Bourbaki

Definition 2

Tag 047T, permanent. Cite it as [Bourbaki, Tag 047T].

Commutative Algebra, chapter II, § 3. Local rings. Passage from the local to the global

Let A,B\mathbf{A}, \mathbf{B} be two local rings and m,nm, n their respective maximal ideals. A homomorphism u:A→Bu : \mathbf{A} \to \mathbf{B} is called local if u(m)⊂nu(m) \subset n.

This amounts to saying that uˉ−1(n)=m\bar{u}^{-1}(n) = m, for uˉ−1(n)\bar{u}^{-1}(n) is then an ideal containing mm and not containing 1 and hence equal to mm. Taking quotients, we then derive canonically from uu an injective homomorphism A/m→B/n\mathbf{A}/m \to \mathbf{B}/n from the residue field of A\mathbf{A} to that of B\mathbf{B}.

Examples (1) A field is a local ring. A ring which is reduced to 0 is not a local ring. (2) Let A\mathbf{A} be a local ring and kk its residue field. The ring of formal power series B=A[[X1,…,Xn]]B = \mathbf{A}[[X_1, \ldots, X_n]] is a local ring, for the non-invertible elements of BB are the formal power series whose constant terms are not invertible in A\mathbf{A} (Algebra, Chapter IV, §5, no. 6, Proposition 4). The canonical injection of A\mathbf{A} into BB is a local homomorphism and the corresponding injection of residue fields is an isomorphism. (3) Let bb be an ideal of a ring A\mathbf{A} which is only contained in a single maximal ideal mm; then A/bA/b is a local ring with maximal ideal m/bm/b and residue field canonically isomorphic to A/mA/m. This applies in particular to the case b=mkb = m^k, mm being any maximal ideal of AA (§1§ 1, no. 1, Corollary to Proposition 1). If AA itself is a local ring with maximal ideal mm, then for every ideal b≠Ab \neq A of AA, A/bA/b is a local ring, the canonical homomorphism A→A/bA \to A/b a local homomorphism and the corresponding homomorphism of residue fields an isomorphism.

(4) Let XX be a topological space, x0x_0 a point of XX and AA the ring of germs at the point x0x_0 of real-valued functions continuous in a neighbourhood of x0x_0 (General Topology, Chapter I, § 6, no. 10). Clearly, for the germ at x0x_0 of a continuous function ff to be invertible in AA, it is necessary and sufficient that f(x0)≠0f(x_0) \neq 0, as that implies that f(x)≠0f(x) \neq 0 in a neighbourhood of x0x_0. The ring AA is therefore a local ring whose maximal ideal mm is the set of germs of functions which are zero at x0x_0; taking quotients, the mapping g↦g(x0)g \mapsto g(x_0) of AA to R\mathbf{R} gives an isomorphism of the residue field A/mA/m onto R\mathbf{R}.

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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