts out the maximal abelian extension inside the field fixed by the kernel? $\endgroup$– dragoboyCommented Dec 17, 2022 at 12:37
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1$\begingroup$ There is another complication if $f$ has trivial nebentype character, but $k-1$ is divisible by $p-1$; in this case, the the mod $p$ representation actually lands in $\operatorname{SL}_2$, so you will never see roots of unity appearing. $\endgroup$– David LoefflerCommented Dec 17, 2022 at 17:00
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$\begingroup$ Oops, my counterexample does not exist, you are right (if $p \ne 2$ and $p-1 \mid k-1$ then $k$ is odd so the nebentype cannot be trivial). But you can certainly cook up examples this way with $det \rho_{f, p}$ being a quadratic character. $\endgroup$– David LoefflerCommented Mar 16, 2023 at 14:42
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