Let $f:\mathbb{N}\to\mathbb{N}$ be the map sending $n$ to the number of isomorphism classes of elliptic curves over $\mathbb{Q}$ with conductor $n$.
Is $f(\mathbb{N})$ finite?
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$\begingroup$ Does this answer your question? Finiteness of elliptic curves of a given conductor $\endgroup$– WojowuCommented Aug 24, 2021 at 20:16
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$\begingroup$ No that's a different one. There it's said that $f$ is well-defined while I'm asking if $f(\mathbb{N})$ is finite. $\endgroup$– Matias2Commented Aug 24, 2021 at 20:18
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1$\begingroup$ Ah I see, I wasn't careful in reading. I retracted my close vote but will keep the comment above for reference $\endgroup$– WojowuCommented Aug 24, 2021 at 20:24
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$\begingroup$ Related question, but with no conclusive answer: mathoverflow.net/q/350291/30186 $\endgroup$– WojowuCommented Aug 24, 2021 at 20:28
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1 Answer
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The answer is no. Indeed, counterexamples are given by Mordell curves $E_d:y^2=x^3+d$, where $d$ is a sixth-power-free integer. These curves are pairwise non-isomorphic over $\mathbb Q$. Furthermore, this Weierstrass equation is minimal, which implies $E_d$ has additive reduction at all primes dividing $d$. Therefore the conductor of $E_d$ has the form $2^a3^b\prod_{p\mid d,p>3}p^2$, where $a,b$ range over finitely many possibilities (at most $24$ in this case if memory serves me right). But if we take $d$ to range over integers of the form $p_1^{e_1}\dots p_k^{e_k}$ where $1\leq e_i\leq 5$, we get $5^k$ curves with the same conductor give or take powers of $2,3$, and hence in total at least $5^k/24$ curves with the same conductor.
Of course, these curves all arise as twists of a single curve, and it could be interesting to ask for nontrivial examples where this doesn't hold. Alternatively we could restrict to squarefree conductors, that is semistable elliptic curves.
