Today, We call the Kronecker's Jugendtraum Hilbert's 12th problem. But, Hilbert's interpretation of the "Jugendtraum" was not that intended by Kronecker.


And Weber missed his chance to disprove Hilbert's claim in the third volume of his *Lehrbuch der Algebra*.


I read Schappacher's "[On the History of Hilbert's Twelfth Problem: A Comedy of Errors](https://www.emis.de/journals/SC/1998/3/pdf/smf_sem-cong_3_243-273.pdf)" But I can't understand well What Weber was overlooking.



What is Weber's mistake about Hilbert's 12th problem?
It would help to be clearer about where in Schappacher's paper this issue is brought up. I think you are asking about pages 256-258, yes?


The end of the 2nd paragraph on p. 257 says "Translating back to the characterization by ray class groups, Weber overlooked precisely the possibility of choosing different signs in $\pm 1$ modulo different prime factors of $\mathfrak m$." Is that the step you're asking about?


As an elementary analogue, if you want to solve $x^2 \equiv 1 \bmod m$ in integers where $m$ is an odd integer bigger than $1$, the answer is $\pm 1 \bmod m$ when $m$ is an odd prime power, but there are more answers than that mod $m$ when $m$ has more than one odd prime factor: if $m = p\_1^{e\_1}\cdots p\_ke^{e\_k}$, for distinct odd primes $p\_i$, solving $x^2 \equiv 1 \bmod m$ is equivalent to solving $x^2 \equiv 1 \bmod p\_i^{e\_i}$ for all $i$, and that is equivalent to solving $x \equiv \pm 1 \bmod p\_i^{e\_i}$ for all $i$. There are a total of $2^k$ different solutions mod $m$ to that system of congruences, which is more than the trivial solutions $x \equiv \pm 1 \bmod m$ if $k \geq 2$ ($m$ has more than one prime factor).


