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 $\iota:X\rightarrow F$ such that for all functions $f:X\rightarrow A$ where $A$ is a commutative $K$-algebra determine a unique algebra homomorphism $\hat{f}:F\rightarrow A$ such that $\iota\hat{f}=f$. 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:

 $\xymatrix{&X\ar[ld]_{\iota}\ar[rd]^{f}&\\ F\ar[rr]^{\hat{f}}&&A.}$

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 $K[X]$.

Proof.

Let $A$ be any commutative associative $K$-algebra and $f:X\rightarrow A$. Recall $K\langle X\rangle$ is the free associative $K$-algebra on $X$ and so by the universal mapping property of this free object there exists a map $\hat{f}:K\langle X\rangle\rightarrow A$ such that $\iota_{K\langle X\rangle}\hat{f}=f$.

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

Title free commutative algebra FreeCommutativeAlgebra 2013-03-22 16:51:22 2013-03-22 16:51:22 Algeboy (12884) Algeboy (12884) 5 Algeboy (12884) Theorem msc 08B20 PolynomialRing free commutative algebra