The Symmetric Algebra

Introduction

This article assembles the symmetric powers of a module into a single graded algebra. The result, the symmetric algebra $\operatorname{Sym}(M)$, is the free commutative algebra on the module $M$. It is the commutative counterpart of the tensor algebra, and when $M$ is free it is a polynomial algebra. The treatment is introductory and purely mathematical.

The base structure is a commutative ring $R$ with identity $1 \neq 0$. Tensor products and tensor powers are taken from Modules (§13), as in Symmetric Powers. That article constructed the symmetric power $\operatorname{Sym}^n M$ and established its universal property, its monomial basis for free $M$, and its behaviour under direct sums. This article takes those results as its starting point.

The tensor algebra $T(M)$ and its general quotients belong to category 05; the systematic treatment of $T(M)$ is in the companion article Tensor Powers and the Free Algebra. Here the tensor algebra appears only in the one role that defines the symmetric algebra: it is the associative algebra on $M$, and the symmetric algebra is its quotient by the two-sided ideal generated by $x \otimes y - y \otimes x$. The anti-symmetric analogue, the exterior algebra, belongs to category 07 and is not treated here beyond a remark.

The Graded Algebra

Definition

Let $M$ be an $R$-module. As a module, the symmetric algebra is the direct sum of the symmetric powers:

$$ \operatorname{Sym}(M) = \bigoplus_{n \geq 0} \operatorname{Sym}^n M, $$

with $\operatorname{Sym}^0 M = R$ and $\operatorname{Sym}^1 M = M$. An element of $\operatorname{Sym}(M)$ is a finite sum $u_0 + u_1 + \cdots + u_N$ with $u_n \in \operatorname{Sym}^n M$; the summand $u_n$ is the degree-$n$ component of the element, and the elements of $\operatorname{Sym}^n M$ are homogeneous of degree $n$. The augmentation map is the projection

$$ \operatorname{Sym}(M) \longrightarrow \operatorname{Sym}^0 M = R $$

onto the degree-zero part; its kernel $\operatorname{Sym}^{+}(M) = \bigoplus_{n \geq 1} \operatorname{Sym}^n M$ is the augmentation ideal.

The Graded Product

The product is built degree by degree. For each pair $m, n$ define a map

$$ \operatorname{Sym}^m M \times \operatorname{Sym}^n M \longrightarrow \operatorname{Sym}^{m+n} M $$

on generating elements by

$$ (x_1 \cdots x_m,\ y_1 \cdots y_n) \longmapsto x_1 \cdots x_m y_1 \cdots y_n , $$

where the right-hand side is the image under $\pi_{m+n} : M^{m+n} \to \operatorname{Sym}^{m+n} M$. This is well defined: keeping the first argument fixed and letting the second vary, the right-hand side is symmetric multilinear in $y_1, \ldots, y_n$, so by the universal property of $\operatorname{Sym}^n M$ it induces a map $\operatorname{Sym}^m M \times \operatorname{Sym}^n M \to \operatorname{Sym}^{m+n} M$ linear in the second variable; varying the first argument, the same argument applied to $\operatorname{Sym}^m M$ gives linearity in the first variable. Hence it is bilinear, and therefore induces a linear map

$$ \mu_{m,n} : \operatorname{Sym}^m M \otimes_R \operatorname{Sym}^n M \longrightarrow \operatorname{Sym}^{m+n} M. $$

Summing over all $m,n$ and extending bilinearly gives the multiplication

$$ \mu : \operatorname{Sym}(M) \otimes_R \operatorname{Sym}(M) \longrightarrow \operatorname{Sym}(M). $$

We write $uv = \mu(u \otimes v)$. The element $1 \in \operatorname{Sym}^0 M = R$ acts as a two-sided identity, since the product $\operatorname{Sym}^0 M \times \operatorname{Sym}^m M \to \operatorname{Sym}^m M$ sends the empty tuple paired with $(y_1, \ldots, y_m)$ to $y_1 \cdots y_m$, that is, it is the canonical identification $R \otimes_R \operatorname{Sym}^m M \cong \operatorname{Sym}^m M$.

