Algebras

Let R be any ring. An algebra over R (or R-algebra) is a module $\modmap A \from R \to R$ together with module morphisms

$$\map \mu \from A \xR A \to A\quad, \quad \map \eta \from R \to A$$

such that

$$\index{Associativity} \hbox to\hsize{\hbox{(Associativity)}\... ... \mu\;} & A \xR A \rto^-{\mu\;} & A \enddiagram\qquad\hfill}$$

$$\index{Identity} \hbox to\hsize{\hbox{(Identity)}\hfill \spr... ...{\omit} & A \xR A \rto^-{\mu\;} & A\;\;. \enddiagram\hfill}$$

Notice that A becomes a ring with multiplication $a\,b = \mu (a \xO b)$ and identity $1 = \eta (1)$.

For R-algebras $\mmap A\,, B \from R \to R$ an algebra morphism $\map f \from A \to B$ is a module morphism satisfying

$$\spreaddiagramcolumns{1pc}\spreaddiagramrows{.5pc}\diagram A ... ...} \drto^-{\eta} & \\ A \rrto^-{f\;} & & B\rlap{~~.}\enddiagram$$

We write $\Alg_R(A\,,B)$ for the set of algebra morphisms from A to B.

Example 7.1  

For any module $\mmap M\from R\to S$, the endomorphism algebras, over S and R respectively, are given by

$$\begin{eqnarray*} \End_R(M)&=& \hbox{\modmap \Hom_R(M,M)\;\from S\; \to {\;S}} \\ \End^S(M)&=& \hbox{\modmap \Hom^S(M,M)\;\from R\; \to {\;R}} \end{eqnarray*}$$

In each case the multiplication is given by composition.

A module morphism

$$\relax\twomodmap{\hat{\mu}}\from A\to\to\End^S(M)\;\from R\;\to{\;R}$$

corresponds to a module morphism

$$\relax\twomodmap{\mu}\from A\xR M \to\to M \;\from{R\;}\to{\;S}\;\;.$$

To say that $\hat{\mu}$ is an algebra morphism is precisely to say that $\mu$ is a scalar multiplication enriching M with the structure of left A-module.

Example 7.2  

For any module $\mmap M \from R \to R$, write

$$\relax M^{\xo n}_{}\;=\; M\xR\cdots\;\xR M\rlap{$\qquad${\em ($n$\ terms) .}}$$

The tensor algebra on M is defined by the ``geometric series"

$$\relax T(M)\;\;=\;\;\sum_{n=0}^{\infty}\;M^{\xo n}_{}$$

with multiplication $\relax \map\mu\from T(M)\xR T(M)\to{T(M)}$ induced by the canonical isomorphisms

$$\diagram \relax M^{\xo p}_{}\xR M^{\xo q}_{}\rto^-{\isom\;}& M^{\xo (p+q)}_{}\enddiagram$$

and unit $\relax \map\eta\from R\to{T(M)}$ equal to the injection

$$\relax \map\inj_0 \from R \;=\; M^{\xo 0}\;\to {\;\sum _{n=0}^{\infty} M^{\xo n}_{}}\;.$$

Composition with the injection $\relax \map \inj_1^{}\from M\to{T(M)}$ gives a bijection

$$\Alg_R(T(M)\,,A)\;\;\isom\;\;\Hom_R^R(M,A)$$

for any algebra A. The inverse takes $\relax \map f\from M\to A$ to $\relax \map g\from T(M)\to A$ given by $\relax g(m_1^{} \xO\,\cdots\,\xO m_r^{}) = f(m_1^{})\,\cdots\, f(m_r^{})\,$. In particular, if we take M = A and $\relax f = 1_{\!A}^{}$, we obtain an algebra morphism

$$\map\mu\from T(A)\to{S}\quad\mbox{with}\quad \mu(a_1^{}\xO\,\cdots\,\xO a_r^{})\;=\; a_1^{}\,\cdots\,a_r^{}\; .$$

