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Difference between revisions of "Symmetric algebra"

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For any homomorphism $f:M \to N$ of $A$-modules, the $p$-th tensor power $T^p(f)$ induces a homomorphism $S^p(f) : S^p(M) \to S^p(N)$ (the $p$-th symmetric power of the homomorphism $f$). A homomorphism $S(f) : S(M) \to S(N)$ of $A$-algebras is obtained. The correspondences $f \mapsto S^p(f)$ and $f \mapsto S(f)$ are, respectively, covariant functors from the category of $A$-modules into itself and into the category of $A$-algebras. For any two $A$-modules $M$ and $N$ there is a natural isomorphism $S(M\oplus N) = S(M) \otimes_A S(N)$.
 
For any homomorphism $f:M \to N$ of $A$-modules, the $p$-th tensor power $T^p(f)$ induces a homomorphism $S^p(f) : S^p(M) \to S^p(N)$ (the $p$-th symmetric power of the homomorphism $f$). A homomorphism $S(f) : S(M) \to S(N)$ of $A$-algebras is obtained. The correspondences $f \mapsto S^p(f)$ and $f \mapsto S(f)$ are, respectively, covariant functors from the category of $A$-modules into itself and into the category of $A$-algebras. For any two $A$-modules $M$ and $N$ there is a natural isomorphism $S(M\oplus N) = S(M) \otimes_A S(N)$.
If <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159043.png" /> is a vector space over a field of characteristic 0, then the symmetrization <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159044.png" /> (cf. [[Symmetrization (of tensors)|Symmetrization (of tensors)]]) defines an isomorphism from the symmetric algebra <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159045.png" /> onto the algebra <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159046.png" /> of symmetric contravariant tensors over <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159047.png" /> relative to symmetric multiplication:
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If $M$ is a vector space over a field of characteristic 0, then the [[Symmetrization (of tensors)|symmetrization]] $\sigma : T(M) \to T(M)$ defines an isomorphism from the symmetric algebra $S(M)$ onto the algebra $\tilde S(M) \subset T(M)$ of symmetric contravariant tensors over $M$ relative to symmetric multiplication:
 
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$$
<table class="eq" style="width:100%;"> <tr><td valign="top" style="width:94%;text-align:center;"><img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159048.png" /></td> </tr></table>
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x \vee y = \sigma(x \otimes y)\,,\ \ \ x \in \tilde S^p(M)\,,\ \ y \in \tilde S^q(M) \ .
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$$
  
 
====References====
 
====References====
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====Comments====
 
====Comments====
The functor <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159049.png" /> from <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159050.png" />-modules to commutative unitary <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159051.png" />-algebras solves the following universal problem. Let <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159052.png" /> be an <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159053.png" />-module and <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159054.png" /> a commutative unitary <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159055.png" />-algebra. For each homomorphism <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159056.png" /> of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159057.png" />-modules there is a unique homomorphism <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159058.png" /> of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159059.png" />-algebras such that <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159060.png" /> restricted to <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159061.png" /> coincides with <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159062.png" />. Thus, <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159063.png" /> is a left-adjoint functor of the underlying functor from the category of commutative unitary <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159064.png" />-algebras to the category of <img align="absmiddle" border="0" src="https://www.encyclopediaofmath.org/legacyimages/s/s091/s091590/s09159065.png" />-modules.
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The functor $S$ from $A$-modules to commutative unitary $A$-algebras solves the following universal problem. Let $M$ be an $A$-module and $B$ a commutative unitary $A$-algebra. For each homomorphism $f : M \to B$ of $A$-modules there is a unique homomorphism $g : S(M) \to B$ of $A$-algebras such that $g$ restricted to $S^1(M)$ coincides with $f$. Thus, $S$ is a left-adjoint functor of the underlying functor from the category of commutative unitary $A$-algebras to the category of $A$-modules.

Latest revision as of 18:21, 11 April 2017

A generalization of a polynomial algebra. If $M$ is a unital module over a commutative associative ring $A$ with an identity, then the symmetric algebra of $M$ is the algebra $S(M) = T(M)/I$, where $T(M)$ is the tensor algebra of $M$ and $I$ is the ideal generated by the elements of the form $x \otimes y - y \otimes x$ ($x,y \in M$). A symmetric algebra is a commutative associative $A$-algebra with an identity. It is graded: $$ S(M) = \bigoplus_{p \ge 0} S^p(M) $$ where $S^p(M) = T^p(M)/(T^p(M)\cap I)$, and $S^0(M) = A$, $S^1(M) = M$. The module $S^p(M)$ is called the $p$-th symmetric power of the module $M$. If $M$ is a free module with finite basis $x_1,\ldots,x_n$, then the correspondence $x_i \mapsto X_i$ ($i=1,\ldots,n$) extends to an isomorphism of $S(M)$ onto the polynomial algebra $A[X_1,\ldots,X_n]$ (see Ring of polynomials).

For any homomorphism $f:M \to N$ of $A$-modules, the $p$-th tensor power $T^p(f)$ induces a homomorphism $S^p(f) : S^p(M) \to S^p(N)$ (the $p$-th symmetric power of the homomorphism $f$). A homomorphism $S(f) : S(M) \to S(N)$ of $A$-algebras is obtained. The correspondences $f \mapsto S^p(f)$ and $f \mapsto S(f)$ are, respectively, covariant functors from the category of $A$-modules into itself and into the category of $A$-algebras. For any two $A$-modules $M$ and $N$ there is a natural isomorphism $S(M\oplus N) = S(M) \otimes_A S(N)$. If $M$ is a vector space over a field of characteristic 0, then the symmetrization $\sigma : T(M) \to T(M)$ defines an isomorphism from the symmetric algebra $S(M)$ onto the algebra $\tilde S(M) \subset T(M)$ of symmetric contravariant tensors over $M$ relative to symmetric multiplication: $$ x \vee y = \sigma(x \otimes y)\,,\ \ \ x \in \tilde S^p(M)\,,\ \ y \in \tilde S^q(M) \ . $$

References

[1] N. Bourbaki, "Eléments de mathématique" , 2. Algèbre , Hermann (1964) pp. Chapt. IV-VI
[2] A.I. Kostrikin, Yu.I. Manin, "Linear algebra and geometry" , Gordon & Breach (1989) (Translated from Russian)


Comments

The functor $S$ from $A$-modules to commutative unitary $A$-algebras solves the following universal problem. Let $M$ be an $A$-module and $B$ a commutative unitary $A$-algebra. For each homomorphism $f : M \to B$ of $A$-modules there is a unique homomorphism $g : S(M) \to B$ of $A$-algebras such that $g$ restricted to $S^1(M)$ coincides with $f$. Thus, $S$ is a left-adjoint functor of the underlying functor from the category of commutative unitary $A$-algebras to the category of $A$-modules.

How to Cite This Entry:
Symmetric algebra. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Symmetric_algebra&oldid=40937
This article was adapted from an original article by A.L. Onishchik (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article