free objects in the category of commutative algebras
Let be a commutative ring and let be the category of all commutative algebras over 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 ). Therefore we can talk about free objects in (see this entry (http://planetmath.org/FreeObjectsInConcreteCategories2) for definitions).
there exists a unique algebra homomorphism such that
for any .
with . Define by putting
Of course is well defined and obviously . We leave as a simple exercise that is an algebra homomorphism. The uniqueness of again follows from the explicit form of . It is easily seen that depends only on for . This completes the proof.
In particular we can treat as a subalgebra of .
Proof. We have a well-defined function , . By the theorem we have an extension
such that . It remains to show, that is ,,1-1”. Indeed, assume that for some polynomial . But if we recall the expression of as in proof of the theorem and remember that is an algebra homomorphism, then it is easy to see that implies that
In particular , which completes the proof.
Corollary 2. If is an -algebra, then there exists a set such that
for some ideal .
Proof. Let as a set. Define
by . By the theorem we have an algebra homomorphism
such that for . In particular is ,,onto” and thus by the First Isomorphism Theorem for algebras we have
which completes the proof.
|Title||free objects in the category of commutative algebras|
|Date of creation||2013-03-22 19:18:13|
|Last modified on||2013-03-22 19:18:13|
|Last modified by||joking (16130)|