§ 6. Symmetric algebras

1. Definition of the symmetric algebra of a module

Definition 1

Let A be a commutative ring and M an A-module. The symmetric algebra of M, denoted by S(M), or Sym(M), or S_A(M), is the quotient algebra over A of the tensor algebra T(M) by the two-sided ideal $\mathfrak{J}'$ (also denoted by $\mathfrak{J}'_M$ ) generated by the elements $xy - yx = x \otimes y - y \otimes x$ of T(M), where x and y run through M.

Since the ideal $\mathfrak{J}'$ is generated by homogeneous elements of degree 2, it is a graded ideal (II, § 11, no. 3, Proposition 2); we write $\mathfrak{J}'_n = \mathfrak{J}' \cap T^n(M)$ ; the algebra S(M) is then graded by the graduation (called canonical) consisting of the $S^n(M) = T^n(M)/\mathfrak{J}'_n$ . Now $\mathfrak{J}'_0 = \mathfrak{J}'_1 = \{0\}$ and hence $S^0(M)$ is canonically identified with A and $S^1(M)$ with $T^1(M) = M$ ; we shall always make these identifications in what follows and denote by $\phi'$ or $\phi'_M$ the canonical injection $M \to S(M)$ .

Proposition 1

The algebra S(M) is commutative.

By definition $\phi'(x)\phi'(y) = \phi'(y)\phi'(x)$ for $x, y$ in M and, as the elements $\phi'(x)$ , where x runs through M, generate S(M), the conclusion follows from § 1, no. 7.

Proposition 2

Let E be an A-algebra and $f : M \to E$ an A-linear mapping such that*

(1)

f(x)f(y) = f(y)f(x) \text{ for all } x, y \text{ in } M.

*There exists one and only one A-algebra homomorphism

g : S(M) \to E

such that

f = g \circ \phi'

In other words, $(S(M), \phi')$ is a solution of the universal mapping problem (Set Theory, IV, § 3, no. 1), where $\Sigma$ is the species of A-algebra structure, the $\alpha$ -mappings being the linear mappings of the A-module M to an A-algebra satisfying (1).

The uniqueness of g follows from the fact that $\phi'(M) = M$ generates S(M). To prove the existence of g, note that by virtue of § 5, No. 1, Proposition 1 there exists an A-algebra homomorphism $g_1 : T(M) \to E$ such that $f = g_1 \circ \phi;$

all that needs to be proved is that $g_1$ is zero on the ideal $\mathfrak{J}'$ , for then, if $p : T(M) \to S(M) = T(M)/\mathfrak{J}'$ is the canonical homomorphism, we can write $g_1 = g \circ p$ , where $g : S(M) \to E$ is an algebra homomorphism, and the conclusion will follow from the fact that $p \circ \phi = \phi'$ . Now the kernel of $g_1$ is a two-sided ideal which, by virtue of (1) and the relation $g_1 \circ \phi = f$ , contains the elements $x \otimes y - y \otimes x$ for $x, y$ in $M$ . This completes the proof.

Remark

(1) Suppose that $E$ is a graded $A$ -algebra of type $\mathbf{Z}$ , with graduation $(E_n)$ , and suppose also that the linear mapping $f$ (assumed to satisfy (1)) is such that (2)

f(M) \subset E_1.

Then the relation

g(x_1 x_2 \ldots x_p) = f(x_1) f(x_2) \ldots f(x_p)

with the

x_i \in M

shows that

g(S^p(M)) \subset E_p

for all

p \geq 0

and hence

g

is a graded algebra homomorphism.

(2) Every element of $S(M)$ is a sum of products of the form $x_1 x_2 \ldots x_n$ , where the $x_i$ belong to $M$ ; care should be taken not to confuse such products taken in $S(M)$ with the analogous products taken in $T(M)$ .

(3) If $n!.1$ is invertible in $A$ , the $A$ -module $S^n(M)$ is generated by the elements of the form $x^n$ , where $x \in M$ ; this follows from the above remark and I, § 8, no. 2, Proposition 2.

2. Functorial properties of the symmetric algebra

Proposition 3

Let $A$ be a commutative ring, $M$ and $N$ two $A$ -modules and $u : M \to N$ an $A$ -linear mapping. There exists one and only one $A$ -algebra homomorphism $u' : S(M) \to S(N)$ such that the diagram

\begin{array}{ccc} M & \xrightarrow{u} & N \\ \downarrow \phi'_M & & \downarrow \phi'_N \\ S(M) & \xrightarrow{u'} & S(N) \end{array}

is commutative. Moreover, $u'$ is a graded algebra homomorphism.

