Let $(a,b,c)$ be a hypothetical nontrivial integer solution to the Fermat equation $x^p + y^p + z^p = 0$, where $p \geq 5$ is prime, and assume $a, b, c$ are (pairwise) coprime. From this solution, we construct the Frey curve
$$y^2 = x(x - a^p)(x + b^p)$$
which is evidently a Weierstrass model for an elliptic curve $E_F/\mathbb Q$. I would like to prove this curve is semistable, and all proofs I have encountered, including those in, for example, Frey's paper and Serre's paper, seem to leave one question unanswered.
I am able to prove that if $\ell \neq 2$ is a prime dividing $abc$, then $E_F$ has multiplicative reduction at $\ell$. I would like to prove the converse. I can prove that if $\ell \neq 2$ i a prime at which $E_F$ has multiplicative reduction, then $\ell$ divides $abc$. The problem is that I do not know how to show that if $E_F$ has bad reduction at an odd prime $\ell$, then that reduction cannot be additive. The given model for the curve is not global minimal, so we cannot use it to reliably detect additive reduction at $\ell$ (at least, not to my knowledge). Serre says we can use this model to see that any bad reduction is multiplicative, but I do not see why.
I would be able to finish easily if I could prove a certain statement whose truth value I am unsure of: suppose that $E/\mathbb Q$ is an elliptic curve. Let $\ell \neq 2$ be a prime. If $C$ is a Weierstrass model for $E$ with integer coefficients which reduces mod $\ell$ to a cuspidal cubic, then $E$ has additive reduction at $\ell$. Is this true?
I know the statement is false when $\ell = 2$, and I know similar statements are true for good/multiplicative reduction (and I can prove them), but this eludes me. And if this is false, how do we conclude that $E_F$ cannot have additive reduction at an odd prime $\ell$ without consulting a global minimal model?
