In other words, $(\bigwedge(M), \phi'')$ 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 from the $A$-module $M$ to an $A$-algebra satisfying (1).

The uniqueness of $g$ follows from the fact that $\phi''(M) = M$ generates $\bigwedge(M)$. To prove the existence of $g$, we note that by § 5, no. 1, Proposition 1 there exists an $A$-algebra homomorphism $g_1 : T(M) \to E$ such that $f = g_1 \circ \phi$; we need to prove that $g_1$ is zero on the ideal $\mathfrak{J}''$, for then if

is the canonical homomorphism, we can write $g_1 = g \circ p$, where $g : \bigwedge(M) \to E$ 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 $x \otimes x$ for $x \in M$. This completes the proof.

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

The existence and uniqueness of $u'$ follow from no. 1, Proposition 1 applied to the algebra $\bigwedge(N)$ and $f = \phi_N'' \circ u : M \to \bigwedge(N)$; for $f(M) \subset N$ and hence $f$ satisfies condition (1) by definition of $\mathfrak{g}_N''$: as $f(M) \subset \bigwedge^1(N) = N$, the fact that $u''$ is a graded algebra homomorphism follows from Remark 1 of no. 1.

The homomorphism $u''$ of Proposition 2 will henceforth be denoted by $\bigwedge(u)$. If $P$ is an $A$-module and $v : N \to P$ is an $A$-linear mapping, then

for $\bigwedge(v) \circ \bigwedge(u)$ is an algebra homomorphism which renders commutative the diagram

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

with the $x_i \in M$, since $\Lambda(u)$ is an algebra homomorphism and $\Lambda^1(u) = u$; the restriction $\Lambda^0(u)$ to $A$ is the identity mapping. Note that $\Lambda^n(u)$ is obtained from $T^n(u): T^n(M) \to T^n(N)$ by passing to the quotients.

The proof is derived from that of § 6, no. 2, Proposition 4, replacing $S$ by $\Lambda$ and $\mathfrak{g}'$ by $\mathfrak{g}''$.

If $u: M \to N$ is an injective linear mapping, it is not always true that $\Lambda(u)$ is an injective mapping (§ 6, Exercise 3) (see however below no. 9, Corollary to Proposition 12). However this is so when $u$ is an injection such that $u(M)$ is a direct factor of $N$ and then the image of $\Lambda(u)$ (isomorphic to $\Lambda(M)$) is a direct factor of $\Lambda(N)$; the proof is the same as that for the analogous assertions for $T(u)$ (§ 5, no. 2) replacing $T$ by $\Lambda$.

The proof is deduced from that of § 5, no. 2, Corollary to Proposition 4 replacing $T$ by $\Lambda$ throughout.

$N$ is called the vector subspace of $M$ associated with the element $z$ of $\Lambda(M)$.

(ii) If there exist two distinct indices $i, j$ such that $x_i = x_j$, the product

is zero.

(i) First of all, since $x \wedge x = 0$ for all $x \in M$ by definition of the ideal $\mathfrak{g}'$, also, for $x, y$ in $M$,

This establishes (6) in the case $n = 2$. The general case then follows from § 4, no. 6, Lemma 3.

(ii) Under the hypothesis of (ii), there exists a permutation $\sigma \in S_n$ such that $\sigma(1) = i$ and $\sigma(2) = j$; then the left hand side of (6) is zero for this permutation and hence so is the right hand side.

By Proposition 5 this reduces to proving

For $1 \leq h < h' \leq p$ or $p + 1 \leq h < h' \leq n$, $\sigma(h') > \sigma(h)$ and the number of ordered pairs $(h, h')$ such that $1 \leq h \leq p < h' \leq n$ and $\sigma(h) > \sigma(h')$ is equal to $\nu$.

It suffices to apply Proposition 13 of $§ 4$, no. 9 to $\bigwedge(M)$, taking as generating system the set $M$ and using Proposition 5.

Let $(x_i)_{1 \leq i \leq n}$ be a generating system of M. Every element of $\bigwedge^p(M)$ is a linear combination of elements of the form

where the indices $i_k$ belong to $\{1, n\}$; by Proposition 5, these indices can be assumed to be distinct (otherwise the corresponding element is zero). If $p > n$, there is no such sequence of indices and hence $\bigwedge^p(M) = \{0\}$. If $p \leq n$, these sequences are finite in number, which completes the proof.

We consider the 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

(II, § 3, no. 9). On the other hand, the A-linear mappings $g : \bigwedge^n(M) \to N$ are in one-to-one correspondence with the A-linear mappings $f : T^n(M) \to N$ such that f is zero on $\mathfrak{g}_n''$, by associating with g the mapping $f = g \circ p_n$, where

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

$(x, u_i, v_j$ in $M$), to say that $f$ is of the form $g \circ p_n$ means that the corresponding $n$-linear function $\bar{f}$ satisfies (8), in other words it is alternating.

The A-module $\bigwedge^n(M)$ is called the n-th exterior power of M. For every A-module homomorphism $u : M \to N$, the mapping

which coincides with $\bigwedge(u)$ on $\bigwedge^n(M)$ is called the n-th exterior power of u.

It suffices to prove that the $n$-linear mapping

of $M^n$ to $M$ is alternating. Now, if $x_i = x_{i+1}$, all the terms in the sum on the right hand side have zero coefficients except $x_i$ and $x_{i+1}$, since $f$ is alternating; on the other hand, the coefficient of $x_i$ is

and that of $x_{i+1}$ is $(-1)^{i+1} f(x_1, \ldots, x_i, x_{i+2}, \ldots, x_n)$ and they are inverses by hypothesis.

of the $B$-linear mapping $1_B \otimes \phi_M'' : B \otimes_A M \to B \otimes_A \bigwedge_A(M)$ is a graded $B$-algebra isomorphism.

The proof is derived from that of § 5, no. 4, Proposition 5 replacing $T$ by $\bigwedge$ and $\phi_M$ by $\phi_M''$.

The proof is the same as that of § 5, no. 4, Proposition 6 replacing throughout T by $\Lambda$ and $\phi$ by $\phi''$ and taking account of the fact that a direct limit of alternating algebras is alternating.

To prove that g is bijective, we define a graded algebra homomorphism

such that $g \circ h$ and $h \circ g$ are respectively the identity mappings on $\Lambda(M)$ and $\bigotimes_{\lambda \in L}^{\mathfrak{g}} \Lambda(M_\lambda)$. For each $\lambda \in L$, consider the composite linear mapping

There exists one and only one A-linear mapping $u : M \to \bigotimes_{\lambda \in L}^{\mathfrak{g}} \Lambda(M_\lambda)$ such that $u \circ j_\lambda = u_\lambda$ for all $\lambda \in L$. The skew tensor product $\bigotimes_{\lambda \in L}^{\mathfrak{g}} \Lambda(M_\lambda)$ is an alternating algebra: for, for L finite, this follows from § 4, no. 9, Proposition 14 and, for L arbitrary, this follows from the definition of this product given in § 4, no. 8, Remark 1 and the fact that a direct limit of alternating graded algebras is alternating. As also $u(M)$ is contained in the submodule of elements of degree 1 in the graded algebra $\bigotimes_{\lambda \in L} \Lambda(M_\lambda)$, it follows from no. 1, Proposition 1 and Remark 1 that there exists a unique graded algebra homomorphism.

such that $h \circ \phi''_M = u$. The verification of the fact that $g \circ h$ and $h \circ g$ are the identity mappings is then performed as in § 6, no. 6, Proposition 9 replacing $S$ by $\Lambda$ and $\phi'$ by $\phi''$.

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