The existence and uniqueness of $u'$ follow from no. 1, Proposition 2 applied to the commutative algebra $S(N)$ and $f = \phi'_N \circ u : M \to S(N)$ ; as

f(M) \subset S^1(N) = N,

the fact that

u'

is a graded algebra homomorphism follows from no. 1, Remark 1.

The homomorphism $u'$ of Proposition 3 will henceforth be denoted by $S(u)$ . If $P$ is an $A$ -module and $v : N \to P$ an $A$ -linear mapping, then

S(v \circ u) = S(v) \circ S(u)

for $S(v) \circ S(u)$ is an algebra homomorphism rendering commutative the diagram

\begin{array}{ccc} M & \xrightarrow{v \circ u} & P \\ \downarrow \phi'_M & & \downarrow \phi'_P \\ S(M) & \xrightarrow{S(v) \circ S(u)} & S(P) \end{array}

As $S(M)$ contains $M = S^1(M)$ , $S(u)$ is sometimes called the canonical extension of $u$ to $S(M)$ . The restriction $S^n(u): S^n(M) \to S^n(N)$ is such that

S^n(u)(x_1 x_2 \ldots x_n) = u(x_1) u(x_2) \ldots u(x_n)

where the

x_i \in M

, since

S(u)

is an algebra homomorphism and

S^1(u) = u

; the restriction

S^0(u)

A

is the identity mapping. Note that

S^n(u)

can be obtained from

T^n(u): T^n(M) \to T^n(N)

by passing to the quotients.

Proposition 4

If $u: M \to N$ is a surjective $A$ -linear mapping, the homomorphism $S(u): S(M) \to S(N)$ is surjective and its kernel is the ideal of $S(M)$ generated by the kernel $P \subset M \subset S(M)$ of $u$ .

We write $v = T(u): T(M) \to T(N)$ ; we know (§ 5, no. 2, Proposition 3) that $v$ is surjective and hence it follows from the definitions that $v(\mathfrak{J}'_M) = \mathfrak{J}'_N$ ; if $\mathfrak{R}$ is the kernel of $v$ , then $v^{-1}(\mathfrak{J}'_N) = \mathfrak{R} + \mathfrak{J}'_M$ . As $S(u): T(M)/\mathfrak{J}'_M \to T(N)/\mathfrak{J}'_N$ is derived from $v$ by passing to the quotients, it is a surjective homomorphism whose kernel is $\mathfrak{R}' = (\mathfrak{R} + \mathfrak{J}'_M)/\mathfrak{J}'_M$ . As $\mathfrak{R}$ is generated by the kernel $P$ of $u$ (§ 5, no. 2), so is $\mathfrak{R}'$ .

If $u: M \to N$ is an injective linear mapping, it is not always true that $S(u)$ is an injective mapping (Exercise 1). However it is so when $u$ is an injection such that $u(M)$ is a direct factor in $N$ and then the image of $S(u)$ (isomorphic to $S(M)$ ) is a direct factor of $S(N)$ ; the proof is the same as that for the analogous assertions for $T(u)$ (§ 5, no. 2) replacing $T$ by $S$ .

Proposition 5

Let $N$ and $P$ be two submodules of an $A$ -module $M$ such that their sum $N + P$ is a direct factor in $M$ and their intersection $N \cap P$ is a direct factor in $N$ and in $P$ . Then the homomorphisms $S(N) \to S(M)$ , $S(P) \to S(M)$ and*

S(N \cap P) \to S(M),

canonical extensions of the canonical injections, are injective; if $S(N)$ , $S(P)$ and $S(N \cap P)$ are identified with subalgebras of $S(M)$ by means of these homomorphisms, then

S(N \cap P) = S(N) \cap S(P).

(4)

The proof reduces to that of § 5, no. 2, Proposition 4 replacing $T$ by $S$ throughout. The hypotheses of Proposition 5 always hold for arbitrary submodules $N, P$ of $M$ when $A$ is a field.

Corollary

Let K be a commutative field and M a vector space over K. For every element $z \in S(M)$ there exists a smallest vector space N of M such that $z \in S(N)$ and N is finite-dimensional.

