Thread: K-Algebra - Meaning and background of the concept

I am trying to understand the concept of a k-algebra without much success.

Can someone please give me a clear explanation of the background, definition and use of the concept.

Also I would be extremely grateful of some examples of k-algebras

Peter

ok, you know how fields can be considered "two groups in one" (one is the additive group, and one is the multiplicative group of non-zero elements)? well a k-algebra is a similar idea of "two structures in one".

on one hand, we have that a k-algebra is a vector space over k. this is the same thing as a k-module (which is FREE over any basis).

on the other hand, we have that a k-algebra forms a ring (usually associative and with unity). the ring structure and the vector space structure have to be COMPATIBLE.

this means: the ring multiplication is bilinear.

another way to look at this is that you have a ring, A, together with a ring-homomorphism η:k-->Z(A) (the center of A).

if A is not a trivial ring (just 0), then η is injective (because k is a field, and field-homomorphisms are always monomorphisms), so normally n(k) is identified with k.

this allows us to define a vector space on A by:

vector addition is just the ring addition (for any ring A, (A,+) is an abelian group).

scalar multiplication is defined like so: for c in k, and a in A:

ca = η(c)a (where the RHS is the ring multiplication of A).

******************

some typical examples:

let k be a field, and let E be any field containing k as a subfield. define for c in k, and a in E: ca to be the product in E. for example, the complex numbers C are an R-algebra. this is a 2-dimensional R-algebra.

let k be a field, and let k[x] be the ring of polynomials over k. then k[x] is a k-algebra. this is an infinite-dimensional k-algebra.

let k be a field, and let V be any vector space over k. then Hom(V,V) = End(V), the set of all k-linear mappings from V to V, is a k-algebra. if V is finite-dimensional, of dimension n, then End(V) has dimension n². for the finite-dimensional case, this can be identified with Mat_nxn(k), the set of all nxn matrices with entries in k.

let G be any group, and let k be any field. the then group-algebra k[G], consisting of formal k-linear combinations of elements of G (together with a "polynomial-like" multiplication) is a k-algebra.

the fact that k-algebras are vector spaces over k, lets us use the tools of linear algebra to investigate them. the fact the k-algebras are also rings, lets us use ring-concepts in investigating them.

for example, for any k-algebra, we can speak of its group of units. for the k-algebra k[x], this is just k. for the k-algebra End_k(V), this is GL_k(V), the general linear group of V. again, if V is finite-dimensional, this corresponds to the INVERTIBLE nxn matrices.

one can also form "function algebras". for example, one has the R-algebra C[a,b], consisting of continuous functions f:[a,b]--->R, where the ring-structure is inherited from R:

(f+g)(x) = f(x) + g(x)
(fg)(x) = f(x)g(x)
(cf)(x) = c(f(x)) <--this is the "scalar multiplication".

this example can easily be generalized to functions f:S-->k, where S is any set, and k is the field k. often we are interested in some sub-ring of k^S (continuous, differentiable, linear, etc.) and often S has additional properties (such as a topology, or is a vector space itself).

historically, k-algebras (and not specific instances of them) are relatively recent, being studied in their own right only since around the beginning of the 20th century. some of the development of k-algebras probably came about as an attempt to realize various structures as matrix algebras, for example it is well-known that the complex numbers can be realized as a sub-algebra of Mat_2x2(R) as:



in a similar vein, the quaternions form a 4-dimensional algebra over R, and a 2-dimensional algebra over C, a 2x2 complex matrix that describes a quaternion is:



where A* = the complex conjugate of A. each entry can be viewed as a 2x2 "block matrix", yielding a 4x4 real matrix.

a nifty feature of End(V), for finite-dimensional V: we have a monoid-homomorphism det:End(V)-->k, to the multiplicative monoid (k,*), which preserves units: T is in U(End(V)) (that is: GL(V)) iff det(T) is in U(k) = k* (that is: if det(T) ≠ 0).

in general, we can investigate the sub-structure of a k-algebra A, by considering the ideals of A as a ring. such ideals are automatically subspaces of A since:

u in J and v in J means u+v is in J (ideals are closed under addition)

u in J and c in k means cu is in J (here we are implicitly considering c in A via the monomorphism η).