Example 7.3  

Let G be any monoid. There is an R-algebra R(G) which is just the free module $\Free_R^R(G)$ on the underlying set of G together with the multiplication $\mu$ which extends that of G in the sense that

$$\spreaddiagramcolumns{2pc}\spreaddiagramrows{1pc}\diagram G \... ...^R(G)$}\dto^-{\mu} \\ G \, \rinto & \Free_R^R(G)\;.\enddiagram$$

This R(G) is called the monoid R-algebra of G; or when G is a group, the group R-algebra of G. Each monoid morphism $\map \from G \to A$ into the multiplicative monoid of A extends uniquely to an R-algebra morphism $\map \from R(G) \to A$.

A representation of G on M is an R(G)-module. Scalar multiplication $\map \from R(G) \xR M \to M$ can be viewed as a monoid morphism

$$\map\from G\; \to {\;\End_R(M)} \;.$$

The subset of M given by $\{ gm-m \;\vert\; g \in G\,,\,m \in M \}$ generates a submodule $(gm-m \>\vert\> g \in G\,,\, m \in M)$ and we write M/G for the quotient module $M/(gm-m \;\vert\; g \in G\,,\,m \in M)$.

An ideal in an algebra A is a submodule I such that $\relax a\,x\,b \in I$ for all $\relax x \in I$ and $\relax a\,,b \in A$. There is a unique structure of algebra on the quotient module A/I for which the canonical $\relax \map\rho\from A\to{A/I}$ is an algebra morphism. The kernel of any algebra morphism $\map f \from A \to B$ is an ideal in A.

If X is a subset of an algebra A, we write (X) for the smallest ideal of A containing X. This should not cause confusion with the module notation; the ideal (X) is precisely the submodule (A X A) generated by the subset $A\,X\,A = \{ a\,x\,b\;\vert\;a\,,b \in A\,,\, x \in X \}$ of A. Given any algebra morphism $\map g \from A \to B$ satisfying g(x) = 0 for all $x \in X$, then an algebra morphism $\map f \from A/(X) \to B$ is uniquely determined via the equation $f \circ \rho = g$.

Now suppose that R is a commutative ring. Then left R-modules are ``the same thing" as right R-modules. Moreover, each left R-module M can be naturally regarded as a module $\modmap M \from R \to R$ by defining

$$r\,m\,s \;=\; (r\,s)\,m \qquad\hbox{for all}\quad r\,,s \in R \quad\hbox{and}\quad m \in M\;\,.$$

In dealing with modules over a commutative ring, we happily regard left modules as two-sided via this process. Thus for R-modules $M_1^{},\,\ldots\,, M_n^{}$ we have a tensor product R-module

$$M_1^{} \xR \;\cdots\; \xR M_n^{} \;.$$

Furthermore every permutation $\xi$ on the set $\{1\,,\,\ldots\,,\,n \}$ induces a canonical module isomorphism

$$\spreaddiagramcolumns{1pc}\spreaddiagramrows{-2pc}\diagram\re... ...d m_{\xi(1)}^{} \xO \;\cdots\; \xO m_{\xi(n)}^{}\;. \enddiagram$$

Given an algebra A over R with multiplication $\mu$ and unit $\relax\eta$, we obtain an opposite algebra $\relax A^\op$ on the same module A, with multiplication

$$\spreaddiagramcolumns{.5pc}\diagram \mu^\op: A \xR A \rto^-{\sigma\;} & A \xR A \rto^-{\mu\;} & A \enddiagram$$

and with the same unit $\map \eta \from R \to A$. Call A commutative when $\relax A^\op = A$ as algebras. It follows that the composite

$$\spreaddiagramcolumns{.5pc}\diagram A \xR \;\cdots\; \xR A \r... ...}^{}\;} & A \xR \;\cdots\; \xR A \rto^-{\mu\;} & A \enddiagram$$

is independent of the permutation $\xi$.

