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In Kummer's theorem, it is possible to write ideal factorization as- $$\langle x + y\rangle\langle x + y\zeta\rangle···\langle x + y\zeta^{p−1}\rangle = \langle z^p\rangle \cdots (1)$$ in which all factors are interpreted as principal ideals, here, $\zeta = e^{\frac{2πi}{p}}$, $\langle a \rangle$ is an ideal. which is dreied from $$(x + y)(x + y\zeta)···(x + y\zeta^{p−1}) = z^p \cdots (2)$$

But if $\langle a \rangle$ is an ideal, it represents all multiple of $a$ which makes the ideal $\langle a \rangle$ an infinite set, then how can we write an equation (1) from equation (2)?

Doesn't the left hand and right hand side of equation (1) represent infinite set?

What am I missing? Please explain.

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    $\begingroup$ Note that you could think of $2$, $3$, and $6$ as representatives of infinite sets (congruence classes) in the congruence relation $2\times3\equiv6\pmod7$ $\endgroup$ – J. W. Tanner Jan 29 at 20:53
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    $\begingroup$ Related! math.stackexchange.com/questions/1267777/… $\endgroup$ – eduard Jan 29 at 20:59
  • $\begingroup$ Well if you have that z=ab, multiplying ideals (a) and (b) is equivalent to considering ambn where m,n span over every element of the ring. I’m assuming you only care about communitive rings with multiplicative identity so we obviously know that ambn=abmn. Since m and n span over every element of the ring, it’s obvious that mn will also span over the ring, so we can just replace mn with a single variable that spans over the ring, lets call that k. We have that abmn=ab*k, which is exactly how we would write (ab), which is equal to (z) $\endgroup$ – uhhhhidk Jan 29 at 21:06

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