Bourbaki

Exercise 13, § 2

Algebra, III, 622

Bourbaki marks this one as harder than the rest.

Let A be a commutative ring and E an A-module. An A-module EnE_n is defined inductively by writing

E1=E,En=⨁p+q=n(Ep⊗AEq)for n≥2. E_1 = E, \quad E_n = \bigoplus_{p+q=n} (E_p \otimes_A E_q) \quad \text{for } n \geq 2.

We write ME=⨁n=1∞EnME = \bigoplus_{n=1}^\infty E_n and an A-algebra structure is defined on ME by setting, for xp∈Epx_p \in E_p and xq∈Eqx_q \in E_q, xpxq=xp⊗xq∈Ep+qx_p x_q = x_p \otimes x_q \in E_{p+q}.

(a) Show that for every A-algebra B and every A-linear mapping f:E→Bf : E \to B, there exists one and only one A-homomorphism of algebras ME→BME \to B extending ff. ME is called the free algebra of the A-module E.

(b) An integer a(n)a(n) is defined by induction on nn by the formulae

a(1),a(n)=∑p+q=na(p)a(q)if n≥2. a(1), \quad a(n) = \sum_{p+q=n} a(p)a(q) \quad \text{if } n \geq 2.

Show that EnE_n is isomorphic to the direct sum of a(n)a(n) modules isomorphic to E⊗nE^{\otimes n}.

(c) Let f(T)f(T) be the formal power series ∑n=1∞a(n)Tn\sum_{n=1}^\infty a(n)T^n. Show that

f(T)=T+(f(T))2. f(T) = T + (f(T))^2.

Deduce the formulae

f(T)=1−1−4T2anda(n)=2n−11.3.5…(2n−1)n!. f(T) = \frac{1 - \sqrt{1 - 4T}}{2} \quad \text{and} \quad a(n) = 2^{n-1} \frac{1.3.5 \ldots (2n-1)}{n!}.

(d) If I is any set and E=A(I)E = A^{(I)}, show that LibA(I)Lib_A(I) is identified with ME. There are canonical homomorphisms M(I)→ωN(I)→lNM(I) \xrightarrow{\omega} N^{(I)} \xrightarrow{l} N whose composition is length in M(I)M(I); ω(m)\omega(m) is called the multidegree of m∈M(I)m \in M(I); the mapping ω\omega defines on LibA(I)Lib_A(I) a graduation of type N(I)N^{(I)}. Show that the set (LibA(I))α(Lib_A(I))_\alpha of homogeneous elements of multidegree α∈N(I)\alpha \in N^{(I)} under this graduation is a free A-module of rank equal to

a(α)=a(n)n!α!=2n−11.3.5…(2n−1)α! a(\alpha) = a(n) \frac{n!}{\alpha!} = 2^{n-1} \frac{1.3.5 \ldots (2n-1)}{\alpha!}

where α!=∏i∈I(α(i))!\alpha! = \prod_{i \in I} (\alpha(i))! and n=∑i∈Iα(i)=l(α)n = \sum_{i \in I} \alpha(i) = l(\alpha) is the length of α\alpha.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.