Remark

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

Algebra, chapter III, § 2. Examples of algebras

Let I and J be two sets and (P_j)_{j \in J} a family of elements of $\mathbf{Z}[(X_i)_{i \in I}]$ ; if Q is an element of $\mathbf{Z}[(X_j)_{j \in J}]$ such that $Q((P_j)_{j \in J}) = 0$ , then, for every family $(b_i)_{i \in I}$ of pairwise permutable elements of a ring B,

Q((P_j((b_i)_{i \in I}))_{j \in J}) = 0.

The relations which hold of the form

Q((P_j)_{j \in J}) = 0

are sometimes called polynomial identities. For example

(X_1 + X_2)^2 - X_1^2 - X_2^2 - 2X_1X_2 = 0

with

Q = Y_1^2 - Y_2, \quad P_1 = X_1 + X_2, \quad P_2 = X_1^2 + X_2^2 + 2X_1X_2

X_1^n - X_2^n - (X_1 - X_2)(X_1^{n-1} + X_1^{n-2}X_2 + \cdots + X_2^{n-1}) = 0

with

Q = Y_1 - Y_2Y_3, \quad P_1 = X_1^n - X_2^n, \quad P_2 = X_1 - X_2,

P_3 = X_1^{n-1} + X_1^{n-2}X_2 + \cdots + X_2^{n-1}

are polynomial identities.

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

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