The proof is derived from that of § 5, no. 2, Corollary to Proposition 4 replacing T by S throughout.

N is called the vector subspace of M associated with z.

3. n-th symmetric power of a module and symmetric multilinear mappings

Let X, Y be two sets and $n$ an integer $\geqslant 1$ . A symmetric mapping of $X^n$ into Y is any mapping $f : X^n \to Y$ such that, for every permutation $\sigma \in \mathscr{S}_n$ and every element $(x_i) \in X^n$ ,

f(x_{\sigma(1)}, x_{\sigma(2)}, \ldots, x_{\sigma(n)}) = f(x_1, x_2, \ldots, x_n).

As the transpositions which exchange two consecutive integers generate the group $\mathscr{S}_n$ (I, § 5, no. 7), it suffices that condition (5) hold when $\sigma$ is such a transposition.

When Y is a module over a commutative ring A, clearly the set of symmetric mappings of $X^n$ into Y is a submodule of the A-module $Y^{X^n}$ of all mappings of $X^n$ into Y.

Proposition 6

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.

Remark

Let $\sigma$ be a permutation in $\mathfrak{S}_n$ ; as the mapping

(x_1, x_2, \ldots, x_n) \mapsto x_{\sigma^{-1}(1)} \otimes x_{\sigma^{-1}(2)} \otimes \cdots \otimes x_{\sigma^{-1}(n)}

of $M^n$ into $T^n(M)$ is A-multilinear, it may be written uniquely as

(x_1, \ldots, x_n) \mapsto u_\sigma(x_1 \otimes x_2 \otimes \cdots \otimes x_n),

where $u_\sigma$ is an endomorphism of the A-module $T^n(M)$ , also denoted by $z \mapsto \sigma . z$ . Clearly, if $\sigma$ is the identity element of $\mathfrak{S}_n$ , $u_\sigma$ is the identity; on the other hand, writing $y_i = x_{\sigma^{-1}(i)}$ , we obtain, for every permutation $\tau \in \mathfrak{S}_n, y_{\tau^{-1}(i)} = x_{\sigma^{-1}(\tau^{-1}(i))}$ and hence $\tau . (\sigma . z) = (\tau \sigma) . z$ ; in other words, the A-module $T^n(M)$ is a left $\mathfrak{S}_n$ -*set* under the operation $(\sigma, z) \mapsto \sigma . z$ (I, § 5, no. 1). The elements of $T^n(M)$ such that $\sigma . z = z$ for all $\sigma \in \mathfrak{S}_n$ are called (contravariant) symmetric tensors of order n; they form a sub-A-module $S'_n(M)$ of $T^n(M)$ .

For all $z \in T^n(M)$ , we write $s . z = \sum_{\sigma \in \mathfrak{S}_n} \sigma . z$ and call $s . z$ the symmetrization of the tensor $z$ ; clearly $s . z$ is a symmetric tensor and therefore $z \mapsto s . z$ is an endomorphism of $T^n(M)$ whose image $S''_n(M)$ is contained in $S'_n(M)$ ; in general, $S''_n(M) \neq S'_n(M)$ (Exercise 5). If $z$ is a symmetric tensor, then $s . z = n! z$ ; it follows that when n! is invertible in A, the endomorphism $z \mapsto (n!)^{-1} s . z$ is a projector of $T^n(M)$ (II, § 1, no. 8), whose image is $S'_n(M) = S''_n(M)$ ; moreover the kernel on this projector is just $\mathfrak{J}'_n$ . For, obviously $\sigma(\mathfrak{J}'_n) \subset \mathfrak{J}'_n$ for all $\sigma \in \mathfrak{S}_n$ and $\mathfrak{J}'_n$ is by definition generated by the tensors $z - \rho . z$ , where $\rho$ is a transposition exchanging two consecutive numbers in $\{1, n\}$ ; also, if $\sigma, \tau$ are two permutations in $\mathfrak{S}_n$ , then $z - (\sigma \tau) . z = z - \sigma . z + \sigma . (z - \tau . z)$ , whence it follows (since every permutation in $\mathfrak{S}_n$ is a product of transpositions exchanging two consecutive numbers) that $z - \sigma . z \in \mathfrak{J}'_n$ for all $z \in T^n(M)$ and $\sigma \in \mathfrak{S}_n$ . Therefore (always supposing that n! is invertible in A), it is seen that

z - (n!)^{-1} s . z = \sum_{\sigma \in \mathfrak{S}_n} (n!)^{-1} (z - \sigma . z) \in \mathfrak{J}'_n

for all $z \in T^n(M)$ , which proves our assertion.

