Let $\tau'$ be the unique homomorphism of $T$ into $L$ which extends $\sigma$ and maps 1 to 1. Then, for $x, y$ in $g$,

hence $\tau'$ is zero on $J$ and defines on passing to the quotient a homomorphism $\tau$ of $U$ into $L$, mapping 1 to 1, such that $\sigma = \tau \circ \sigma_0$. The uniqueness of $\tau$ is immediate since $\sigma_0(g)$ and 1 generate the algebra $U$.

Let $g'$ be another Lie algebra over $K$, $U'$ its enveloping algebra and $\sigma'_0$ the canonical mapping of $g'$ into $U'$. Let $\phi$ be a homomorphism of $g$ into $g'$. Then $\sigma'_0 \circ \phi$ is an $\alpha$-mapping of $g$ into $U'$; hence there exists one and only one homomorphism $\tilde{\phi}$ of $U$ into $U'$ mapping 1 to 1 and such that the diagram

is commutative. This homomorphism maps the elements of $U$ whose constant term is zero to elements of $U'$ whose constant term is zero. If $g''$ is another Lie algebra over $K$ and $\phi'$ is a homomorphism of $g'$ into $g''$, then $(\phi' \circ \phi)^{\sim} = \tilde{\phi}' \circ \tilde{\phi}$.

The mapping $\sigma': (x_1, x_2) \mapsto \sigma_1(x_1) \otimes 1 + 1 \otimes \sigma_2(x_2)$ ($x_1 \in g_1, x_2 \in g_2$) is an $\alpha$-mapping of $g$ into $U_1 \otimes_K U_2$ and hence there exists (no. 1, Proposition 1) a unique homomorphism $\tau$ of $U$ into $U_1 \otimes_K U_2$ mapping 1 to 1, such that:

Now $\phi \circ \tau \circ \sigma = \phi \circ \sigma' = \sigma$ and $\tau \circ \phi \circ \sigma' = \tau \circ \sigma = \sigma'$, hence $\phi \circ \tau$ and $\tau \circ \phi$ are the identity mappings of $U$ and $U_1 \otimes_K U_2$ respectively. Hence the proposition.

$U_1 \otimes_K U_2$ is identified with $U$ under the isomorphism $\phi$. Then the canonical mapping of $g$ into $U$ is identified by (1) with the mapping:

Analogously, if $g_1, \ldots, g_n$ are Lie algebras over $K$ with enveloping algebras $U_1, \ldots, U_n$, the enveloping algebra $U$ of $g_1 \times \cdots \times g_n$ is canonically identified with $U_1 \otimes_K \cdots \otimes_K U_n$ and the canonical mapping of $g_1 \times \cdots \times g_n$ into $U$ is identified with the mapping:

($\sigma_i$ denoting the canonical mapping of $g_i$ into $U_i$).

Let $\sigma''$ be the canonical mapping of $g/\mathfrak{h}$ into W. The commutative diagram:

proves that $\tilde{p}$ is zero on $\sigma(\mathfrak{h})$ and hence on R. Let $\psi$ be the canonical homomorphism of U onto U/R. There exists a homomorphism $\phi$ of U/R

into W such that $\tilde{p} = \phi \circ \psi$. The mapping $\psi \circ \sigma$ of g into U/R is an $\alpha$-mapping and is zero on $\mathfrak{h}$ and hence defines an $\alpha$-mapping $\theta$ of $g/\mathfrak{h}$ into U/R such that $\theta \circ p = \psi \circ \sigma$. Then $\phi \circ \theta \circ p = \phi \circ \psi \circ \sigma = \sigma'' \circ p$. Hence $\phi \circ \theta = \sigma''$. There exists (no. 1, Proposition 1) one and only one homomorphism $\phi'$ of W into U/R mapping 1 to 1 and such that $\theta = \phi' \circ \sigma''$. Then $\phi' \circ \phi \circ \theta = \phi' \circ \sigma'' = \theta$ and $\phi \circ \phi' \circ \sigma'' = \phi \circ \theta = \sigma''$ and hence $\phi' \circ \phi$ and $\phi \circ \phi'$ are the identity mappings of U/R and W respectively. This completes the proof.

U/R is identified with W under the isomorphism $\phi$. Then the canonical mapping $\sigma''$ of $g/\mathfrak{h}$ into W is identified with $\theta$, that is with the mapping of $g/\mathfrak{h}$ into U/R derived from $\sigma$ by taking quotients.

