Bourbaki

Theorem 7 (decomposition into prime factors)

Tag 05A8, permanent. Cite it as [Bourbaki, Tag 05A8].

Algebra, chapter I, § 4. Groups and groups with operators

Let aa be a strictly positive integer. There exists one and only one family (vp(a))p∈P(v_p(a))_{p \in \mathfrak{P}} of integers >0> 0 such that the set of p∈Pp \in \mathfrak{P} with vp(a)≠0v_p(a) \neq 0 is finite and

a=∏p∈Ppvp(a). a = \prod_{p \in \mathfrak{P}} p^{v_p(a)}.

As the group Z/aZ\mathbf{Z}/a\mathbf{Z} is finite, it admits a Jordan-Hölder series. Lemma 2 then implies that aa is a product of prime integers, whence the existence of the family (vp(a))(v_p(a)); further, for every family (vp(a))p∈P(v_p(a))_{p \in \mathfrak{P}} satisfying the conditions of Theorem 7, the integer vp(a)v_p(a) is, for all p∈Pp \in \mathfrak{P}, equal to the number of factors of a Jordan-Hölder series of Z/aZ\mathbf{Z}/a\mathbf{Z} isomorphic to Z/pZ\mathbf{Z}/p\mathbf{Z} (Lemma 2). The uniqueness of the family (vp(a))(v_p(a)) therefore follows from the Jordan-Hölder theorem (no. 7, Theorem 6).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed edition, 1998, Springer. Not translated.