[proofplan] We compare the Euler products prime by prime. At every good prime $p\nmid N$, the equality of Frobenius traces makes the quadratic Euler factors identical, and at every bad prime $p\mid N$, equality is assumed. Hasse's bound for $E$ and Deligne's bound for the weight-$2$ newform give absolute convergence of the two Euler products on the half-plane $\operatorname{Re}(s)>3/2$. Thus the Euler-product definition of $L(E,s)$ agrees there with $L(f,s)$, and $L(E,s)$ inherits its meromorphic continuation from the already known meromorphic continuation of the newform $L$-function. [/proofplan]

[step:Identify the good-prime Euler factors from the trace equality]For a prime $p\nmid N$, the elliptic curve $E$ has good reduction at $p$. Its local Euler polynomial is \begin{align*} P_p(E,T) := 1 - a_p(E)T + pT^2 \in \mathbb{Q}[T]. \end{align*} The normalized newform $f$ has good-prime local Euler polynomial \begin{align*} P_p(f,T) := 1 - a_p(f)T + pT^2 \in \mathbb{Q}[T]. \end{align*} By hypothesis, $a_p(E)=a_p(f)$ for every prime $p\nmid N$. Hence \begin{align*} P_p(E,T) = P_p(f,T) \end{align*} for every prime $p\nmid N$.[/step]

[guided]Fix a prime $p\nmid N$. Since $p$ does not divide the conductor $N$, the curve $E$ has good reduction at $p$, and the good Euler factor of its Hasse-Weil $L$-function is determined by the Frobenius trace: \begin{align*} P_p(E,T) := 1 - a_p(E)T + pT^2. \end{align*} For a normalized weight-$2$ newform of level $N$, the corresponding unramified local Euler polynomial at such a prime is \begin{align*} P_p(f,T) := 1 - a_p(f)T + pT^2. \end{align*} The two polynomials have the same constant term, the same quadratic coefficient, and, by hypothesis, the same linear coefficient. Therefore \begin{align*} P_p(E,T)=P_p(f,T) \end{align*} for every prime $p\nmid N$. This is exactly the point of comparing Frobenius traces: at good primes, the trace is the only coefficient of the local quadratic factor not already fixed by $p$.[/guided]

[step:Use the bad-prime hypothesis to obtain equality of every local factor] For each prime $p\mid N$, define $P_p(E,T)\in\mathbb{Q}[T]$ and $P_p(f,T)\in\mathbb{Q}[T]$ to be the denominator polynomials of the local Euler factors at $p$ in the variable $T=p^{-s}$; that is, the local factors are $P_p(E,p^{-s})^{-1}$ and $P_p(f,p^{-s})^{-1}$. These polynomials may have degree smaller than $2$ at bad primes. The theorem assumes equality of the bad-prime local Euler factors, hence \begin{align*} P_p(E,T)=P_p(f,T) \end{align*} for every prime $p\mid N$. Together with the equality at primes $p\nmid N$, this gives \begin{align*} P_p(E,T)=P_p(f,T) \end{align*} for every prime $p$. [/step]

[step:Compare the Euler products on $\operatorname{Re}(s)>3/2$]Define the open half-plane \begin{align*} \Omega := \{s\in\mathbb{C}: \operatorname{Re}(s)>3/2\}. \end{align*} For every good prime $p\nmid N$, write \begin{align*} P_p(E,T)=(1-\alpha_pT)(1-\beta_pT), \end{align*} where $\alpha_p,\beta_p\in\mathbb{C}$ are the reciprocal roots of the good Euler polynomial. Hasse's bound gives $|\alpha_p|=|\beta_p|=p^{1/2}$. For $f$, Deligne's bound for weight-$2$ newforms gives the same estimate for the reciprocal roots of $P_p(f,T)$. Since there are only finitely many primes $p\mid N$, the bad Euler factors do not affect convergence of the infinite product. Thus, for every $s\in\Omega$, both Euler products converge absolutely: \begin{align*} L(E,s) &= \prod_p P_p(E,p^{-s})^{-1},\\ L(f,s) &= \prod_p P_p(f,p^{-s})^{-1}. \end{align*} Indeed the good-prime logarithmic majorants are bounded by a constant multiple of $p^{1/2-\operatorname{Re}(s)}$, and $\sum_p p^{1/2-\operatorname{Re}(s)}$ converges when $\operatorname{Re}(s)>3/2$. Since $P_p(E,T)=P_p(f,T)$ for every prime $p$, substituting $T=p^{-s}$ gives equal finite partial products over every finite set of primes. Passing to the absolutely convergent infinite products gives \begin{align*} L(E,s)=L(f,s) \end{align*} for every $s\in\Omega$.[/step]

[guided]We first make the convergence region explicit. Define \begin{align*} \Omega := \{s\in\mathbb{C}: \operatorname{Re}(s)>3/2\}. \end{align*} At a good prime $p\nmid N$, factor the elliptic-curve Euler polynomial as \begin{align*} P_p(E,T)=(1-\alpha_pT)(1-\beta_pT), \end{align*} where $\alpha_p,\beta_p\in\mathbb{C}$ are the reciprocal roots. Hasse's bound gives $|\alpha_p|=|\beta_p|=p^{1/2}$. The analogous factorization of the newform Euler polynomial has reciprocal roots of absolute value $p^{1/2}$ by Deligne's bound for weight-$2$ newforms. Hence, if $\sigma:=\operatorname{Re}(s)>3/2$, each good-prime factor differs from $1$ by terms controlled by a constant multiple of $p^{1/2-\sigma}$, and the series \begin{align*} \sum_p p^{1/2-\sigma} \end{align*} converges. The finitely many bad primes $p\mid N$ cannot affect convergence of the infinite product. Therefore both Euler products converge absolutely on $\Omega$: \begin{align*} L(E,s) &= \prod_p P_p(E,p^{-s})^{-1},\\ L(f,s) &= \prod_p P_p(f,p^{-s})^{-1}. \end{align*} Now use the local comparison. The equality $P_p(E,T)=P_p(f,T)$ is an identity in the auxiliary variable $T$. Substituting \begin{align*} T=p^{-s} \end{align*} for each prime $p$ gives equality of the corresponding local factors. Therefore the finite partial products over any finite set of primes agree. Because the infinite products converge absolutely on $\Omega$, taking the limit of these equal finite partial products is justified, and we obtain \begin{align*} L(E,s)=L(f,s) \end{align*} for every $s\in\Omega$.[/guided]

[step:Transfer the newform continuation to the elliptic curve $L$-function] The newform $L$-function $L(f,s)$ has a meromorphic continuation to $\mathbb{C}$. On the nonempty open half-plane $\Omega=\{s\in\mathbb{C}:\operatorname{Re}(s)>3/2\}$, the Euler-product-defined function $L(E,s)$ agrees with $L(f,s)$. Therefore the [meromorphic function](/page/Meromorphic%20Function) $L(f,s)$ is a meromorphic continuation of the Euler-product-defined $L(E,s)$ from $\Omega$ to $\mathbb{C}$. With this continuation, we have \begin{align*} L(E,s)=L(f,s) \end{align*} as meromorphic functions on $\mathbb{C}$. [/step]

[step:Conclude modularity with associated newform $f$] By definition, an elliptic curve $E/\mathbb{Q}$ is modular with associated newform $f$ when its Hasse-Weil $L$-function agrees with the newform $L$-function attached to $f$, including the local factors at all primes. We have proved precisely that \begin{align*} L(E,s)=L(f,s) \end{align*} as meromorphic functions, with equality of every local Euler factor. Therefore $E$ is modular with associated newform $f$. [/step]