When $n!$ is invertible in $A$ , the submodules $S'_n(M)$ and $\mathfrak{g}'_n$ of $T^n(M)$ are therefore supplementary and the restriction to $S'_n(M)$ of the canonical homomorphism $T^n(M) \to S^n(M) = T^n(M)/\mathfrak{g}'_n$ is an $A$ -module isomorphism, which allows us in the case envisaged to identify symmetric tensors of order $n$ with the elements of the $n$ -th symmetric power of $M$ . Note however that this identification is not compatible with multiplication, the product (in $T(M)$ ) of two symmetric tensors not being symmetric in general and not therefore having as image in $S(M)$ the product of the images of the symmetric tensors considered.

4. Extension of the ring of scalars

Let $A, A'$ be two commutative rings, $\rho : A \to A'$ a ring homomorphism, $M$ an $A$ -module, $M'$ an $A'$ -module and $f : M \to M'$ an $A$ -homomorphism (relative to $\rho$ ) of $M$ into $M'$ . The composite mapping $M \xrightarrow{f} M' \xrightarrow{\phi_{M'}} S_{A'}(M')$ is an $A$ -linear mapping of $M$ into the commutative algebra $\rho_*(S_A(M'))$ ; then there exists no. 1, Proposition 2) one and only one $A$ -homomorphism of algebras $f' : S_A(M) \to S_{A'}(M')$ rendering commutative the diagram

\begin{array}{ccc} M & \xrightarrow{f} & M' \\ \downarrow \phi_M' & & \downarrow \phi_{M'}' \\ S_A(M) & \xrightarrow{f'} & S_{A'}(M') \end{array}

It follows immediately that if $\sigma : A' \to A''$ is another ring homomorphism, $M''$ an $A''$ -module, $g : M' \to M''$ an $A'$ -homomorphism (relative to $\sigma$ ) and $g' : S_{A'}(M') \to S_{A''}(M'')$ the corresponding $A'$ -homomorphism of algebras, then the composite $A$ -homomorphism of algebras

S_A(M) \xrightarrow{f'} S_{A'}(M') \xrightarrow{g'} S_{A''}(M'')

corresponds to the composite $A$ -homomorphism $g \circ f : M \to M''$ (relative to $\sigma \circ \rho$ ).

Proposition 7

Let $A, B$ be two commutative rings, $\rho : A \to B$ a ring homomorphism and $M$ an $A$ -module. The canonical extension

\psi : S_B(B \otimes_A M) \to B \otimes_A S_A(M)

of the $B$ -linear mapping $l_B \otimes \phi_M' : B \otimes_A M \to B \otimes_A S_A(M)$ is a graded $B$ -algebra isomorphism.

The proof is derived from that of § 5, no. 3, Proposition 5 replacing $T$ by $S$ and $\phi_M$ by $\phi_M'$ .

5. Direct limit of symmetric algebras

Let $(A_\alpha, \phi_{\beta\alpha})$ be a directed direct system of commutative rings, $(M_\alpha, f_{\beta\alpha})$ a direct system of $A_\alpha$ -modules, $A = \lim \rightarrow A_\alpha$ and $M = \lim \rightarrow M_\alpha$ . For $\alpha \leq \beta$ , we derive canonically from the $A_\alpha$ -homomorphism $f_{\beta\alpha}: M_\alpha \to M_\beta$ an $A_\alpha$ -algebra homomorphism (no. 4, formula (8)) $f'_{\beta\alpha}: S_{A_\alpha}(M_\alpha) \to S_{A_\beta}(M_\beta)$ and it follows from (9) (no. 4) that $(S_{A_\alpha}(M_\alpha), f'_{\beta\alpha})$ is a direct system of $A_\alpha$ -algebras. On the other hand, $f_\alpha: M_\alpha \to M$ be the canonical $A$ -homomorphism; we derive (no. 4, formula (8)) an $A_\alpha$ -algebra homomorphism

f'_\alpha: S_{A_\alpha}(M) \to S_A(M)

and it also follows from (9) that the $f'_\alpha$ constitute a direct system of $A_\alpha$ -homomorphisms.

