# Minimal discriminant of an elliptic curve in terms of its Galois representation

From the Galois representation of an elliptic curve $E$ we can read the conductor of $E$, and further some information about the minimal discriminant. So is there any more information about the minimal discriminant we can read from the Galois representation?

For example, it is natural to consider the differences between isogenous (equivalently having the same Galois representation) elliptic curves , like in this paper LOCAL INVARIANTS OF ISOGENOUS ELLIPTIC CURVES, it seems there are indeed some invariance on the Kodaira types this isogeny condition, which implies some invariance on the minimal discriminant.

• I don't think the minimal discriminant is isogeny invariant, so I guess that's a proof that you can't read it off from the Galois representation. The discriminant is a rather coarse invariant really. You can check to see if a prime divides it, by checking that the prime is ramified in the ell-adic representation, but do you want more than this?
– eric
Nov 3, 2015 at 18:38
• Sure, I want more than this. Though it varies in isogenous families, as the paper shows, there could be restrictions. Nov 3, 2015 at 19:24
• As eric points, since the minimal discriminant is not isogeny invariant, you can't see it in the $\mathbb Q_\ell$ Galois representation of the elliptic curve. But maybe you can see it in the $\mathbb Z_\ell$-rep? Nov 5, 2015 at 10:50
• Is there example of isogenous elliptic curves with non isomorphic $\mathbb{Z}_l$ rep? Nov 5, 2015 at 16:47
• Sure. Isomorphic Z_l reps would imply isomorphic mod ell reps, but if a curve has a point of order ell and you quotient out by the subgroup generated by that point there's no reason the quotient would have a point of order ell (by Weil pairing it would have a mu_ell, so a Z/ell quotient but there's no reason things should be split). For an explicit example see the three ell curves of conductor 11 and set ell=5; one Galois rep is split Z/ell +mu_ell, the other two are non-split (they must be because if the others were split then they'd each possess two degree 5 isogenies to other curves).
– eric
Nov 7, 2015 at 16:08