Bourbaki

Definition 2

Tag 028N, permanent. Cite it as [Bourbaki, Tag 028N].

Théories spectrales, chapter I, § 3. Algèbres de Banach commutatives

The image of the mapping X(h)\mathsf{X}(h) is called the simultaneous spectrum of xx, and denoted by SpAΛ(x)^{\Lambda}_A(x) or SpΛ(x)^{\Lambda}(x).

The simultaneous spectrum of xx is a compact subset of CΛ\mathbf{C}^{\Lambda}. An element c=(cλ)∈CΛc= (c_{\lambda})\in \mathbf{C}^{\Lambda} belong to SpAΛ(x)^{\Lambda}_A(x) if and only if the elements xλ−cλx_{\lambda}-c_{\lambda} belong to one and the same maximal ideal of A, in other words if the family (xλ−cλ)λ∈Λ(x_{\lambda}-c_{\lambda})_{\lambda\in\Lambda} does not generate the algebra A.

If Λ contains only one element, so that the family xx reduces to a single element x∈Ax\in A, we have SpAΛ(x)=^{\Lambda}_A(x) = SpA(x)_A(x) (I, p. 37, prop. 6, b)). If Λ′⊂Λ\Lambda '\subset \Lambda, then SpAΛ′((xλ)λ∈Λ′)^{\Lambda'}_A((x_{\lambda})_{\lambda\in\Lambda'}) is the image of SpAΛ((xλ)λ∈Λ)^{\Lambda}_A((x_{\lambda})_{\lambda\in\Lambda}) under the canonical projection mapping of CΛ\mathbf{C}^{\Lambda} onto CΛ′\mathbf{C}^{\Lambda'}. In particular, we have

SpAΛ(x)⊂∏λ∈Λ^{\Lambda}_A(x)\subset \prod_{\lambda\in\Lambda} SpA(xλ)_A(x_{\lambda}).

Let zλz_{\lambda}, for λ∈Λ\lambda \in \Lambda, denote the coordinate functions on CΛ\mathbf{C}^{\Lambda}. If χ∈\chi \in X(A)\mathsf{X}(A), the value at χ\chi of zλ∘X(h)z_{\lambda}\circ \mathsf{X}(h) is χ(xλ)\chi (x_{\lambda}), hence zλ∘X(h)=G(xλ)z_{\lambda}\circ \mathsf{X}(h) =\mathscr{G}(x_{\lambda}).

Let A and B be commutative unital Banach algebras, φ\varphi a unital morphism from A into B, and x=(xλ)λ∈Λx= (x_{\lambda})_{\lambda\in\Lambda} a family of elements of A. Let φ(x)\varphi (x) denote the family (φ(xλ))λ∈Λ(\varphi (x_{\lambda}))_{\lambda\in\Lambda} of elements of B. We have, for every χ∈X(B)\chi \in \mathsf{X}(B), and every λ∈Λ\lambda \in \Lambda

χ(φ(xλ))=(X(φ)(χ))(xλ) \chi (\varphi (x_{\lambda})) = (\mathsf{X}(\varphi )(\chi ))(x_{\lambda})

hence SpBΛ(φ(x))⊂^{\Lambda}_B(\varphi (x))\subset SpAΛ(x)^{\Lambda}_A(x). The diagram

←X(φ) \leftarrow^{\mathsf{X}(\varphi)}
X(B)→X(A) \mathsf{X}(B)\rightarrow \mathsf{X}(A)
→←→←(1) \rightarrow \leftarrow \rightarrow \leftarrow \tag{1}
Λ←iΛ _{\Lambda}\leftarrow_{i\Lambda}

SpB(φ(x))→_B(\varphi (x))\rightarrow SpA(x)_A(x)

where ii denotes the inclusion, and where the vertical arrows denote the mappings defined by the families φ(x)\varphi (x) and xx, is therefore commutative.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Machine translation of the English, by gpt-5.4, gpt-5-mini, against glossary version 34. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.