Proposition 8

The $A$ -homomorphism* $f' = \lim \rightarrow f'_\alpha : \lim \rightarrow S_{A_\alpha}(M_\alpha) \to S_A(M)$ *is a graded algebra isomorphism.

The proof is the same as that of § 5, no. 5, Proposition 6 replacing throughout $T$ by $S$ and $\phi$ by $\phi'$ and taking account of the fact that a direct limit of commutative algebras is commutative.

6. Symmetric algebra of a direct sum. Symmetric algebra of a free module. Symmetric algebra of a graded module

Let $A$ be a commutative ring, $M = \bigoplus_{\lambda \in L} M_\lambda$ the direct sum of a family of $A$ -modules and $j_\lambda: M_\lambda \to M$ the canonical injection; we derive an $A$ -homomorphism of algebras $S(j_\lambda): S(M_\lambda) \to S(M)$ . Since $S(M)$ is commutative, Proposition 8 of § 4, no. 5, can be applied to the homomorphisms $S(j_\lambda)$ and there therefore exists a unique algebra homomorphism

g: \bigotimes_{\lambda \in L} S(M_\lambda) \to S(M)

(also denoted by

g_M

) such that

S(j_\lambda) = g \circ f_\lambda

for all

\lambda \in L

, where

f_\lambda: S(M_\lambda) \to \bigotimes_{\lambda \in L} S(M_\lambda)

denotes the canonical homomorphism.

Proposition 9

The canonical homomorphism* $g$ (formula (10)) is a graded algebra isomorphism (cf. § 4, no. 8, Remark 1).

To prove that $g$ is bijective, we define an algebra homomorphism

h: S(M) \to \bigotimes_{\lambda \in L} S(M_\lambda)

such that $g \circ h$ and $h \circ g$ are respectively the identity mappings on $S(M)$ and $\bigotimes_{\lambda \in L} S(M_\lambda)$ . For each $\lambda \in L$ , let $u_\lambda$ be the composite linear mapping

M_\lambda \xrightarrow{\phi'_M} S(M_\lambda) \xrightarrow{f_\lambda} \bigotimes_{\lambda \in L} S(M_\lambda).

There exists one and only one

A

-linear mapping

u : M \to \bigotimes_{\lambda \in L} S(M_\lambda)

such that

u \circ j_\lambda = u_\lambda

for all

\lambda \in L

. As the

S(M_\lambda)

are commutative, so is their tensor product (§ 4, no. 5) and hence (no. 1, Proposition 2) there exists a unique algebra homomorphism

h : S(M) \to \bigotimes_{\lambda \in L} S(M_\lambda)

such that

h \circ \phi'_M = u

; on the other hand, it is immediate that

u(M)

is contained in the submodule of elements of degree 1 of the graded algebra

\bigotimes_{\lambda \in L} S(M_\lambda)

and hence

h

is a graded algebra homomorphism. For

x_\lambda \in M_\lambda

h(g(u_\lambda(x_\lambda))) = h(g(f_\lambda(\phi'_M(x_\lambda)))) = h(S(j_\lambda)(\phi'_M(x_\lambda))) = h(\phi'_M(j_\lambda(x_\lambda))) = u_\lambda(x_\lambda);

as the

u_\lambda(x_\lambda)

generate the algebra

\bigotimes_{\lambda \in L} S(M_\lambda)

(§ 4, no. 5, Proposition 8),

h \circ g

is certainly the identity mapping. Similarly,

g(h(\phi'_M(j_\lambda(x_\lambda)))) = g(u_\lambda(x_\lambda)) = g(f_\lambda(\phi'_M(x_\lambda))) = S(j_\lambda)(\phi'_M(x_\lambda)) = \phi'_M(j_\lambda(x_\lambda))

and, as the elements

\phi'_M(j_\lambda(x_\lambda))

generate the algebra

S(M)

g \circ h

is certainly the identity mapping.