Associativity and Commutativity

Theorem. The product on $\operatorname{Sym}(M)$ is associative and commutative, and unital with unit $1 \in \operatorname{Sym}^0 M$.

Proof. For homogeneous elements $u = x_1 \cdots x_m$, $v = y_1 \cdots y_n$, $w = z_1 \cdots z_k$, associativity reads

$$ (u v) w = x_1 \cdots x_m y_1 \cdots y_n z_1 \cdots z_k = u (v w), $$

which is immediate from the definition, since both sides are the image of the concatenated tuple under $\pi_{m+n+k}$. On general elements both sides are linear. Commutativity likewise reduces to

$$ (x_1 \cdots x_m)(y_1 \cdots y_n) = x_1 \cdots x_m y_1 \cdots y_n = y_1 \cdots y_n x_1 \cdots x_m = (y_1 \cdots y_n)(x_1 \cdots x_m), $$

where the middle equality is the symmetry of $\pi_{m+n}$: the tuple with the $y$'s moved to the front is a permutation of the original, so the two have the same image. Unitality is the definition of $\pi_0$.

Remark. The product is not merely commutative but graded-commutative with no sign: for homogeneous $u$ of degree $m$ and $v$ of degree $n$, $uv = vu$ with the sign $+1$, because the grading is by $\mathbb{Z}$ concentrated in non-negative degrees and the symmetry group acts without signs. The sign appears in the exterior algebra, where $uv = (-1)^{mn} vu$, and that is category 07.

The Universal Property

Theorem. Let $A$ be a commutative unital $R$-algebra and let $f : M \to A$ be an $R$-linear map. There is a unique $R$-algebra homomorphism $F : \operatorname{Sym}(M) \to A$ with $F|_{\operatorname{Sym}^1 M} = f$; equivalently, $F(x_1 \cdots x_n) = f(x_1)\cdots f(x_n)$.

Proof. The map $M^n \to A$, $(x_1, \ldots, x_n) \mapsto f(x_1) \cdots f(x_n)$, is symmetric multilinear, because $A$ is commutative (so the product does not depend on the order) and bilinear. By the universal property of $\operatorname{Sym}^n M$ it induces a linear map $F_n : \operatorname{Sym}^n M \to A$ with $F_n(x_1 \cdots x_n) = f(x_1)\cdots f(x_n)$. Summing the $F_n$ gives a linear map $F : \operatorname{Sym}(M) \to A$, and it is multiplicative on homogeneous generating elements:

$$ F\bigl((x_1\cdots x_m)(y_1\cdots y_n)\bigr) = F(x_1\cdots x_m y_1 \cdots y_n) = \prod_i f(x_i)\prod_j f(y_j) = F(x_1\cdots x_m) F(y_1\cdots y_n). $$

Since these elements span, $F$ is an algebra homomorphism; it extends $f$ because $F_1 = f$ under the identification $\operatorname{Sym}^1 M = M$. Uniqueness holds because $M = \operatorname{Sym}^1 M$ generates $\operatorname{Sym}(M)$ as an algebra, so any two homomorphisms agreeing on $M$ agree everywhere.

This is the precise sense in which the symmetric algebra is free: it is the free commutative unital algebra on the module $M$. In the language of adjunctions, the functor

$$ \operatorname{Sym} : \mathsf{Mod}_R \longrightarrow \mathsf{CAlg}_R , \qquad M \longmapsto \operatorname{Sym}(M), $$

from $R$-modules to commutative unital $R$-algebras is left adjoint to the forgetful functor $U : \mathsf{CAlg}_R \to \mathsf{Mod}_R$:

$$ \operatorname{Hom}_{\mathsf{CAlg}_R}\bigl(\operatorname{Sym}(M), A\bigr) \cong \operatorname{Hom}_{\mathsf{Mod}_R}\bigl(M, U(A)\bigr), $$

naturally in $M$ and $A$. The natural transformation $M \to U(\operatorname{Sym}(M))$ is the inclusion of the degree-one part.

