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Let $f$ be a newform of weight $k$ and level $N$ with integer coefficients. Deligne-Serre theorem theorem says there exist a nice associated representation $\rho_{f}^{(\ell)}:\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\to \text{GL}_2(\mathbb{Q}_{\ell}),$ for any prime $\ell$ not dividing $N$. In particular, this induces a representation $\rho_{f,\ell}:\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\to \text{GL}_2(\mathbb{F}_{\ell}).$ Due to Serre's open image theorem, both adelic and mod $\ell$ representations are surjective for all but finitely many $\ell$, when $f$ has weight $2$ and comes from an elliptic curve. For the general case (especially when the weight is more than $2$ ?) is the mod $\ell$ representation coincides with the matrices having determinant a $(k-1)^{th}$ power in $\mathbb{F}_{\ell}^*$, for all but finitely many $\ell$ ? If not, what's stopping it to be ? and what we can say about the images ?

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In the setting that $f$ has integer Hecke eigenvalues, then the result is true, so long as $f$ has weight $k\ge 2$ and does not have complex multiplication. See Theorem 3.1 of this paper by Ribet: in this setting, $R=\mathbb Z$.

In general, the Hecke eigenvalues of $f$ will all lie in a finite extension $E$ of $\mathbb Q$, and $\rho_f$ will be valued in $\mathrm{GL}_2(E_{\lambda})$ for a prime $\lambda$ above $\ell$. Let $\mathbb F_\lambda$ be the residue field of $E_\lambda$. If $k\ge 2$ and $f$ does not have CM, then, for all but finitely many primes, the image of the mod $\lambda$ representation $\overline{\rho}_f$ in $\mathrm{GL}_2(\mathbb F_\lambda)$ will contain a subgroup conjugate to $\mathrm{SL}_2(\mathbb F_\ell)$ (but not necessarily $\mathrm{SL}_2(\mathbb F_\lambda)$).

To say anything further, we have to be a lot more careful: I'd suggest looking at the answers to this question.

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  • $\begingroup$ Yes, I am happy when eigenvalues are integer.s In the Ribet's paper, are you talking about (2) at page 186 ? but that's only said for level 1, no ? $\endgroup$ – dragoboy May 20 '20 at 12:42
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    $\begingroup$ @dragoboy I'm referring to Theorem 3.1 on p191. You're right that the result on p186 only applies to level $1$ forms. $\endgroup$ – Ariel Weiss May 20 '20 at 12:44
  • $\begingroup$ Great! thanks... $\endgroup$ – dragoboy May 20 '20 at 13:04

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