There exists a homomorphism $\phi'$ of U into $V^0$ mapping 1 to 1 and such that $\sigma_0 = \phi' \circ \sigma$. $\phi'$ can be considered as a homomorphism of $U^0$ into V. Then $\sigma_0 = \phi' \circ \phi \circ \sigma_0$ and $\sigma = \phi \circ \phi' \circ \sigma$ and hence $\phi' \circ \phi$ and $\phi \circ \phi'$ are the identity mappings of V and U. Hence the proposition.

V is identified with $U^0$ under the isomorphism $\phi$. Then $\sigma_0$ is identified with $\sigma$.

With this identification, the isomorphism $\theta : x \mapsto -x$ of $g$ onto $g^0$ defines an isomorphism $\tilde{\theta}$ of U onto $V = U^0$. This isomorphism can be considered as an antiautomorphism of U. It is called the principal antiautomorphism of U. If $x_1, \ldots, x_n$ are in $g$, then:

If $t \in T^n$ and $t' \in T^p$, then $\phi(t)\phi(t') = \phi(tt')$ by definition of the multiplication on $G$. Hence $\phi$ is an algebra homomorphism and clearly $\phi(1) = 1$. If $x, y$ are in $g$, then $x \otimes y - y \otimes x \in T^2$ and the canonical image of this element in $U_2$ is equal to that of $[x, y]$ and therefore belongs to $U_1$. Hence $\phi(x \otimes y - y \otimes x) = 0$, which proves the proposition.

Let $S$ be the symmetric algebra of the $K$-module $g$ and $\tau$ the canonical homomorphism of $T$ onto $S$. Proposition 5 proves that there exists a unique homomorphism $\omega$, called canonical, of the algebra $S$ onto the algebra $G$, mapping 1 to 1, such that $\phi = \omega \circ \tau$. We have $\omega(S^n) = \phi(T^n) = G^n$. Let $\tau_n$ be the restriction of $\tau$ to $T^n$, $\omega_n$ the restriction of $\omega$ to $S^n$, $\psi_n$ the canonical mapping of $T^n$ into $U_n$ and $\theta_n$ the canonical mapping of $U_n$ onto $G^n$. The definition of $\omega_n$ proves that the following diagram is commutative:

S is a finitely generated algebra over K and hence a Noetherian ring (Commutative Algebra, Chapter III, § 2, no. 10, Corollary 3 to Theorem 2). Hence G, which is isomorphic to a quotient ring of S, is Noetherian. Hence U is right and left Noetherian (Commutative Algebra, Chapter III, § 2, no. 10, Remark 2).

By induction on m it suffices to consider the case of, for example, two right ideals. The right U-module I₁ is generated by a finite number of elements u₁, ..., uₚ (Proposition 6). Let v₁, ..., vₐ be elements of U whose classes modulo I₂ generate the vector space U/I₂. Then the canonical images in I₁/I₁I₂ of the uᵢvⱼ generate the vector space I₁/I₁I₂, which is therefore finite-dimensional. Hence dim_K(U/I₁I₂) = dim_K(U/I₁) + dim_K(I₁/I₁I₂) < +∞.

Let $(x_\lambda)_{\lambda \in \Lambda}$ be a basis of the K-module g; we give $\Lambda$ a total ordering (Set Theory, Chapter III, § 2, no. 3, Theorem 1). Let P be the polynomial algebra $K[z_\lambda]_{\lambda \in \Lambda}$ in indeterminates $z_\lambda$ in one-to-one correspondence with the $x_\lambda$. For every sequence $M = (\lambda_1, \lambda_2, ..., \lambda_n)$ of elements of $\Lambda$, let $z_M$ denote the monomial $z_{\lambda_1}z_{\lambda_2}...z_{\lambda_n}$ and $x_M$ the tensor $x_{\lambda_1} \otimes x_{\lambda_2} \otimes \cdots \otimes x_{\lambda_n}$.

The $z_M$, for M increasing, form a basis of the K-module P (we make the convention that $\varnothing$ is an increasing sequence and that $z_\varnothing = 1$). Let $P_p$ be the submodule of polynomials of degree $\leq p$. We shall first prove several lemmas. (To abbreviate, we write $\lambda \leq M$ if $\lambda \leq \mu$ for every index $\mu$ of the sequence M.)

Moreover, the restriction of $f_p$ to $g \otimes P_{p-1}$ coincides with $f_{p-1}$.

The last assertion follows from the others since the restriction of $f_p$ to $g \otimes_K P_{p-1}$ satisfies conditions (A_{p-1}), (B_{p-1}) and (C_{p-1}). We shall prove the existence and uniqueness of $f_p$ by induction on $p$. For $p = 0$, condition (A_0) gives $f_0(x_\lambda \otimes 1) = z_\lambda$ and conditions (B_0) and (C_0) are then obviously satisfied. Suppose now that the existence and uniqueness of $f_{p-1}$ are proved. We show that $f_{p-1}$ admits a unique extension $f_p$ to $g \otimes_K P_p$ satisfying conditions (A_p), (B_p) and (C_p).