The Symmetric Algebra as a Quotient of the Tensor Algebra

The Tensor Algebra

The tensor algebra on $M$ is the direct sum

$$ T(M) = \bigoplus_{n \geq 0} M^{\otimes n}, \qquad M^{\otimes 0} = R, $$

with the product defined on elementary tensors by concatenation

$$ (x_1 \otimes \cdots \otimes x_m)(y_1 \otimes \cdots \otimes y_n) = x_1 \otimes \cdots \otimes x_m \otimes y_1 \otimes \cdots \otimes y_n $$

and extended bilinearly. This makes $T(M)$ an associative unital $R$-algebra, generally non-commutative. It is characterised by a universal property dual to the one above: every $R$-linear map $M \to A$ into an associative unital $R$-algebra $A$ extends uniquely to an algebra homomorphism $T(M) \to A$. We use only this property and the explicit description of the product; the systematic development is in Tensor Powers and the Free Algebra.

The Commutator Ideal

Let $I \subseteq T(M)$ be the two-sided ideal generated by all the elements

$$ x \otimes y - y \otimes x, \qquad x, y \in M. $$

Equivalently, $I$ is the smallest two-sided ideal containing the image of the linear map $M \otimes M \to T(M)$, $x \otimes y \mapsto x \otimes y - y \otimes x$. Because the generators are homogeneous of degree $2$, the ideal $I$ is graded: $I = \bigoplus_n I_n$ with $I_n = I \cap M^{\otimes n}$, and $I_0 = I_1 = 0$.

The Quotient

Theorem. There is a natural isomorphism of graded algebras

$$ \operatorname{Sym}(M) \cong T(M) / I . $$

Proof. The quotient $T(M)/I$ is generated as an algebra by the image of $M = T^1(M)$, and it is commutative: its generators commute because the class of $x \otimes y - y \otimes x$ is zero. Hence by the universal property of $\operatorname{Sym}(M)$ the identity map $M \to T(M)/I$ extends to a unique algebra homomorphism $\Phi : \operatorname{Sym}(M) \to T(M)/I$ with $\Phi(x_1 \cdots x_n) = x_1 \otimes \cdots \otimes x_n + I$.

In the other direction, the identity $M \to \operatorname{Sym}(M)$ extends by the universal property of $T(M)$ to an algebra homomorphism $T(M) \to \operatorname{Sym}(M)$ sending $x_1 \otimes \cdots \otimes x_n$ to $x_1 \cdots x_n$. This map kills each generator $x \otimes y - y \otimes x$, since $\operatorname{Sym}(M)$ is commutative, so it descends to $\Psi : T(M)/I \to \operatorname{Sym}(M)$. On the generators $\Phi$ and $\Psi$ are mutually inverse, hence they are inverse isomorphisms.

This is the quotient referred to in the category description: the symmetric algebra is the tensor algebra modulo the two-sided ideal generated by $x \otimes y - y \otimes x$, and by nothing else. The exterior algebra is the quotient by the ideal generated instead by $x \otimes x$, or equivalently, when $2$ is invertible, by $x \otimes y + y \otimes x$; that is category 07.

Polynomial Algebras

Free Modules and Monomials

Let $M = R^d$ be free with basis $e_1, \ldots, e_d$. By Symmetric Powers, each $\operatorname{Sym}^n M$ is free with basis the monomials $e^a = e_1^{a_1} \cdots e_d^{a_d}$, $|a| = n$. Summing over $n$ gives the following statement.

Theorem. The symmetric algebra on a free module of rank $d$ is the polynomial algebra:

$$ \operatorname{Sym}(R^d) \cong R[x_1, \ldots, x_d], $$

as graded $R$-algebras, with $x_i \leftrightarrow e_i$. Consequently a general polynomial algebra is a symmetric algebra, and the free commutative algebra on a set of $d$ generators is $\operatorname{Sym}(R^d)$.

Proof. The universal property of $\operatorname{Sym}(R^d)$ applied to the linear map $e_i \mapsto x_i$ gives a homomorphism $\operatorname{Sym}(R^d) \to R[x_1,\ldots,x_d]$, and the universal property of the polynomial algebra gives the inverse; both are isomorphisms on the monomial bases, hence inverse isomorphisms.