Example 7.4  

For any set X, the set R^X of all functions from X into the commutative ring R becomes a commutative R-algebra after defining addition, scalar multiplication and multiplication as acting pointwise. The unit $\map \eta \from R \to {R^X}$ is given by $\eta (r)(x) = r$ for all $r \in R$ and $x \in X$.

Example 7.5  

Let M be any module over the commutative ring R. There is a natural representation of the symmetric group $\Symm_n$ on $M^{\xo n}$ given by $\map\sigma_{\!\slot}\from\Symm_n^{}\to{\End_R(M^{\xo n})}$; that is $\xi \cdot (m_1^{} \xO \cdots \xO m_n^{}) = m_{\xi(1)}^{} \xO \cdots \xO m_{\xi(n)}^{}$.

The symmetric R-algebra on M is given by the ``exponential series"

$$\Symm(M) \;=\; \sum^{\infty}_{n = 0} M^{\xo n} / \Symm_n \;.$$

Another way of constructing this is as follows. For any R-algebra A we can form a commutative R-algebra by taking the quotient of A by the ideal $(a\,b-b\,a \;\vert\; a\,,b \in A\,)$. Applying this construction to the tensor algebra T(M) gives $\Symm(M)$.

For every commutative R-algebra A, we have that

$$\Alg_R(\Symm(M)\mkern2mu, A) \;\;\isom\;\; \Hom_R(M\mkern2mu,A) \;.$$

In particular, corresponding to the identity map $\map 1_{\!A}^{} \from A \to A$ there is an R-algebra morphism $\map \mu \from \Symm(A) \to A$.

The following diagram of ``forgetful'' and ``free'' constructions summarizes some of the above.

$$\spreaddiagramcolumns{1pc}\spreaddiagramrows{1pc}\diagram & \... ...5>\qquad\Text<5.1pc>{forget\\ module~structure}}{'} \enddiagram$$

Skew commutativity a b + b a = 0 for an R-algebra is too strong as a requirement for all $\relax a\,,b \in A$. For example taking b = 1, it would give (1 + 1)a = 0. Hence if R is a field of characteristic other than 2 (meaning $\relax 1 + 1 \ne 0$ in R), we would get a = 0, and so $\relax A=\{ 0\}$.

An R-algebra A is said to be skew commutative when for all $a \in A$ either a² = 0 or $a \in \eta(R)$. Then, provided none of a ,b and a+b are in the image of $\map \eta \from R \to A$, we have

a b + b a = (a+b)² - a² - b² = 0 .

Example 7.6  

For any R-algebra A we can form the quotient by the ideal $(a^2 \,\vert\, a \notin \eta(R))$ to obtain a skew commutative algebra. If we do this to the tensor algebra T(M) we obtain the exterior algebra $\Lambda (M)$. Alternatively, let $\Lambda_n (M)$ be the quotient module of $M^{\xo n}$ by the submodule generated by the elements $m_1^{} \xO \cdots \xO m_n^{}$ with m_i = m_j for some $i \neq j$ (this submodule is { 0 } when n = 0 or 1); then

$$\Lambda(M) \;=\; \sum^{\infty}_{n=0} \Lambda_n(M) \;.$$

We write $m_1^{}\wedge\cdots\wedge m_n^{}$ for the image of $m_1^{} \xO \cdots \xO m_n^{}$ in $\Lambda (M)$. For all $x\,,y\,,z \in M$ we have

$$\begin{eqnarray*} x \wedge x &=& 0 \qquad\hbox{\em therefore }\quad x \wedge ... ...e x \\ (rx + sy) \wedge z &=& r(x \wedge z) + s(y \wedge z) \;. \end{eqnarray*}$$