Remark (1) Let $N = \bigoplus_{\lambda \in L} N_\lambda$ be the direct sum of another family of $A$ -modules with $L$ as indexing set and, for all $\lambda \in L$ , let $v_\lambda : M_\lambda \to N_\lambda$ be an $A$ -linear mapping, whence there is an $A$ -linear mapping $v = \bigoplus_\lambda v_\lambda : M \to N$ (II, § 1, no. 6, Proposition 6). Then the diagram

\begin{array}{ccc} \bigotimes_{\lambda \in L} S(M_\lambda) & \xrightarrow{g_M} & S(M) \\ \downarrow & & \downarrow S(v) \\ \bigoplus_{\lambda \in L} S(v_\lambda) & \xrightarrow{g_N} & S(N) \end{array}

is commutative, as follows from the definitions (§ 4, no. 5, Corollary to Proposition 8).

The sub- $A$ -module of $\bigotimes_{\lambda \in L} S(M_\lambda)$ with which $S^n(M)$ is identified by means of the isomorphism $g$ can be described more precisely. For every finite subset $J$ of $L$ , we write $E_J = \bigotimes_{\lambda \in J} S(M_\lambda)$ , so that $\bigotimes_{\lambda \in L} S(M_\lambda) = \varprojlim E_J$ relative to the directed set $\mathscr{F}(L)$ of finite subsets of $L$ , by definition (§ 4, no. 5). For every family $\nu = (n_\lambda) \in \mathbf{N}^{(L)}$ (thus having finite support) such that $\sum_{\lambda \in L} n_\lambda = n$ and every finite subset $J$ of $L$ containing the support of the family $\nu$ , we write

S^{J,\nu}(M) = \bigotimes_{\lambda \in J} S^{n_\lambda}(M_\lambda)

so that the submodule $E_{J,n}$ of elements of degree $n$ in $E_J$ is the direct sum of the $S^{J,\nu}(M)$ over all the families $\nu$ of support contained in $J$ and such that $\sum_{\lambda \in L} n_\lambda = n$ (§ 4, no. 7, Proposition 10 and § 4, no. 8). As a convention we write $S^{J,\nu}(M) = \{0\}$ for the families $\nu$ whose support is not contained in $J$ ; then $E_{J,n}$ can also be called the direct sum of all the $S^{J,\nu}(M)$ , where $\nu$ runs through the set $H_n$ of all families $\nu = (n_\lambda)_{\lambda \in L}$ such that $\sum_{\lambda \in L} n_\lambda = n$ . Since $S^0(M_\lambda)$ is identified with $A$ , clearly also, for two finite subsets $J \subset J'$ of $L$ and a family $\nu$ of support contained in $J$ , the canonical mapping $S^{J,\nu}(M) \to S^{J',\nu}(M)$ (restriction of the canonical mapping $E_J \to E_J$ to $S^{J,\nu}(M)$ ) is bijective. If we write, for all $\nu \in H_n$ ,

S^\nu(M) = \varprojlim S^{J,\nu}(M)

it is seen that, taking account of II, § 6, no. 2, Proposition 5:

Corollary

The A-module* $S^n(M)$ is the image under isomorphism (10) of the submodule of $\bigotimes_{\lambda \in L} S(M_\lambda)$ the direct sum of the submodules $S^\nu(M)$ for all the families $\nu = (n_\lambda) \in \mathbf{N}^{(L)}$ such that $\sum_{\lambda \in L} n_\lambda = n$ ; if $J$ is the support of $\nu$ , $S^\nu(M)$ is canonically isomorphic to $\bigotimes_{\lambda \in J} S^{n_\lambda}(M_\lambda)$ .

In general $S^\nu(M)$ , $\bigotimes_{\lambda \in J} S^{n_\lambda}(M_\lambda)$ and their image in $S^n(M)$ are identified.

Theorem 1

Let* $A$ be a commutative ring and $M$ a free $A$ -*module with basis* $(e_\lambda)_{\lambda \in L}$ . For every mapping $\alpha : L \to \mathbf{N}$ *of finite support, we write

e^\alpha = \prod_{\lambda \in L} e_\lambda^{\alpha(\lambda)}

(product in the commutative algebra $S(M)$ ). Then, when $\alpha$ runs through the set $\mathbf{N}^{(L)}$ of mappings of $L$ into $\mathbf{N}$ , of finite support, the $e^\alpha$ form a basis of the $A$ -*module* $S(M)$ .

