[proofplan] The argument is local at each prime $p$. We use the [minimal-discriminant criterion for elliptic curves](/page/Minimal%20Discriminant) to identify primes dividing $\Delta_E$ with primes of bad reduction, and we use the [local Artin conductor formula](/page/Artin%20Conductor) for the $\ell$-adic Tate module to identify the conductor exponent with the inertia action. Good reduction gives inertia acting as the identity and exponent $0$; multiplicative reduction gives a one-dimensional space of inertia invariants and exponent $1$; for additive reduction at $p \ge 5$, the [Tate algorithm](/page/Tate%20Algorithm) gives the corresponding tame conductor exponent $2$. [/proofplan] [step:Localize the conductor calculation at a prime] Fix a prime number $p$. Choose a prime number $\ell \ne p$, and define the $\ell$-adic Tate module representation \begin{align*} \rho_{E,\ell}: G_{\mathbb{Q}_p} &\to \operatorname{GL}(V_\ell(E)), \end{align*} where $G_{\mathbb{Q}_p}:=\operatorname{Gal}(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)$ and $V_\ell(E):=T_\ell(E)\otimes_{\mathbb{Z}_\ell}\mathbb{Q}_\ell$ is the two-dimensional $\mathbb{Q}_\ell$-[vector space](/page/Vector%20Space) attached to $E/\mathbb{Q}_p$. Let $I_p\subset G_{\mathbb{Q}_p}$ denote the inertia subgroup. By definition, the local conductor exponent $f_p(E)$ is the Artin conductor exponent of $\rho_{E,\ell}$; it is independent of the chosen prime $\ell \ne p$. [/step] [step:Identify divisibility of the minimal discriminant with bad reduction] Let $\mathcal{E}_p$ be a minimal Weierstrass model of $E$ over $\mathbb{Z}_p$, let $\Delta_p\in \mathbb{Z}_p$ denote its minimal discriminant, and let $v_p:\mathbb{Q}_p^\times\to \mathbb{Z}$ denote the normalized $p$-adic valuation with $v_p(p)=1$. Since $\Delta_E$ is the global minimal discriminant, $p\mid \Delta_E$ if and only if the valuation $v_p(\Delta_p)$ is positive. The [minimal-discriminant criterion for elliptic curves](/page/Minimal%20Discriminant) says that $v_p(\Delta_p)=0$ if and only if the special fibre of $\mathcal{E}_p$ is smooth over $\mathbb{F}_p$, that is, if and only if $E$ has good reduction at $p$. Hence $p\mid \Delta_E$ if and only if $E$ has bad reduction at $p$. [guided] We first translate the global divisibility statement into a local reduction statement. For the fixed prime $p$, let $\mathcal{E}_p$ be a minimal Weierstrass model of $E$ over $\mathbb{Z}_p$, let $\Delta_p\in \mathbb{Z}_p$ be its minimal discriminant, and let $v_p:\mathbb{Q}_p^\times\to \mathbb{Z}$ denote the normalized $p$-adic valuation with $v_p(p)=1$. The global minimal discriminant satisfies \begin{align*} p\mid \Delta_E \quad \Longleftrightarrow \quad v_p(\Delta_p)>0. \end{align*} The [minimal-discriminant criterion for elliptic curves](/page/Minimal%20Discriminant) states that $v_p(\Delta_p)=0$ exactly when the reduction of the minimal model modulo $p$ is nonsingular. This nonsingularity is precisely good reduction at $p$. Therefore \begin{align*} p\mid \Delta_E \quad \Longleftrightarrow \quad E \text{ has bad reduction at } p. \end{align*} This step uses minimality: without a minimal model, the discriminant of a chosen Weierstrass equation could contain extraneous powers of $p$ introduced by a non-minimal change of variables. [/guided] [/step] [step:Use the local conductor formula to compare bad reduction and conductor divisibility] The [local Artin conductor formula](/page/Artin%20Conductor) for elliptic curves gives \begin{align*} f_p(E)=2-\dim_{\mathbb{Q}_\ell} V_\ell(E)^{I_p}+\delta_p(E), \end{align*} where $V_\ell(E)^{I_p}:=\{v\in V_\ell(E):\rho_{E,\ell}(\sigma)v=v\text{ for all }\sigma\in I_p\}$ and $\delta_p(E)\ge 0$ is the Swan conductor, the wild conductor contribution. By the [Néron-Ogg-Shafarevich criterion](/page/Neron-Ogg-Shafarevich%20Criterion), $E$ has good reduction at $p$ if and only if $I_p$ acts as the identity on $V_\ell(E)$. In the good reduction case, $\dim_{\mathbb{Q}_\ell}V_\ell(E)^{I_p}=2$ and $\delta_p(E)=0$, so $f_p(E)=0$. Conversely, if $f_p(E)=0$, then both non-negative terms in the conductor formula vanish, so $\dim_{\mathbb{Q}_\ell}V_\ell(E)^{I_p}=2$; hence inertia acts as the identity, and the [Néron-Ogg-Shafarevich criterion](/page/Neron-Ogg-Shafarevich%20Criterion) gives good reduction. Therefore $p\mid N_E$ if and only if $f_p(E)>0$, if and only if $E$ has bad reduction at $p$. [/step] [step:Compute the tame conductor exponents for $p\ge 5$] Assume $p\ge 5$. If $p\nmid \Delta_E$, then $E$ has good reduction by the previous step, and therefore $f_p(E)=0$. If $E$ has multiplicative reduction at $p$, the [Tate curve description of multiplicative reduction](/page/Tate%20Curve) gives a unipotent inertia action on $V_\ell(E)$ that is not the identity and has \begin{align*} \dim_{\mathbb{Q}_\ell}V_\ell(E)^{I_p}=1. \end{align*} Multiplicative reduction is semistable, so the Swan conductor satisfies $\delta_p(E)=0$. The conductor formula gives \begin{align*} f_p(E)=2-1+0=1. \end{align*} If $E$ has additive reduction at $p\ge 5$, the [Tate algorithm](/page/Tate%20Algorithm), equivalently the Kodaira-Néron classification in residue characteristic at least $5$, gives the local conductor formula \begin{align*} f_p(E)=2 \end{align*} for every additive Kodaira type. In the potentially good additive cases $II, III, IV, I_0^*, IV^*, III^*, II^*$, the inertia action is tame for $p\ge 5$, has no nonzero invariant vector on $V_\ell(E)$, and has Swan conductor $\delta_p(E)=0$. In the potentially multiplicative additive cases $I_n^*$, the same Tate-algorithm conductor formula gives no nonzero inertia invariant on $V_\ell(E)$ and no wild contribution for $p\ge 5$. Thus in all additive cases at $p\ge 5$, \begin{align*} \dim_{\mathbb{Q}_\ell}V_\ell(E)^{I_p}=0,\qquad \delta_p(E)=0, \end{align*} and the Artin conductor formula yields \begin{align*} f_p(E)=2-0+0=2. \end{align*} [/step] [step:Combine the local statements] From the discriminant criterion, $p\mid \Delta_E$ is equivalent to bad reduction at $p$. From the conductor calculation, bad reduction at $p$ is equivalent to $f_p(E)>0$, which is equivalent to $p\mid N_E$. Hence the primes dividing $N_E$ are precisely the primes dividing $\Delta_E$. The explicit cases for $p\ge 5$ were computed above, and at $p=2$ and $p=3$ the Swan conductor $\delta_p(E)$ can be positive for additive reduction, so the additive conductor exponent may be larger than $2$. [/step]