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In an introductory note of local Langlands correspondence http://wwwf.imperial.ac.uk/~buzzard/maths/research/notes/old_introductory_notes_on_local_langlands.pdf, section $11$ describes a recipe to elliptic curves:

If $E$ is a modular elliptic curve over $\mathbb Q$ coming from a modular form $f$, then one can ask how the possible local behaviours of $E$ at a prime $p$ match up with the admissible representation $V_{f,p}$ associated to $f$. Here’s how it works. Choose a prime $p$.

1) $V_{f,p}$ is unramified principal series iff $E$ has good reduction at $p$. This is because these two cases are the only cases where the conductor is $0$, so they must match up.

2) $V_{f,p}$ is special iff E has potentially multiplicative reduction at $p$. This is because these are the only cases where the image of inertia in the associated Galois representation is infinite, so they must match up.

3) $V_{f,p}$ is special associated to an unramified character iff E has multiplicative reduction. This is because these are the only two subcases of case $2)$ where the conductor is $1$.

4) $V_{f,p}$ is ramified principal series or supercuspidal iff E has bad, but potentially good, reduction. This is because these are the only cases left.

This is a easy case-by-case verification e.g we know the associated $GL_2(\mathbb Q_p)$ representation has same conductor as $E / \mathbb Q_p$, and $p||N$ iff $E$ has multiplicative reduction at $p$. More interestingly,

5) $V$ is ramified principal series iff E attains good reduction over an abelian extension of $\mathbb Q_p$.

6) $V$ is supercuspidal iff E attains good reduction over a non-abelian extension of $\mathbb Q_p$.

Why are $5)$ and $6)$ true? Do above discussions hold if $p=2,3$?

In general, given a elliptic curve over a $p$-adic field $F$, assume we know the associated irreducible $GL_2(F)$-representation of $E$, can we tell the reduction type of $E$ over all finite extensions over $F$ in an explicit way as above (and even more precisely)? I am also interested in the case $p=2,3$.

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  • $\begingroup$ If it's ramified principal series then restricting to inertia it is sum of characters and character has abelian kernel. So basically the question becomes whether there is a two diml irrep r of $A\rtimes\mathbb{Z}/k$ for an abelian group A. Then because frobenius acts on $r|A$, which has to be sum of two characters, by raising to the pth power, $r|A$ is isom to its $p^k$th power, so it means the order of $A$ divides $(p^{2k} -1)$ but this means it's tamely ramified which never happens for supercuspidal unless $p=2$. $\endgroup$
    – GTA
    Commented Apr 10, 2020 at 8:27
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    $\begingroup$ Perhaps the easiest way to understand this is not in terms of the GL2 rep'n $\pi_{E, p}$ but the associated Weil--Deligne rep $WD_p(E)$. It is kind of hard to see what the base-change maps for extensions of $\mathbf{Q}_p$ look like on GL2 reps, but they are totally transparent on the Weil--Deligne reps. $\endgroup$ Commented Apr 10, 2020 at 8:41
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    $\begingroup$ For your last question, the most automorphic approach would probably be to use base change for automorphic representations to see what the associated representation is over every finite extension, and then apply this list. $\endgroup$
    – Will Sawin
    Commented Sep 9, 2020 at 0:55
  • $\begingroup$ @WillSawin Thank you and David, this solves the question. One of my initial motivation is to under base change by geometry e.g how the conductor changes, as I am more comfortable with geometry... $\endgroup$
    – Zhiyu
    Commented Sep 9, 2020 at 2:18
  • $\begingroup$ Yeah, to understand how the conductor changes you certainly want to look at Wel-Deligne representations. For a curve with potentially good reduction, the action of inertia has to lie inside the automorphism group of the curve mod $p$, meaning it is cyclic of order at most $6$ (acting by the sum of a character and its dual) for $p=2,3$ and contained in the group of automorphisms of a supersingular elliptic curve in that characteristic (en.wikipedia.org/wiki/Supersingular_elliptic_curve#Examples). $\endgroup$
    – Will Sawin
    Commented Sep 9, 2020 at 12:15

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