The identification organises the classical vocabulary of polynomials in the language of the symmetric algebra:

Example. For $d = 1$, $\operatorname{Sym}(R) \cong R[x]$, the polynomial algebra in one variable, and $\operatorname{Sym}^n R \cong R$ for all $n$, spanned by $x^n$. For $d = 2$, $\operatorname{Sym}(R^2) \cong R[x,y]$, with $\operatorname{Sym}^2(R^2)$ spanned by $x^2, xy, y^2$.

Evaluation and Substitution

The universal property is exactly the formal statement of evaluation. An $R$-algebra homomorphism $\operatorname{Sym}(R^d) \to A$ is determined by the images of $x_1, \ldots, x_d$, which are arbitrary elements $a_1, \ldots, a_d \in A$; the corresponding homomorphism is

$$ F \longmapsto F(a_1, \ldots, a_d), $$

the evaluation map. Thus

$$ \operatorname{Hom}_{\mathsf{CAlg}_R}\bigl(R[x_1, \ldots, x_d], A\bigr) \cong A^d, $$

the functor of points of affine $d$-space. This is the cleanest form of the statement that a polynomial is determined by its values and that substituting is the only operation on polynomials compatible with the algebra structure. Substitution of one polynomial tuple into another is composition of homomorphisms.

Functoriality

The Functor $\operatorname{Sym}$

Let $u : M \to N$ be $R$-linear. The map $\operatorname{Sym}^n u$ of Symmetric Powers is linear for each $n$ and compatible with the products, since

$$ \operatorname{Sym}^{m+n}u\bigl(x_1\cdots x_m y_1\cdots y_n\bigr) = u(x_1)\cdots u(x_m) u(y_1)\cdots u(y_n) = \operatorname{Sym}^m u(x_1\cdots x_m)\operatorname{Sym}^n u(y_1\cdots y_n), $$

so the $\operatorname{Sym}^n u$ assemble into an algebra homomorphism

$$ \operatorname{Sym}(u) : \operatorname{Sym}(M) \longrightarrow \operatorname{Sym}(N). $$

The construction is functorial: $\operatorname{Sym}(\operatorname{id}_M) = \operatorname{id}_{\operatorname{Sym}(M)}$ and $\operatorname{Sym}(v \circ u) = \operatorname{Sym}(v) \circ \operatorname{Sym}(u)$. On free modules, $\operatorname{Sym}(u)$ is the substitution homomorphism determined by $e_i \mapsto u(e_i)$.

The Symmetric Algebra of a Direct Sum

Theorem. For all $R$-modules $M, N$ there is a natural isomorphism of graded algebras

$$ \operatorname{Sym}(M \oplus N) \cong \operatorname{Sym}(M) \otimes_R \operatorname{Sym}(N). $$

Proof. The inclusions $M \hookrightarrow M \oplus N \hookrightarrow \operatorname{Sym}(M \oplus N)$ and $N \hookrightarrow M \oplus N \hookrightarrow \operatorname{Sym}(M \oplus N)$ give, by the universal property, algebra homomorphisms $\operatorname{Sym}(M) \to \operatorname{Sym}(M \oplus N)$ and $\operatorname{Sym}(N) \to \operatorname{Sym}(M \oplus N)$ whose images commute (all elements of $\operatorname{Sym}(M \oplus N)$ commute). By the universal property of the tensor product of algebras there is therefore a homomorphism

$$ \Phi : \operatorname{Sym}(M) \otimes_R \operatorname{Sym}(N) \longrightarrow \operatorname{Sym}(M \oplus N). $$

On the other side, the linear map $M \oplus N \to \operatorname{Sym}(M) \otimes_R \operatorname{Sym}(N)$ sending $(x, y) \mapsto x \otimes 1 + 1 \otimes y$ induces by the universal property a homomorphism $\Psi$ in the reverse direction. On generators both composites are the identity, and the generators span, so $\Phi$ and $\Psi$ are inverse isomorphisms. Alternatively, the isomorphism is the degree-wise form $\operatorname{Sym}^n(M \oplus N) \cong \bigoplus_k \operatorname{Sym}^k M \otimes \operatorname{Sym}^{n-k} N$ established in Symmetric Powers.

