Bourbaki

Remark

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

Algebra, chapter II, § 1. Modules

When E and F are two commutative groups considered as modules over the ring Z\mathbf{Z} (no. 1), every homomorphism uu of the group E (without operators) into the group F (without operators) is also a linear mapping of E into F, since for n an integer > 0, the relation u(n.x)=n.u(x)u(n . x) = n . u(x) follows from u(x+y)=u(x)+u(y)u(x + y) = u(x) + u(y) by induction on n and for n=−m<0n = -m < 0,

u(n.x)=u(−(m.x))=−u(m.x)=−(m.u(x))=n.u(x). u(n . x) = u(-(m . x)) = -u(m . x) = -(m . u(x)) = n . u(x).

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

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Cites

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