[proofplan] The proof is a direct combination of two ingredients. First, the characteristic polynomial identity for a linear endomorphism of a two-dimensional [vector space](/page/Vector%20Space) expresses $\det(I_2-XA)$ in terms of the trace and determinant of $A$. Second, Deligne's construction of $\rho_{f,\lambda}$ identifies the trace and determinant of Frobenius at every prime $p \nmid N\ell$ with $a_p(f)$ and $\varepsilon(p)p^{k-1}$. Substituting these two values gives the displayed polynomial, and then setting $X=p^{-s}$ gives the stated good Euler factor. [/proofplan] [step:Express the determinant polynomial of a two-dimensional endomorphism through trace and determinant] Let \begin{align*} A := \rho_{f,\lambda}(\operatorname{Frob}_p) \in GL_2(E_{f,\lambda}). \end{align*} Since $A$ is a $2 \times 2$ matrix over the field $E_{f,\lambda}$, write its entries as \begin{align*} A= \begin{pmatrix} \alpha & \beta\\ \gamma & \delta \end{pmatrix}, \qquad \alpha,\beta,\gamma,\delta \in E_{f,\lambda}. \end{align*} Then \begin{align*} \det(I_2-XA) &= \det \begin{pmatrix} 1-\alpha X & -\beta X\\ -\gamma X & 1-\delta X \end{pmatrix} \\ &=(1-\alpha X)(1-\delta X)-\beta\gamma X^2\\ &=1-(\alpha+\delta)X+(\alpha\delta-\beta\gamma)X^2\\ &=1-\operatorname{tr}(A)X+\det(A)X^2. \end{align*} [guided] We first isolate the linear algebra from the arithmetic input. Define \begin{align*} A := \rho_{f,\lambda}(\operatorname{Frob}_p) \in GL_2(E_{f,\lambda}). \end{align*} The representation is two-dimensional, so $A$ is a $2 \times 2$ matrix after choosing an $E_{f,\lambda}$-basis. Write \begin{align*} A= \begin{pmatrix} \alpha & \beta\\ \gamma & \delta \end{pmatrix}, \qquad \alpha,\beta,\gamma,\delta \in E_{f,\lambda}. \end{align*} Now compute the determinant polynomial directly: \begin{align*} \det(I_2-XA) &= \det \begin{pmatrix} 1-\alpha X & -\beta X\\ -\gamma X & 1-\delta X \end{pmatrix} \\ &=(1-\alpha X)(1-\delta X)-(-\beta X)(-\gamma X)\\ &=(1-\alpha X)(1-\delta X)-\beta\gamma X^2\\ &=1-(\alpha+\delta)X+(\alpha\delta-\beta\gamma)X^2. \end{align*} By definition of trace and determinant for a $2 \times 2$ matrix, \begin{align*} \operatorname{tr}(A)=\alpha+\delta, \qquad \det(A)=\alpha\delta-\beta\gamma. \end{align*} Therefore \begin{align*} \det(I_2-XA)=1-\operatorname{tr}(A)X+\det(A)X^2. \end{align*} This is the exact polynomial identity that will convert Deligne's Frobenius trace and determinant formulas into the Euler factor. [/guided] [/step] [step:Insert Deligne's trace and determinant formulas at good primes] Because $p \nmid N\ell$, the representation $\rho_{f,\lambda}$ is unramified at $p$, so $\rho_{f,\lambda}(\operatorname{Frob}_p)$ is defined up to conjugacy. The determinant polynomial is conjugacy-invariant. By Deligne's trace and determinant formulas for the Galois representation attached to a normalized cuspidal eigenform (citing a result not yet in the wiki: Deligne's construction of the Galois representation attached to a modular eigenform), \begin{align*} \operatorname{tr}\left(\rho_{f,\lambda}(\operatorname{Frob}_p)\right)&=a_p(f),\\ \det\left(\rho_{f,\lambda}(\operatorname{Frob}_p)\right)&=\varepsilon(p)p^{k-1}. \end{align*} Substituting these two identities into the two-dimensional determinant formula gives \begin{align*} \det\left(I_2-X\rho_{f,\lambda}(\operatorname{Frob}_p)\right) &=1-\operatorname{tr}\left(\rho_{f,\lambda}(\operatorname{Frob}_p)\right)X +\det\left(\rho_{f,\lambda}(\operatorname{Frob}_p)\right)X^2\\ &=1-a_p(f)X+\varepsilon(p)p^{k-1}X^2. \end{align*} [guided] The arithmetic input enters exactly at the good primes. Since $p \nmid N\ell$, the prime $p$ is outside both the level of the modular form and the residue characteristic of the coefficient field place $\lambda$. Deligne's representation is therefore unramified at $p$, so the conjugacy class of \begin{align*} \rho_{f,\lambda}(\operatorname{Frob}_p) \in GL_2(E_{f,\lambda}) \end{align*} is well-defined. The expression $\det(I_2-XA)$ is unchanged when $A$ is replaced by a conjugate matrix, because determinant is invariant under conjugation: \begin{align*} \det(I_2-XBAB^{-1}) &=\det\left(B(I_2-XA)B^{-1}\right)\\ &=\det(B)\det(I_2-XA)\det(B^{-1})\\ &=\det(I_2-XA) \end{align*} for every $B \in GL_2(E_{f,\lambda})$. Thus the polynomial attached to Frobenius is independent of the chosen representative of the Frobenius conjugacy class. Now use Deligne's trace and determinant formulas for the Galois representation attached to a normalized cuspidal eigenform (citing a result not yet in the wiki: Deligne's construction of the Galois representation attached to a modular eigenform). These formulas apply because $f$ is a normalized cuspidal Hecke eigenform and $p \nmid N\ell$. They give \begin{align*} \operatorname{tr}\left(\rho_{f,\lambda}(\operatorname{Frob}_p)\right)&=a_p(f),\\ \det\left(\rho_{f,\lambda}(\operatorname{Frob}_p)\right)&=\varepsilon(p)p^{k-1}. \end{align*} Combining these identities with the two-dimensional determinant formula from the previous step yields \begin{align*} \det\left(I_2-X\rho_{f,\lambda}(\operatorname{Frob}_p)\right) &=1-\operatorname{tr}\left(\rho_{f,\lambda}(\operatorname{Frob}_p)\right)X +\det\left(\rho_{f,\lambda}(\operatorname{Frob}_p)\right)X^2\\ &=1-a_p(f)X+\varepsilon(p)p^{k-1}X^2. \end{align*} This is precisely the asserted Frobenius polynomial. [/guided] [/step] [step:Substitute $X=p^{-s}$ to recover the good Euler factor] By the definition of the good local Euler factor attached to $\rho_{f,\lambda}$ at a prime $p \nmid N\ell$, \begin{align*} L_p(f,s):=\det\left(I_2-p^{-s}\rho_{f,\lambda}(\operatorname{Frob}_p)\right)^{-1}. \end{align*} This is obtained from the polynomial identity just proved by substituting $X=p^{-s}$. Hence \begin{align*} L_p(f,s)=\left(1-a_p(f)p^{-s}+\varepsilon(p)p^{k-1}p^{-2s}\right)^{-1} =\det\left(I_2-p^{-s}\rho_{f,\lambda}(\operatorname{Frob}_p)\right)^{-1}. \end{align*} This proves both the Frobenius polynomial formula and the asserted expression for the good Euler factor. [/step]