If $M = \Free_R \{x_1^{},\ldots, x_k ^{}\}$ is a free module on a k-element set then $\Lambda_n (M)$ is a free module on a $\relax k\choose n$-element set; so $\Lambda (M)$ is a free module on a set with $\relax 2^k$-elements. In particular $\relax \Lambda_k (M)$ is free on the singleton set $\relax\{x_1^{} \wedge \cdots \wedge x_k^{} \}$, so if

$$y_i^{} \;=\; \sum^{k}_{j=1}\; r_{ij}^{}\, x_j^{}$$

then $\relax y_1^{} \wedge \cdots \wedge y_k^{}$ must be a unique scalar multiple

$$y_1^{} \wedge \cdots \wedge y_k^{} \;=\; \det (r_{ij}^{})\; x_1^{} \wedge \cdots \wedge x_k^{}$$

of $\relax x_1^{} \wedge \cdots \wedge x_k^{}$. This can be taken as a definition of the determinant of $\relax (r_{ij}^{}) \in \Mat(k\,,R)$.

If A is a skew-commutative algebra then we have a bijection

$$\Alg_R(\Lambda (M)\,,A) \;\;\isom\;\; \Hom_R(M,A)\;.$$

An R- Lie algebra is an R-module L together with a module morphism $\map\beta\from L\xR L\to{L\,}$ satisfying the conditions

$$\begin{eqnarray*} \index{Jacobi identity} && \beta (x\,,x) \;=\; 0 \\ \hbox{(Ja... ...ta (z\,,x)\,,y\big) + \beta\big(\beta (y\,,z)\,,x\big)\; =\; 0 \end{eqnarray*}$$

Call such a $\beta$ a Lie bracket on the module L.

Example 7.7  

For any R-algebra A the commutator [ a ,b ] = a b - b a defines a Lie bracket on the underlying R-module of A:

$$\begin{eqnarray*} \lefteqn{[\,[\,a\,, b\,]\,,c\,] + [\,[\,c\,,a\,]\,,b\,] + [\,[... ...ad + \,(b\,c\,a - c\,b\,a) - (a\,b\,c - a\,c\,b) \;\; =\;\; 0\;. \end{eqnarray*}$$

So A becomes a Lie algebra, denoted by A_L. It turns out (at least when R is a field) that every Lie algebra is a submodule, closed under commutator, of such an example.

Example 7.8  

Let A be an R-algebra and $\modmap M \from A \to A$ a module. Then a  derivation $\map D \from A \to M$ is an R-module morphism satisfying

$$\hbox to\hsize{{\rm (Leibniz rule)}\hfill $D(a\,b) \;\;=\;\; D(a)\,b \;+\; a\,D(b)\,$\,. \qquad\qquad\hfill}$$

Notice that a = b = 1 gives D(1) = 2 D(1), so D(1) = 0. Let $\Der_R(A\,,M)$ denote the submodule of $\Hom_R(A,M)$ consisting of the derivations. We write $\Der_R(A)$ for $\relax \Der_R(A\,, A)$. It is easy to check that $\relax \Der_R(A)$ is closed under commutator in the algebra $\relax \End_R(A)$; that is, if $\relax \map D_1^{}\,, D_2^{} \from A \to A$ are derivations then so is $\relax [\,D_1^{}\,,D_2^{}\,] = D_1^{} \circ D_2^{} - D_2^{} \circ D_1^{}$.

Example 7.9  

The tangent space at the identity of each Lie group is a Lie algebra. The pioneering work of Sophus Lie and Eli Cartan showed how much information about the Lie group is obtainable from the Lie algebra (especially in the compact case).

