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Finitely Generated k-algebra - Nature of the finite generation - basic question

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
In Dummit and Foote Chapter 15 on page 657 we find the following definition of a k-algebra:

Let k be a field. A ring R is a k-algebra if k is contained in the centre of R and the identity of k is the identity of R.

This defintion is followed by the definition of a finitely generated k-algebra:

The ring R is a finitely generated k-algebra if R is generated as a ring by k together with some finite set [TEX] r_1, r_2, ... ... , r_n [/TEX] of elements of R.

I wish to make sure that I understand the idea of the finite generation of the ring R.

I am assuming that we have as generators, the set of elements of k, namely [TEX] k_1, k_2, k_3, ... ... [/TEX] (may be infinite!)

and also the ring elements [TEX] r_1, r_2, ... ... , r_n [/TEX]

So the generated elements would be the those elements themselves plus the elements formed by combining these elements using (possibly repeated) the ring operations of addition and multiplication ...

So typical generated elements would include elements like

[TEX] r_1 + r_2, r_1 + k_1, r_1k_1, r_1k_2 + r_2k_3, r_1r_2 + k_1k_4 + r_3k_4, r_1r_2k_1, r_2,r_3k_2 + k_1k_2r_3, ... ... [/TEX] and so on ...

Does this look right? Can someone please confirm that I am understanding this correctly.

Peter

[Note: This has also been posted on MHF]
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Yes, but....

The subscripts on the k's are a bit misleading, the field k can be uncountably infinite (in fact, this is often the case, with k-algebras of practical interest, such as a suitable complex-valued function space defined on some compact (or perhaps open) subset of the complex plane).

We not only have all k-linear combinations of the generating elements, but also any n-fold product of such k-linear combinations such as (for example):


\(\displaystyle (k_1r_1r_3 + k_2r_4)^2(k_3r_5)^3(k_2r_1r_3 + r_2)\)

In practice, one seeks to get a "simple way of describing" such elements by finding a BASIS (as a k-vector space) for \(\displaystyle k[r_1,r_2,\dots,r_n]\) (in other words, you should think of polynomial spaces over k as "the prototypical example").

Another "guiding example" is the space \(\displaystyle \text{Mat}_k(n) \cong \text{End}_k(k^n)\) which uses the natural ring structure of endomorphisms of an abelian group to create the algebra structure (\(\displaystyle k^n\) has a natural addition as the n-fold direct product of the additive group of k).

A finite-dimension k-algebra is of necessity a finitely-generated k-algebra (any basis is a generating set). The converse is NOT true: the polynomial ring \(\displaystyle k[x]\) is infinite-dimensional over k, but as a k-algebra is clearly generated by {x}, which is a finite subset.

A third important example is afforded by \(\displaystyle k[G]\) the group ring of a finite group G over the field k. In this case, we see that \(\displaystyle k[G]\) is finitely-generated by the elements of G (it is most fortunate that the mapping for any g in G:

\(\displaystyle g(x) = gx, \forall x \in G\)

is a bijection, this is what guarantees we can define a suitable multiplication on \(\displaystyle k[G]\)).

k-algebras have a dual nature:

1) They are rings (in your definition they are UNITAL rings)
2) they are vector spaces over k (this follows from the fact that (R,+) is an abelian group, and that for \(\displaystyle \alpha \in k, r \in R\) we can set:


\(\displaystyle \alpha(r) = \alpha r = r\alpha\) (since \(\displaystyle k \subseteq Z(R)\)).

note that (2) means that R is both a left- and right-k-module, so we can apply the full force of ring theory, module theory, or linear algebra, whichever is convenient.

Unfortunately, k-algebras usually aren't "nice" rings, they are often non-commutative (often due to the fact that functional composition is "one-way") and often contain zero-divisors. They ARE nice k-modules, though, because a field is a very well-behaved ring.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Yes, but....

The subscripts on the k's are a bit misleading, the field k can be uncountably infinite (in fact, this is often the case, with k-algebras of practical interest, such as a suitable complex-valued function space defined on some compact (or perhaps open) subset of the complex plane).

We not only have all k-linear combinations of the generating elements, but also any n-fold product of such k-linear combinations such as (for example):


\(\displaystyle (k_1r_1r_3 + k_2r_4)^2(k_3r_5)^3(k_2r_1r_3 + r_2)\)

In practice, one seeks to get a "simple way of describing" such elements by finding a BASIS (as a k-vector space) for \(\displaystyle k[r_1,r_2,\dots,r_n]\) (in other words, you should think of polynomial spaces over k as "the prototypical example").

Another "guiding example" is the space \(\displaystyle \text{Mat}_k(n) \cong \text{End}_k(k^n)\) which uses the natural ring structure of endomorphisms of an abelian group to create the algebra structure (\(\displaystyle k^n\) has a natural addition as the n-fold direct product of the additive group of k).

A finite-dimension k-algebra is of necessity a finitely-generated k-algebra (any basis is a generating set). The converse is NOT true: the polynomial ring \(\displaystyle k[x]\) is infinite-dimensional over k, but as a k-algebra is clearly generated by {x}, which is a finite subset.

A third important example is afforded by \(\displaystyle k[G]\) the group ring of a finite group G over the field k. In this case, we see that \(\displaystyle k[G]\) is finitely-generated by the elements of G (it is most fortunate that the mapping for any g in G:

\(\displaystyle g(x) = gx, \forall x \in G\)

is a bijection, this is what guarantees we can define a suitable multiplication on \(\displaystyle k[G]\)).

k-algebras have a dual nature:

1) They are rings (in your definition they are UNITAL rings)
2) they are vector spaces over k (this follows from the fact that (R,+) is an abelian group, and that for \(\displaystyle \alpha \in k, r \in R\) we can set:


\(\displaystyle \alpha(r) = \alpha r = r\alpha\) (since \(\displaystyle k \subseteq Z(R)\)).

note that (2) means that R is both a left- and right-k-module, so we can apply the full force of ring theory, module theory, or linear algebra, whichever is convenient.

Unfortunately, k-algebras usually aren't "nice" rings, they are often non-commutative (often due to the fact that functional composition is "one-way") and often contain zero-divisors. They ARE nice k-modules, though, because a field is a very well-behaved ring.
Deveno,

Thanks so much for this guidance and help.

Just working through your post now.

This will enable me to move on with this topic with more confidence

Peter