Bourbaki

Definition 1

Tag 04PJ, permanent. Cite it as [Bourbaki, Tag 04PJ].

Commutative Algebra, chapter V, § 2. The lift of prime ideals

Let A,A′\mathbf{A}, \mathbf{A}' be two rings and h:A→A′h : \mathbf{A} \to \mathbf{A}' a ring homomorphism. An ideal a′∈A′a' \in A' is said to lie above an ideal aa of A\mathbf{A} if a=h‾1(a′)a = \overline{h}^1(a').

To say that a prime ideal p′p' of A′A' lies above an ideal pp of A\mathbf{A} therefore means that pp is the image of p′p' under the continuous mapping "h:Spec(A′)→Spec(A)h : \mathrm{Spec}(A') \to \mathrm{Spec}(A)" associated with hh (Chapter 11, § 4, no. 3).

Note that for there to exist an ideal of A′A' lying above the ideal (0)(0) of A\mathbf{A}, it is necessary and sufficient that h:A→A′h : \mathbf{A} \to \mathbf{A}' be injective.

Let aa be an ideal of A\mathbf{A}; by taking quotients the homomorphism hh gives a homomorphism h1:A/a→A′/a′h_1 : \mathbf{A}/a \to \mathbf{A}'/a'; to say that a′a' is an ideal of A′A' lying above aa is equivalent to saying that aA′⊂a′aA' \subset a' and that a′/aA′a'/aA' is an ideal of A′/aA′\mathbf{A}'/aA' lying above (0)(0).

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 volume. Not translated.