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It is reported in this paper by Zagier, as well as in Sage, that the elliptic curve $E=197A1$ has congruence number 10. (Since $E$ has prime conductor, a theorem of Ribet ensures that the congruence number is equal to the modular degree.)

If $m_E$ is the congruence number of an elliptic curve $E$, and if the newform corresponding to $E$ is $f \in S_2(\Gamma_0(N))$, then there exists another cuspidal eigenform $g \in S_2(\Gamma_0(N))$ with integral Fourier coefficients such that $f \equiv g \mod m_E$. Note that $g$ is orthogonal (with respect to the Petersson inner product) to $f$, so in particular $f \neq g$. [See the linked Zagier paper, Section 5, for equivalent formulations.]

Now we come to my confusion. One quickly checks (on LMFDB or using Sage, for instance) that there are no other cuspidal eigenforms with integral coefficients at weight $2$ and level $197$. But $m_E=10$ implies that such a form does exist, and furthermore, it should be congruent to $f$ modulo $10$. What is going on?

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    $\begingroup$ I'm skeptical of your stated definition of the congruence number. According to Zagier, the congruence number is not the largest positive integer for which there is another integer coefficient eigenform congruent to it. It is the largest positive integer $r$ for which there is a form $g$ with integer coefficients in the $\mathbb{C}$-span of the other eigenforms with $g \equiv f \pmod{r}$. (Essentially, there is no eigenform with integer coefficients congruent to $f$, but there is another eigenform with algebraic integer coefficients, and the same "mod 10" reduction.) $\endgroup$
    – Jeremy Rouse
    Commented Aug 5, 2015 at 19:15
  • $\begingroup$ @JeremyRouse Oh, of course! What a silly mistake on my part. If you'd like to elevate your comment to an answer, I'd be happy to accept it. $\endgroup$
    – Jeff H
    Commented Aug 5, 2015 at 19:17
  • $\begingroup$ Or I can delete this question and pretend I never asked it ;) $\endgroup$
    – Jeff H
    Commented Aug 5, 2015 at 19:19
  • $\begingroup$ It seems the LMFDB is broken again. I clicked on your LMFDB link and got to 197a1 page, but going to Related objects "Modular form 197.2a" gave the excuse: " We are very sorry. The sought space could not be found in the database. " How are you able to use LMFDB to verify this? $\endgroup$
    – ABCDveve
    Commented Aug 5, 2015 at 23:49
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    $\begingroup$ In the case that $r$ is a prime number, this is a consequence of the Deligne-Serre lifting lemma (see Theorem 1.2 and Corollary 1.3 of the document here). In the case that $r$ is composite, maybe what I said is a little bit too naive (there might be a different eigenform $h_{i}$ for each prime factor of $r$ with $f \equiv h_{i} \pmod{\mathfrak{r}_{i}}$). $\endgroup$
    – Jeremy Rouse
    Commented Jun 24, 2023 at 20:55

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I am writing an answer to expand slightly on my comment. One of Zagier's definitions of the congruence number is the largest positive integer $r$ so that there is a cusp form $g$ with integer coefficients (and not necessarily a Hecke eigenform) that is orthogonal to $f$ (under the Petersson inner product).

This is related to the problem of reducing the space of modular forms $S_{k}^{{\rm new}}(\Gamma_{0}(N))$ modulo $p$ and diagonalizing the Hecke action (modulo $p$). It is possible that there may be multiple eigenforms in $S_{k}^{{\rm new}}(\Gamma_{0}(N), \mathbb{F}_{p})$ with the same Hecke eigenvalues. The Deligne-Serre lifting lemma then implies that there are (characteristic zero) eigenforms $f$ and $g$ with coefficients in some number field and a prime ideal of residue degree $1$, $\mathfrak{p}$, so that $f \equiv g \pmod{\mathfrak{p}}$, although the coefficient fields of $f$ and $g$ can be different.

There are many ways this can occur. For example, in level $1$, every Hecke eigenform is congruent modulo $p$ to a Hecke eigenform of weight at most $p^{2}$ or so. Also, given any elliptic curve $E$, the work of Rubin and Silverberg implies the existence of infinitely many $E'$ for which the modular forms $f_{E}$ and $f_{E'}$ are congruent modulo $5$.

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