free commutative algebra

Fix a commutative unital ring $K$ and a set $X$. Then a commutative associative $K$-algebra $F$ is said to be free on $X$ if there exists an injection $\iota :X\to F$ such that for all functions $f:X\to A$ where $A$ is a commutative $K$-algebra determine a unique algebra homomorphism $\widehat{f}:F\to A$ such that $\iota \widehat{f}=f$. This is an example of a universal mapping property for commutative associative algebras and in categorical settings is often explained with the following commutative diagram:

To construct a free commutative associative algebra we observe that commutative associative algebras are a subcategory of associative algebras and thus we can make use of free associative algebras in the construction and proof.

Theorem 1.

Given a set $X$, and a commutative unital ring $K$, the free commutative associative $K$-algebra on $X$ is the polynomial ring $K\mathit{}\mathrm{[}X\mathrm{]}$.

Proof.

Let $A$ be any commutative associative $K$-algebra and $f:X\to A$. Recall $K⟨X⟩$ is the free associative $K$-algebra on $X$ and so by the universal mapping property of this free object there exists a map $\widehat{f}:K⟨X⟩\to A$ such that ${\iota }_{K⟨X⟩}\widehat{f}=f$.

We also have a map $p:K⟨X⟩\to K[X]$ which effectively maps words over $X$ to words over $X$. Only in $K[X]$ the indeterminants commute. Since $A$ is commutative, $\widehat{f}$ factors through $p$, in the sense that there exists a map $\tilde{f}:K[X]\to A$ such that $p\tilde{f}=\widehat{f}$. Thus $\tilde{f}$ is the desired map which proves $K[X]$ is free in the category of commutative associative algebras. ∎

Title	free commutative algebra
Canonical name	FreeCommutativeAlgebra
Date of creation	2013-03-22 16:51:22
Last modified on	2013-03-22 16:51:22
Owner	Algeboy (12884)
Last modified by	Algeboy (12884)
Numerical id	5
Author	Algeboy (12884)
Entry type	Theorem
Classification	msc 08B20
Related topic	PolynomialRing
Defines	free commutative algebra