We must define $f_p(x_\lambda \otimes z_M)$ for an increasing sequence M of $p$ elements.

If $\lambda \leq M$, the value is given by condition (A_p). Otherwise, M can be written uniquely in the form ($\mu, N$), where $\mu < \lambda, \mu \leq N$. Then

by (A_{p-1}), so that the left hand side of (C_p) is $f_p(x_\lambda \otimes z_M)$. Now the right hand side of (C_p) is already defined: for (B_{p-1}) allows us to write: where $w \in P_{p-1}$; hence the right hand side of (C_p) becomes:

Thus $f_p$ is defined uniquely and obviously satisfies conditions (A_p) and (B_p). Condition (C_p) is satisfied if $\mu < \lambda, \mu \leq N$. As $[x_\mu, x_\lambda] = -[x_\lambda, x_\mu]$, condition (C_p) is also satisfied for $\lambda < \mu, \lambda \leq N$. As (C_p) is trivially satisfied for $\lambda = \mu$, (C_p) is therefore satisfied if $\lambda \leq N$ or $\mu \leq N$. If none of these inequalities holds, $N = (\nu, Q)$, where $\nu \leq Q, \nu < \lambda, \nu < \mu$. Writing henceforth to abbreviate $f_p(x \otimes z) = xz$ for $x \in g$ and $z \in P_p$, we have by the induction hypothesis:

Now $z_\mu z_Q$ is of the form $z_\mu z_Q + w$, where $w \in P_{p-2}$. (C_p) can be applied to $x_\lambda(x_\nu(z_\mu z_Q))$, since $\nu \leq Q$ and $\nu < \mu$, and to $x_\lambda(x_\nu w)$ by the induction hypothesis, and hence to $x_\lambda(x_\nu(x_\mu z_Q))$. Hence:

Exchanging $\lambda$ and $\mu$ and subtracting:

and hence by the Jacobi identity

which completes the proof of Lemma 1.

(1) $\sigma(x_{\lambda})z_{M} = z_{\lambda}z_{M}$ for $\lambda \leq M$; (2) $\sigma(x_{\lambda})z_{M} \equiv z_{\lambda}z_{M}$ (mod. $P_{p}$) if $M$ has $p$ elements.

By Lemma 1 there exists a homomorphism $f$ of the K-module $g \otimes_{K} P_{p}$ into $P$ satisfying, for all $p$, conditions (A$_{p}$), (B$_{p}$), (C$_{p}$) (where $f_{p}$ is replaced by $f$). This homomorphism defines a homomorphism $\sigma$ of the K-module $g$ into the K-module $\mathscr{L}_{K}(P)$ and $\sigma$ is an $\alpha$-mapping because of condition (C$_{p}$). Finally, $\sigma$ satisfies properties (1) and (2) of the lemma because of conditions (A$_{p}$) and (B$_{p}$).

We write $t_{n}$ in the form $\sum_{i=1}^{r} x_{M_{i}}$, where the $M_{i}$ are sequences of $n$ elements of $\Lambda$. The mapping $\sigma$ extends to a homomorphism of the algebra $T$ into the algebra $\mathscr{L}_{K}(P)$ (which we shall also denote by $\sigma$), which is zero on $J$. By Lemma 2, $\sigma(t) . 1$ is a polynomial whose terms of highest degree are $\sum_{i=1}^{r} z_{M_{i}}$. As $t \in J$, $\sigma(t) = 0$ and hence $\sum_{i=1}^{r} z_{M_{i}} = 0$ in $P$. Now $P$ is canonically identified with $S$, since $g$ has basis $(x_{\lambda})$. Hence the canonical image of $t_{n}$ in $S$ is zero, that is $t_{n} \in I$.

We can now prove Theorem 1. It is necessary to prove that the canonical homomorphism of $S$ onto $G$ is injective. In other words, if $t \in T^{n}$ and $\psi$ denotes the canonical homomorphism of $T$ onto $U$, it is necessary to show that the condition $\psi(t) \in U_{n-1}$ implies $t \in I$. Now $\psi(t) \in U_{n-1}$ means that there exists a tensor $t' \in T_{n-1}$ such that $t - t' \in J$. The tensor $t - t'$ admits $t$ as homogeneous component of order $n$ and hence $t \in I$ by Lemma 3.