Proposition 6

Tag 062E, permanent. Cite it as [Bourbaki, Tag 062E].

Algebra, chapter III, § 6. Symmetric algebras

Let A be a commutative ring and M and N two A-modules. If with every A-linear mapping $g : S^n(M) \to N$ ( $n \geqslant 1$ ) is associated the n-linear mapping

(x_1, x_2, \ldots, x_n) \mapsto g(x_1 x_2 \ldots x_n)

(where on the right-hand side the product is taken in the algebra $S(M)$ ), a bijective A-linear mapping is obtained of the A-module $\operatorname{Hom}_A(S^n(M), N)$ onto the A-module of symmetric n-linear mappings of $M^n$ into N.

Recall (II, § 3, no. 9) that there is a canonical bijection of the A-module $\operatorname{Hom}_A(T^n(M), N)$ onto the A-module $\mathscr{L}_n(M, \ldots, M; N)$ of all n-linear mappings of $M^n$ into N obtained by associating with every A-linear mapping $f : T^n(M) \to N$ the n-linear mapping

\tilde{f} : (x_1, x_2, \ldots, x_n) \mapsto f(x_1 \otimes x_2 \otimes \cdots \otimes x_n).

On the other hand, the A-linear mappings $g : S^n(M) \to N$ are in one-to-one correspondence with the A-linear mappings $f : T^n(M) \to N$ such that $f$ is zero on $\mathfrak{J}'_n$ , by associating with $g$ the mapping $f = g \circ p_n$ , where

p_n : T^n(M) \to S^n(M) = T^n(M)/\mathfrak{J}'_n

is the canonical homomorphism (II, § 2, no. 1, Theorem 1). But as $\mathfrak{J}'_n$ is a linear combination of elements of the form

(u_1 \otimes u_2 \otimes \cdots \otimes u_p) \otimes (x \otimes y - y \otimes x) \otimes (v_1 \otimes \cdots \otimes v_{n-p-2})

n-th symmetric power

(x, y, u_i, v_j in M), to say that the function f is of the form g \circ p_n means that the corresponding n-linear function $\bar{f}$ satisfies the relation

\bar{f}(u_1, \ldots, u_p, x, y, v_1, \ldots, v_{n-p-2}) = \bar{f}(u_1, \ldots, u_p, y, x, v_1, \ldots, v_{n-p-2});

in other words, by what has been seen above, this means that $\bar{f}$ is symmetric; whence the proposition, taking account of the fact that

p_n(x_1 \otimes x_2 \otimes \cdots \otimes x_n) = x_1 x_2 \ldots x_n

with the $x_i \in M$ .

The A-module $S^n(M)$ is called the n-th symmetric power of M. For every A-module homomorphism $u : M \to N$ , the mapping $S^n(u) : S^n(M) \to S^n(N)$ which coincides with $S(u)$ on $S^n(M)$ is called the n-th symmetric power of u.

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

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Nothing.

Proposition 6

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