Bourbaki

Proposition 3

Tag 06GJ, permanent. Cite it as [Bourbaki, Tag 06GJ].

Algebra, chapter V, § 1. Prime fields. Characteristic

— Let BB be a perfect ring of characteristic pp and AA a subring of BB. Write Ap−∞=⋃f≥0Ap−f\mathbf{A}^{p^{-\infty}} = \bigcup_{f \geq 0} \mathbf{A}^{p^{-f}} and denote by jj the canonical injection of AA in Ap−∞\mathbf{A}^{p^{-\infty}}. Then Ap−∞\mathbf{A}^{p^{-\infty}} is the smallest perfect subring of BB containing AA and (Ap−∞,j)(\mathbf{A}^{p^{-\infty}}, j) is a perfect closure of AA.

For each integer f∈Zf \in \mathbf{Z} denote by πf\pi_f the automorphism b↦bpfb \mapsto b^{p^f} of BB. The sequence of subrings π−f(A)\pi_{-f}(A) of BB (for f≥0f \geq 0) is ascending and its union Ap−∞\mathbf{A}^{p^{-\infty}} is thus a subring of BB. We have π1(Ap−∞)=⋃f≥0π−(f−1)(A)=Ap−m\pi_1(\mathbf{A}^{p^{-\infty}}) = \bigcup_{f \geq 0} \pi_{-(f-1)}(A) = \mathbf{A}^{p^{-m}}, hence Ap−∞\mathbf{A}^{p^{-\infty}} is a perfect subring of BB. Finally let B0B_0 be a perfect subring of BB containing AA; for every integer f≥0f \geq 0 we have π−f(A)⊂π−f(B0)=B0\pi_{-f}(A) \subset \pi_{-f}(B_0) = B_0, whence Ap−∞⊂B0\mathbf{A}^{p^{-\infty}} \subset B_0.

If vv is a homomorphism of AA into a perfect ring B′B' of characteristic pp, then for every integer f≥0f \geq 0 we can define a homomorphism hfh_f of π−f(A)\pi_{-f}(A) into B′B' by hf(π−f(a))=v(a)p−fh_f(\pi_{-f}(a)) = v(a)^{p^{-f}} for all a∈Aa \in A. We see at once that hf+1h_{f+1} agrees with hfh_f on π−f(A)\pi_{-f}(A); thus there exists a homomorphism hh of Ap−∞\mathbf{A}^{p^{-\infty}} into B′B' which induces hfh_f on π−f(A)\pi_{-f}(A) for all f≥0f \geq 0 and in particular, hh extends h0=vh_0 = v. If h′h' is another extension of vv to a homomorphism of Ap−∞A^{p^{-\infty}} into B, the equality h′=hh' = h may be established as in the proof of Th. 3.

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, 2003, Springer. Not translated.