Bourbaki

Exercise 7, § 9

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

Algebra, VIII, p. 166

Bourbaki marks this one as harder than the rest.

Let A be a ring, and let I be the set of ordinals α\alpha such that Card(α\alpha )⩽\leqslant Card(A) (Set Theory, III, §6, p. 241, Exercise 10). Define two-sided ideals Nα\mathfrak{N}_{\alpha} for all α∈I\alpha \in I as follows by transfinite induction. Take N0=N\mathfrak{N}_0=\mathfrak{N} (Exercise 6, b)). If α\alpha admits a successor α+1\alpha + 1, then Nα+1\mathfrak{N}_{\alpha+1} is the two-sided ideal such that Nα+1/Nα\mathfrak{N}_{\alpha+1}/\mathfrak{N}_{\alpha} is the sum of the nilpotent two-sided ideals of A/NαA/\mathfrak{N}_{\alpha}. If α\alpha is a limit ordinal, take Nα\mathfrak{N}_{\alpha} to be the union of the Nβ\mathfrak{N}_{\beta} for β<α\beta < \alpha. Let τ\tau be the smallest ordinal such that Nτ+1=Nτ\mathfrak{N}_{\tau+1}=\mathfrak{N}_{\tau}; the ideal P=Nτ\mathfrak{P}=\mathfrak{N}_{\tau} is called the prime radical of A.

a) Prove that we have N⊂P⊂G\mathfrak{N}\subset \mathfrak{P}\subset \mathfrak{G} and that P\mathfrak{P} is the smallest semiprime nil ideal of A.[^1]

b) Prove that P\mathfrak{P} is the intersection of the semiprime two-sided ideals of A (cf. VIII, p. 149, Exercise 3).

c) Prove that P\mathfrak{P} is the intersection of the prime two-sided ideals of A (loc. cit., c)). d) Prove that P\mathfrak{P} is the set of elements aa of A with the following property: every sequence (an)n⩾0(a_n)_{n\geqslant 0} of elements of A such that a0=aa_0=a and an+1∈anAana_{n+1}\in a_nAa_n for every nn has finite support (same method as in loc. cit.).

[^1]: For an example of a ring such that P≠G\mathfrak{P}\not=\mathfrak{G}, see R. Baer, Radical ideals, Amer. J. of Math. 65 (1943), p. 540.

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