The Lie groups $\relax \GL(n\,,\R)$ and $\SL(n\,,\R)$ and $\relax \OL(n\,,\R)$ consist of those matrices $\relax \xX \in \Mat\!(n\,,\R)$ for which respectively $\relax \xX$ is invertible, $\relax \det \xX = 1$ and $\relax \mbf\xX\,\xX^t = 1$. They have associated Lie algebras

$$\begin{eqnarray*} \gl(n\,,\R) & = & \Mat\!(n\,,\R) \\ \sl (n\,,\R) & = & \big\{... ...g\{ \xX \in \gl(n\,,\R) \;\;\vert\;\; \xX^t\,=\,-\xX\,\big\} \;. \end{eqnarray*}$$

(We shall not stop to prove this here.) The Lie bracket is $\relax [\,\xX\,,\yY\,] = \xX\,\yY - \yY\,\xX$ in each case. As an exercise the reader should check that $\relax \sl (n\,,\R)$ and $\relax \ol(n\,,\R)$ are closed under commutator.

Suppose L and L' are R-Lie algebras and $\map f \from L \to {L'}$ is an R-module morphism. Then f is a Lie algebra morphism when it satisfies

$$f\big(\,\beta(x\,,y)\big) \;\;=\;\; \beta\big(f(x)\,,f(y)\big) \;.$$

Write $\Lie_R(L\,,L')$ for the set of Lie algebra morphisms $\map f \from L \to {L'}$.

We saw in the 1st Example that each R-algebra A gives rise to an R-Lie algebra A_L using the commutator. We shall describe an ``adjoint" for this process: for each R-Lie algebra L we obtain an R-algebra $\scrU(L)$, called the universal enveloping algebra of L, such that there is a natural bijection

$$\hbox to\hsize{\hfill $\Alg_R\big(\scrU(L)\,,A\big) \;\;\isom\;\; \Lie\!(L\,,A_L^{})\;. \qquad$\hfill\llap{ $(*)$}}$$

For this we use the tensor algebra T(L) on the underlying R-module of L, and take the quotient by the appropriate ideal:

$$\scrU(L)\;=\;T(L)/\big(x\xO y-y\xO x-\beta(x\,,y)\;\vert\; x\,,y\in L\,\big) \;.$$

We have a Lie algebra morphism i as in:

$$\spreaddiagramcolumns{1.5pc}\diagram L \drto_-{\inj_1^{}\;}\r... ...;} & & \;\scrU(L)_L^{} \\ & T(L) \urto_-{\;\rho} & \enddiagram$$

and it is composition with i that induces the bijection $\relax(*)$.

The direct sum $L_1^{} \oplus L_2^{}$ of Lie algebras L₁ ,L₂ is their direct sum as modules together with the Lie bracket

$$\beta\big((x_1^{}\,,x_2^{})\,,(y_1^{}\,,y_2^{})\big) \;\;=\... ...g(\beta (x_1^{}\,,y_1^{})\,,\,\beta (x_2^{}\,,y_2^{})\big) \;.$$

Proposition 7.10  

There is an algebra isomorphism

$$\scrU(L_1^{}\oplus L_2^{})\;\;\isom\;\;\scrU(L_1^{})\,\xR\,\scrU(L_2^{})$$

whose composite with $\relax\map i \from L_1^{} \oplus L_2^{} \to {\scrU(L_1^{} \oplus L_2^{})}$ takes the pair $\relax(x_1^{}\,,x_2^{})$ to $\relax x_1^{}\xO 1 + 1\xO x_2^{}$.

A deeper result which we shall not prove here is:

Proposition 7.11   (Poincaré-Birkhoff-Witt) 

If the R-Lie algebra L is free as an R-module then $\map i \from L \to {\scrU(L)_L}$ is injective.

A Lie algebra L is called commutative when $\beta(x\,,y) = 0$ for all $x\,,y \in L$. So an algebra A is commutative iff A_L is commutative.

Notice that, for any module M, we can make M into a commutative Lie algebra. Then the universal enveloping algebra of M is precisely the same as the symmetric algebra of M, that is $\scrU(M) = \Symm(M)$. In particular, we have (see first proposition):

$$\Symm(M \oplus M') \;\;\isom\;\; \Symm(M)\,\xR\,\Symm(M') \;.$$