Bourbaki

Corollary 2

Tag 00A6, permanent. Cite it as [Bourbaki, Tag 00A6].

Algebra, VIII, p. 152, § 9. Radical

Let (Mi)i∈I(M_i)_{i\in I} be a family of A-modules, P its product, and S its direct sum. We have R(P)⊂∏i∈IR(Mi)\mathfrak{R}(P)\subset \prod_{i\in I}\mathfrak{R}(M_i) and R(S)=⨁i∈IR(Mi)\mathfrak{R}(S) =\bigoplus_{i\in I}\mathfrak{R}(M_i).

For i∈Ii\in I, let πi\pi_i be the projection with index ii from P to MiM_i. By Proposition 1, we have πi(R(P))⊂R(Mi)\pi_i(\mathfrak{R}(P))\subset \mathfrak{R}(M_i) for every i∈Ii\in I; this proves the first assertion. We have S⊂PS\subset P, hence

R(S)⊂S∩R(P)⊂S∩∏i∈IR(Mi)=⨁i∈IR(Mi) \mathfrak{R}(S)\subset S\cap \mathfrak{R}(P)\subset S\cap \prod_{i\in I}\mathfrak{R}(M_i) =\bigoplus_{i\in I}\mathfrak{R}(M_i)

Moreover, we have Mi⊂SM_i\subset S and therefore R(Mi)⊂R(S)\mathfrak{R}(M_i)\subset \mathfrak{R}(S) for every i∈Ii\in I, and consequently ⊕i∈IR(Mi)⊂R(S)\oplus_{i\in I}\mathfrak{R}(M_i)\subset \mathfrak{R}(S).

There exist families of modules such that the radical of the product is not isomorphic to the product of the radicals (Exercise 3 of VIII, p. 166).

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Transcribed from the printed edition, 2023, Springer Nature. Not translated.