Corollary. $\operatorname{Sym}(R^d) \cong R[x_1, \ldots, x_d] \cong R[x_1] \otimes_R \cdots \otimes_R R[x_d]$, recovering the familiar fact that polynomials in several variables are obtained by tensoring polynomials in one variable.

The Coproduct Reading

The category $\mathsf{CAlg}_R$ has coproducts, and the coproduct of $A$ and $B$ is $A \otimes_R B$ with the multiplication $(a \otimes b)(a' \otimes b') = aa' \otimes bb'$. The direct-sum theorem above is a special case of the compatibility of a left adjoint with coproducts: $\operatorname{Sym}$ carries the direct sum in $\mathsf{Mod}_R$ (the coproduct there) to the tensor product in $\mathsf{CAlg}_R$ (the coproduct there). The companion articledevelops this.

Gradings and the Augmentation Ideal

The Grading

The decomposition $\operatorname{Sym}(M) = \bigoplus_{n \geq 0} \operatorname{Sym}^n M$ is a grading: $\operatorname{Sym}^m M \cdot \operatorname{Sym}^n M \subseteq \operatorname{Sym}^{m+n} M$. Every element is a sum of homogeneous components, and a homogeneous element of degree $m$ multiplied by one of degree $n$ has degree $m+n$. Consequently each quotient $\operatorname{Sym}(M)/\operatorname{Sym}^{+}(M)^k$ is a graded algebra, and the graded pieces give a filtration

$$ \operatorname{Sym}(M) \supseteq \operatorname{Sym}^{+}(M) \supseteq \operatorname{Sym}^{+}(M)^2 \supseteq \cdots . $$

This descending chain is the $\operatorname{Sym}^{+}(M)$-adic filtration, and the completion it defines is not covered here. The associated graded algebra is again the symmetric algebra: by the proposition of the next subsection,

$$ \operatorname{Sym}^{+}(M)^n \big/ \operatorname{Sym}^{+}(M)^{n+1} \cong \operatorname{Sym}^n M, $$

so $\operatorname{gr}_{\operatorname{Sym}^{+}(M)}\operatorname{Sym}(M) \cong \bigoplus_n \operatorname{Sym}^n M = \operatorname{Sym}(M)$. The completion itself is the direct product of the graded pieces,

$$ \widehat{\operatorname{Sym}(M)} = \varprojlim_n \operatorname{Sym}(M)/\operatorname{Sym}^{+}(M)^n \cong \prod_{n \geq 0} \operatorname{Sym}^n M, $$

because the truncation maps drop one degree at each step and a coherent sequence is an arbitrary family of homogeneous components; for $M = R^d$ free this is the algebra $R[[x_1, \ldots, x_d]]$ of formal power series.

The Augmentation Ideal and Its Powers

The augmentation ideal $\operatorname{Sym}^{+}(M)$ is generated as an ideal by the degree-one part $M$, and it consists exactly of the elements with zero constant term. Its powers are determined by the grading.

Proposition. For every $n \geq 0$,

$$ \operatorname{Sym}^{+}(M)^n = \bigoplus_{k \geq n} \operatorname{Sym}^k M . $$

Proof. An element of $\operatorname{Sym}^{+}(M)^n$ is a sum of products of $n$ elements each of degree at least $1$, so it lies in degrees at least $n$: this gives $\subseteq$. Conversely, a homogeneous element of degree $k \geq n$ is a sum of products $x_1 \cdots x_k$ of degree-one elements, and each such product is a product of $k \geq n$ elements of $\operatorname{Sym}^{+}(M)$, so it lies in $\operatorname{Sym}^{+}(M)^k \subseteq \operatorname{Sym}^{+}(M)^n$. This gives $\supseteq$.

Example. In $R[x_1, \ldots, x_d]$ the augmentation ideal is $(x_1, \ldots, x_d)$, and $\operatorname{Sym}^{+}(R^d)^n = (x_1,\ldots,x_d)^n$ is the ideal of polynomials all of whose monomials have total degree at least $n$. The proposition says that its degree-$k$ homogeneous part is all of $R[x_1,\ldots,x_d]_k$ for $k \geq n$ and zero for $k < n$.

