# free objects in the category of commutative algebras

Let $R$ be a commutative ring and let $\mathcal{ALG}_{c}(R)$ be the category of all commutative algebras over $R$ and algebra homomorphisms. This category together with the forgetful functor is a construct (i.e. it is a concrete category over the category of sets $\mathcal{SET}$). Therefore we can talk about free objects in $\mathcal{ALG}_{c}(R)$ (see this entry (http://planetmath.org/FreeObjectsInConcreteCategories2) for definitions).

For any set $\mathbb{X}$ the polynomial algebra $R[\mathbb{X}]$ (see parent object) is a free object in $\mathcal{ALG}_{c}(R)$ with $\mathbb{X}$ being a basis. This means that for any commutative algebra $A$ and any function

 $f:\mathbb{X}\to A$

there exists a unique algebra homomorphism $F:R[\mathbb{X}]\to A$ such that

 $F(x)=f(x)$

for any $x\in\mathbb{X}$.

Proof. Assume that $f:\mathbb{X}\to A$ is a function. If $W\in R[\mathbb{X}]$, then there are finite subsets $A_{1},\ldots,A_{n}\subseteq\mathbb{X}$ (not necessarily disjoint) and natural numbers $n(x,i)$, $i=1,\ldots,n$ such that $W$ can be uniquely expressed as

 $W=\sum_{i=1}^{n}\bigg{(}\lambda_{i}\cdot\prod_{x\in A_{i}}x^{n(x,i)}\bigg{)}$

with $\lambda_{i}\in R$. Define $F(W)$ by putting

 $F(W)=\sum_{i=1}^{n}\bigg{(}\lambda_{i}\cdot\prod_{x\in A_{i}}f(x)^{n(x,i)}% \bigg{)}.$

Of course $F$ is well defined and obviously $F(x)=f(x)$. We leave as a simple exercise that $F$ is an algebra homomorphism. The uniqueness of $F$ again follows from the explicit form of $W$. It is easily seen that $F(W)$ depends only on $F(x)$ for $x\in\mathbb{X}$. This completes the proof. $\square$

Corollary 1. If $\mathbb{X}$ is a set and $\mathbb{Y}\subseteq\mathbb{X}$, then the inclusion $i:\mathbb{Y}\to\mathbb{X}$ induces an algebra monomorphism

 $I:R[\mathbb{Y}]\to R[\mathbb{X}].$

In particular we can treat $R[\mathbb{Y}]$ as a subalgebra of $R[\mathbb{X}]$.

Proof. We have a well-defined function $i:\mathbb{Y}\to R[\mathbb{X}]$, $i(y)=y$. By the theorem we have an extension

 $I:R[\mathbb{Y}]\to R[\mathbb{X}]$

such that $I(y)=y$. It remains to show, that $I$ is ,,1-1”. Indeed, assume that $I(W)=0$ for some polynomial $W\in R[\mathbb{Y}]$. But if we recall the expression of $W$ as in proof of the theorem and remember that $I$ is an algebra homomorphism, then it is easy to see that $I(y)=y$ implies that

 $I(W)=W.$

In particular $W=0$, which completes the proof. $\square$

Corollary 2. If $A$ is an $R$-algebra, then there exists a set $\mathbb{X}$ such that

 $A\simeq R[\mathbb{X}]/I$

for some ideal $I$.

Proof. Let $\mathbb{X}=A$ as a set. Define

 $f:\mathbb{X}\to A$

by $f(x)=x$. By the theorem we have an algebra homomorphism

 $F:R[\mathbb{X}]\to A$

such that $F(x)=x$ for $x\in\mathbb{X}$. In particular $F$ is ,,onto” and thus by the First Isomorphism Theorem for algebras we have

 $A\simeq R[\mathbb{X}]/\mathrm{Ker}F$

which completes the proof. $\square$

Title free objects in the category of commutative algebras FreeObjectsInTheCategoryOfCommutativeAlgebras 2013-03-22 19:18:13 2013-03-22 19:18:13 joking (16130) joking (16130) 6 joking (16130) Theorem msc 13P05 msc 11C08 msc 12E05