Asmuth-Bloom Secret Sharing $m_0$ public?

Question

i am trying to figure out how Asmuth-Bloom secret sharing works and followed the example on wikipedia and replaced it with own numbers.

Namely, i used:
$S=3$
$k=3$
$n=4$
$m_0 = 13, m_1=45, m_2=46, m_3=47, m_4=49$

I've been able to construct the shares

$s_1 = (17, 45), s_2 = (2, 46), s_3 = (34, 47), s_4=(6,49)$

I then result in $x \equiv 692 \mod 46*47*48$ which is correct, since $S = 692 \equiv 3 \mod 13$ which is my secret.

However, i wonder where $m_0 = 13$ is coming from for the final secret reconstruction, it seems like it is not explained anywhere. The only thing i could find is this stating it would be "kept secret by the dealer unless stated otherwise".

Is $m_0$ just either a public parameter? Or do i need the dealer to store it "secretly" (which would require the dealer to reconstruct the secret)?

Answer 1

According to Dragan&Tiplea,On the Asymptotic Idealness of the Asmuth-Bloom Threshold Secret Sharing Scheme $m_0$ is a public parameter.
The scheme itself does not require $m_0$ to be kept secret. However, by keeping it secret you could eliminiate the possibility to get a probabilistic distribution of secrets as explained here but if $m_0$ for some reason is lost, the secret is not recoverable, which kind of defeats the purpose (or parts of it) of a secret sharing scheme.