free objects in the category of commutative algebras

Let $R$ be a commutative ring and let $𝒜ℒ{𝒢}_c(R)$ 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 $𝒮ℰ𝒯$). Therefore we can talk about free objects in $𝒜ℒ{𝒢}_c(R)$ (see this entry (http://planetmath.org/FreeObjectsInConcreteCategories2) for definitions).

Theorem. For any set $𝕏$ the polynomial algebra $R[𝕏]$ (see parent object) is a free object in $𝒜ℒ{𝒢}_c(R)$ with $𝕏$ being a basis. This means that for any commutative algebra $A$ and any function

$$f:𝕏\to A$$

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

$$F(x)=f(x)$$

for any $x\in 𝕏$.

Proof. Assume that $f:𝕏\to A$ is a function. If $W\in R[𝕏]$, then there are finite subsets $A_1,\mathrm{\dots },A_n\subseteq 𝕏$ (not necessarily disjoint) and natural numbers $n(x,i)$, $i=1,\mathrm{\dots },n$ such that $W$ can be uniquely expressed as

$$W=\sum _{i=1}^n\left({\lambda }_i\cdot \prod _{x\in A_i}{x}^{n(x,i)}\right)$$

with ${\lambda }_i\in R$. Define $F(W)$ by putting

$$F(W)=\sum _{i=1}^n\left({\lambda }_i\cdot \prod _{x\in A_i}f{(x)}^{n(x,i)}\right).$$

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 $x\in 𝕏$. This completes the proof. $\mathrm{\square }$

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

$$I:R[𝕐]\to R[𝕏].$$

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

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

$$I:R[𝕐]\to R[𝕏]$$

such that $I(y)=y$. It remains to show, that $I$ is ,,1-1”. Indeed, assume that $I(W)=0$ for some polynomial $W\in R[𝕐]$. 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. $\mathrm{\square }$

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

$$A\simeq R[𝕏]/I$$

for some ideal $I$.

Proof. Let $𝕏=A$ as a set. Define

$$f:𝕏\to A$$

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

$$F:R[𝕏]\to A$$

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

$$A\simeq R[𝕏]/\mathrm{Ker}F$$

which completes the proof. $\mathrm{\square }$