Bourbaki

Exercise 43, § 1

Commutative Algebra, IX, p. 57

Let E:U(A)→Λ(A)E : U(A) \to \Lambda(A) be the mapping defined by

E(a)=∏n∈J(1−an(−T)n) E(a) = \prod_{n \in J} (1 - a_n(-T)^n)
for a∈U(A)a \in U(A). a) Prove the commutativity of the diagram

AJ↙Φ↘LU(A)→EΛ(A) \begin{array}{ccc} & & A^J \\ & \swarrow_{\Phi} & \searrow^{L} \\ U(A) & \xrightarrow{E} & \Lambda(A) \end{array}

Deduce that if AA is a Q\mathbf{Q}-algebra, one has

∏n∈J(1−anTn)=exp⁡(−∑n∈J(Φn(a)/n)Tn) \prod_{n \in J} (1 - a_n T^n) = \exp(- \sum_{n \in J} (\Phi_n(a)/n) T^n)
for every a∈U(A)a \in U(A). b) Now prove that EE is bijective. (If one puts ∏n∈J(1−an(−T)n)=∑n≥0cn(a)(−T)n\prod_{n \in J} (1 - a_n(-T)^n) = \sum_{n \geq 0} c_n(a)(-T)^n, it is enough to apply Exercise 33, a), p. 51 to the family of polynomials cn(X)c_n(X) in Z[X]\mathbf{Z}[X].) c) Endow Λ(A)\Lambda(A) with the unique ring structure such that EE is a ring isomorphism. Now prove that the addition of Λ(A)\Lambda(A) is the multiplication of series, with identity element 1. The multiplication of the ring Λ(A)\Lambda(A), denoted (f,g)↦f∗g(f, g) \mapsto f * g, is defined by the formula
E(a×b)=E(a)∗E(b) E(a \times b) = E(a) * E(b)
for all a,b∈U(A)a, b \in U(A); its identity element is 1+T1 + T. d) Let τ:A→U(A)\tau : A \to U(A) be the mapping defined in Exercise 37, c), p. 53. One has
E(τ(a))=1+aT(1+aT)∗(1+bT)=1+abT(1+aT)∗f(T)=f(aT)L(1+aT)=(an)n∈J \begin{align*} E(\tau(a)) &= 1 + aT \\ (1 + aT) * (1 + bT) &= 1 + abT \\ (1 + aT) * f(T) &= f(aT) \\ L(1 + aT) &= (a^n)_{n \in J} \end{align*}
for every a,ba, b in AA and f(T)f(T) in Λ(A)\Lambda(A). e) Let ff and gg be two elements of Λ(A)\Lambda(A) which are polynomials in TT, and let mm be the degree of ff. Put
φ(X)=Xmf(T/X) \varphi(X) = X^m f(T/X)
and
γ(X)=g(−X). \gamma(X) = g(-X).
These are polynomials in XX with coefficients in A[T]A[T]; φ\varphi is monic and f∗gf * g is the resultant res⁡(φ,γ)\operatorname{res}(\varphi, \gamma) of the polynomials φ\varphi and γ\gamma (cf. A, IV, p. 75). (One may start from the formula
(1+aT)∗g=g(aT), (1 + aT) * g = g(aT),
deduce the result if A=Z[X1,...,Xm]A = \mathbf{Z}[X_1, ..., X_m] and f=∏i=1m(1+XiT)f = \prod_{i=1}^m (1 + X_i T), and pass from there to the general case.)

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