I seek to correct my proof of the following well-known fact about the Frey curve, and to ask a few questions. References are acceptable answers except for the last bold request; I would like to actually fill in the rest of the proof. If this proof is wrong, please let me know where and how to fix it, or provide an alternative correct proof.
Proposition 1: Let $p \geq 5$ prime and let $a$, $b$, and $c$ be pairwise coprime integers constituting a nontrivial solution to the Fermat equation with exponent $p$, where, without loss of generality, $a \equiv -1 \ \mod \ 4$ and $b$ is even. Then the Frey curve $E_F$ is a semistable elliptic curve whose minimal discriminant and conductor are given by
$$\Delta_m(E_F) = 2^{-8}(abc)^{2p}, \ N_{E_F} = \prod_{\substack{l | abc \\ l \ \text{prime}}} l$$
Furthermore, the field $K_p$ gotten by adjoining the coordinates of the $p$-torsion of $E$ to $\mathbb Q$ is unramified over $\mathbb Q$ outside the divisors of $2p$.
Proof Attempt: The Frey curve is $y^2 = x(x - a^p)(x + b^p)$, where $(a,b,c)$ is a proposed nontrivial solution to the Fermat equation $x^p + y^p + z^p = 0$ over $\mathbb Z$.
One sees directly from reduction of the original equation of $E_F$ that if $l \neq 2$ is prime, then $E_F$ has bad reduction at $l$ iff $l \mid abc$, and that this reduction is multiplicative when it happens. When $l = 2$, reduction of the minimal model shows that $E_F$ has multiplicative reduction at 2. Hence $E$ is semistable and we have found the conductor as desired. No problems here.
Here is where I am not sure what I am doing. Fix a prime $l$ and view $E_F$ as an elliptic curve over $\overline{\mathbb Q}_l$. By the theory of the Tate curve, $K_p$ is just $\mathbb Q_l(\zeta_p, q^{1/p})$, where $q$ satisfies $E_F(\overline{\mathbb Q}_l) \cong \overline{\mathbb Q}^*_l/ q^{\mathbb Z}$ according to Tate Uniformization, and $\zeta_p$ is a primitive $p$th root of unity (Why is this true? Isn't $K_p$ $\mathbb Q$ adjoin $p$-torsion? Doesn't $\mathbb Q_l$ add in a bunch of extra stuff? And where do the root of unity and the $q^{1/p}$ come from? I believe one needs the Weil pairing to see why the root of unity shows up, right?). But $\mathbb Q_l(\zeta_p)$ is unramified over $\mathbb Q_l$ exactly away from $p$ (I think this is true, but why?), and $\mathbb Q_l(q^{1/p})$ is unramified over $\mathbb Q_l$ iff the $l$-adic valuation of $q$ is an integer multiple of $p$ (I am told this is true, but why?). If $l$ is a prime of multiplicative reduction, this happens iff the $l$-adic valuation of $\Delta_m(E_F)$ is an integer multiple of $p$ since $v_l(\Delta_m(E_F)) = -v_l(j(E_F)) = v_l(q)$, where $v_l$ is the $l$-adic valuation. The first equality uses the fact that $j(E_F) = \frac{2^{8}(c^{2p} - (ab)^p)^3}{(abc)^{2p}}$ and $l \nmid (c^{2p} - (ab)^p)^3$ by the multiplicativity of the reduction of $E_F$ at $l$ (see Silverman 1, Chapter 7, page 196). The second equality comes from the fact that, according to the theory of the Tate curve,
$$j(E_F) = \frac{1}{q} + \ \text{power series in $q$ with integer coefficients}$$
and always holds, regardless of $l$. If $l$ is a prime of good reduction, I have no idea what to do.