Symmetric Algebra over a General Ring

Non-Free Modules

Over a general commutative ring a module need not be free, and $\operatorname{Sym}(M)$ need not be a polynomial algebra or even a free module; it is the quotient of the tensor algebra by the commutator ideal, and that description always applies. The universal property, functoriality, the direct-sum isomorphism and the grading are all valid over any commutative ring. Only the monomial-basis statement, and the resulting dimension formula, need $M$ free.

Example. For $R = \mathbb{Z}$ and $M = \mathbb{Z}/m\mathbb{Z}$, the symmetric algebra is

$$ \operatorname{Sym}(\mathbb{Z}/m\mathbb{Z}) \cong \mathbb{Z} \oplus \mathbb{Z}/m\mathbb{Z} \oplus \mathbb{Z}/m\mathbb{Z} \oplus \cdots, $$

with every positive-degree part isomorphic to $\mathbb{Z}/m\mathbb{Z}$ and product induced by the multiplication of $\mathbb{Z}/m\mathbb{Z}$. The augmentation ideal is $\bigoplus_{n\geq1}\mathbb{Z}/m\mathbb{Z}$; its powers are $\operatorname{Sym}^+(M)^n = \bigoplus_{k\geq n}\mathbb{Z}/m\mathbb{Z}$ by the proposition above, so they form a strictly descending chain and the completion is $\prod_{n\geq0}\mathbb{Z}/m\mathbb{Z}$, strictly larger than $\operatorname{Sym}(M)$.

Base Change

The symmetric algebra commutes with base change. For a ring homomorphism $R \to S$ and an $R$-module $M$, the natural map

$$ S \otimes_R \operatorname{Sym}(M) \longrightarrow \operatorname{Sym}_S(S \otimes_R M) $$

is an isomorphism of graded $S$-algebras. Both sides are the free commutative $S$-algebra on $S \otimes_R M$: the left because $\operatorname{Sym}$ is built from tensor powers and tensor products commute with extension of scalars, the right by the universal property. In particular, complexifying a real symmetric algebra gives the symmetric algebra of the complexification, $\mathbb{C} \otimes_{\mathbb{R}} \operatorname{Sym}(M) \cong \operatorname{Sym}_{\mathbb{C}}(\mathbb{C} \otimes_{\mathbb{R}} M)$.

Summary

The symmetric algebra of an $R$-module $M$ is the graded algebra

$$ \operatorname{Sym}(M) = \bigoplus_{n \geq 0} \operatorname{Sym}^n M, \qquad \operatorname{Sym}^0 M = R,\ \operatorname{Sym}^1 M = M, $$

with product induced by concatenation of symmetric tensors. It is associative, commutative and unital, and it is the free commutative unital $R$-algebra on $M$: every $R$-linear map $M \to A$ into a commutative unital algebra extends uniquely to an algebra homomorphism $\operatorname{Sym}(M) \to A$. Equivalently, it is the quotient of the tensor algebra $T(M)$ by the two-sided ideal generated by the commutators $x \otimes y - y \otimes x$. When $M = R^d$ is free of rank $d$, it is the polynomial algebra $R[x_1, \ldots, x_d]$, and its $n$-th graded piece has rank $\binom{n+d-1}{d-1}$. The construction is functorial, is left adjoint to the forgetful functor to modules, carries direct sums to tensor products, and commutes with base change. Its augmentation ideal defines the adic filtration whose completion is treated.

Summary of Notation

Further Reading

• Nicolas Bourbaki, Algebra I: Chapters 1–3 (Springer, 1989), for the symmetric algebra and its universal property.
• Serge Lang, Algebra (Springer, 3rd ed. 2002), for the symmetric and tensor algebras as free objects.
• David S. Dummit and Richard M. Foote, Abstract Algebra (Wiley, 3rd ed. 2004), for polynomial algebras and evaluation.
• Nicolas Bourbaki, Éléments de mathématique: Algèbre commutative (Hermann, 1961), for the symmetric algebra over a commutative ring and its graded structure.
• Saunders Mac Lane, Categories for the Working Mathematician (Springer, 2nd ed. 1998), for the free-forgetful adjunction.