As $M$ is the direct sum of the $M_\lambda = Ae_\lambda$ , it suffices to prove the theorem when $L$ is reduced to a single element and then apply Proposition 9. But when $M = Ae$ (e a free element), then $x \otimes y - y \otimes x = 0$ for all $x, y$ in $M$ ; the ideal $\mathfrak{g}'$ is therefore zero, whence $T(Ae) = S(Ae)$ and the theorem then follows from § 5, no. 5, Theorem 1.

The multiplication table of the basis (14) is given by

(15)

e^\alpha e^\beta = e^{\alpha + \beta}

where

\alpha + \beta

is the mapping

\lambda \mapsto \alpha(\lambda) + \beta(\lambda)

L

into

\mathbf{N}

. In other words, the basis

(e^\alpha)

S(M)

, with the multiplicative law (15), is canonically isomorphic to the free commutative monoid

N^{(L)}

derived from

L

; it follows (§ 2, no. 9) that the symmetric algebra

S(M)

of a free module

M

with basis whose indexing set is

L

, is canonically isomorphic to the polynomial algebra

A[(X_\lambda)_{\lambda \in L}]

, the canonical isomorphism being obtained by mapping

e_\lambda

X_\lambda

. In particular (§ 2, no. 7, Proposition 7), for every mapping

f : L \to E

L

into a commutative

A

-algebra

E

, there exists one and only one

A

-algebra homomorphism

\tilde{f} : S(M) \to E

such that

\tilde{f}(e_\lambda) = f(\lambda)

Remark (2). The above results can equally well be obtained as a consequence of the universal properties of polynomial algebras and symmetric algebras, taking account of II, § 1, no. 11, Corollary 3 to Proposition 17.

Corollary

If $M$ is a projective $A$ -module, $S(M)$ is a projective $A$ -module.

The proof is the same as that for § 5, no. 5, Corollary to Theorem 1, replacing $T$ by $S$ .

Proposition 10

Let $\Delta$ be a commutative monoid, $M$ a graded $A$ -module of type $\Delta$ and $(M_\alpha)_{\alpha \in \Delta}$ its graduation. For every ordered pair $(\alpha, n) \in \Delta \times \mathbf{N}$ , let $S^{\alpha, n}(M)$ be the submodule of $S^n(M)$ the direct sum of the submodules $\bigotimes_{\lambda \in J} S^{n_\lambda}(M_{\alpha_\lambda})$ , where $(n_\lambda)_{\lambda \in L}$ runs through the set of families of integers $\geq 0$ such that $\sum_{\lambda \in L} n_\lambda = n$ , $J$ is its support and, for each $(n_\lambda), (\alpha_\lambda)_{\lambda \in J}$ runs through the set of families of $\Delta^J$ such that $\sum_{\lambda \in J} \alpha_\lambda = \alpha$ . Then $(S^{\alpha, n}(M))_{(\alpha, n) \in \Delta \times \mathbf{N}}$ is the only graduation of type $\Delta \times \mathbf{N}$ compatible with the algebra structure on $S(M)$ and which induces on $M = S^1(M)$ the given graduation.

The fact that $S(M)$ is the direct sum of the $S^{\alpha, n}(M)$ follows from the Corollary to Proposition 9; the rest of the proof is identical with the end of the proof of § 5, no. 5, Proposition 7.

Suppose more particularly that $\Delta = \mathbf{Z}$ and let $S(M)$ be given the total graduation (of type $\mathbf{Z}$ ) (II, § 11, no. 1) corresponding to the graduation of type $\mathbf{Z} \times \mathbf{N}$ (and hence also of type $\mathbf{Z} \times \mathbf{Z}$ ) defined above; the homogeneous elements of degree $n \in \mathbf{Z}$ under this graduation are therefore those of the direct sum of the $S^{q, m}(M)$ for $q + m = n$ . Let $f$ be a homogeneous linear mapping of degree 0 of the graded $A$ -module $M$ into a commutative graded $A$ -algebra $F$ of type $\mathbf{Z}$ ; then the algebra homomorphism $g : S(M) \to F$ such that $f = g \circ \phi_M'$ is a homomorphism of graded algebras of type $\mathbf{Z}$ , as follows from the formula $g(x_1 x_2 \ldots x_n) = f(x_1) f(x_2) \ldots f(x_n)$ for homogeneous $x_i$ in $M$ , from the hypothesis on $f$ and from the definition of the graduation of type $\mathbf{Z}$ on $S(M)$ .

Exercises

See the exercises for § 6.