0 is in any ideal J of A.

note that in the algebra k[x], the ideal generated by x is considerably bigger than the subspace generated by x. so in k-algebras, you sometimes have to be careful specifying "how you're decomposing it". the "dual nature" of k-algebras leads to a rich and varied theory. one of the things k-algebras are useful for is "representation theory". a representation of a k-algebra A consists of two things:

1. a vector space V over k
2. an action on V by A via endomorphisms (equivalently: an algebra homomorphism A --> End_k(V)).

concretely (when dimension V = n), this lets us think of elements of A as nxn matrices, so that instead of doing "abstract algebra" in A, we can do "concrete (matrix) arithmetic" in Mat_nxn(k). this, in turn, is equivalent to turning V into an A-module (you may already be familiar with regarding V as a k[x]-module, by picking a linear transformation T, and setting p(x).v = p(T)(v), or regarding V as a k[G]-module via a homomorphism φ:G-->GL(V) and setting:

(a₁e + a₂g₁ +...+ a_ng_n-1).v = a₁v + a₂φ(g₁)(v) +...+ a_nφ(g_n-1)(v) ).

Thank you for such a helpful post

Just working through it in detail now.

Peter

Deveno,

You write:

"on one hand, we have that a k-algebra is a vector space over k. this is the same thing as a k-module (which is FREE over any basis)."

What do you mean by "FREE over any basis"

Akso, can you explain "compatible" further?

Peter

ok, a free R-module over a ring R is a module M with a subset S such that:

1. S generates M (as finite R-linear combinations of elements of S).
2. S is R-linearly independent

S is called a BASIS for M. for example: consider the group (ZxZ,+). this can be considered a Z-module in a natural way:

n.(a,b) = (na,nb)

the set S = {(1,0),(0,1)} clearly generates ZxZ, since (a,b) = (a,0) + (0,b) = a(1,0) + b(0,1).

moreover, if a(1,0) + b(0,1) = (0,0), then (a,b) = (0,0), whence a = 0, and b = 0.

if R = k, a field, and V is vector space over k (which is what a k-module is), V is uniquely determined (up to isomorphism) by the size of its basis.

in general, the free R-module generated by S, is a module M such that:

S ⊆ M, and given any module N, and any function f:S-->N, there is a unique R-module homomorphism F:M-->N such that F(s) = f(s) for all s in S (F can be thought of as "the unique R-module homomorphism obtained by extending f to all of M via R-linearity").

the following statement is true: the R-module M is free over a subset S iff S is a basis for M.

your original question in this context is: if S is a basis for M, why is M free over S? let's look at that in greater detail:

suppose we are given that S is a basis for an R-module M. since <S> = M, we can write any m in M as:

for some a_j's in R, and s_j's in S (we are only considering FINITE R-linear combinations).

now suppose we are given a function (just an ordinary set-function) f:S-->N to some R-module N.

define F:M-->N by:



it is routine (but tedious) to verify that F(m+m') = F(m) + F(m'), and F(am) = aF(m), so F is R-linear, and thus an R-module homomorphism. clearly for any s in S, F(s) = F(1s) = 1f(s) = f(s). so F is one possible homomorphism that fits the bill. are there any others?

suppose we have an R-module homomorphism G:M-->N with G(s) = f(s), for all s in S. then, given m in M:



(because G is R-linear)

since G(s_j) = f(s_j), by definition of G.

therefore, F = G. now, you may wonder, where did we use the R-linear independence of S? it's sort of "hidden" in the well-definedness of F:

recall that we defined F by defining F on R-linear combinations of the s_j. so, what if:

for two different subsets and of S?

well, we can rewrite each R-linear combination as an R-linear combination of elements in AUB (which is still a finite R-linear combination of elements in S, since A and B are finite), by adding "0-terms" if necessary,

and get:

.

by the R-linear independence of S, each a_u-b_u = 0.

if s_u only occurs in A (that is, we added a "0 term" for b_u) this means that a_u must have been 0 in the first place.

if s_u only occurs in B (that is we padded out our expression in terms of a's by using a "0 term" for a_u), we see that b_u must likewise have been 0.

if, however, s_u is in A∩B, we see the a-terms and the b-terms must match exactly, so that in fact, these are the SAME linear combination, so defining F on R-linear combinations of S uniquely defines it.

if you think about it, we defined F "in the only way possible". R-linear independence removes any ambiguity in our possible definition (this is NOT true for an arbitrary generating set X for M. for example, consider M = Z, and the generating set:

X = {2,3}. it is quite possible to find distinct pairs (k,m) of integers for which n = 2k + 3m for any n. for example 1 = 3 - 2 = 9 - 8 (the first is k = -1, m = 1, the second is k = -4, m= 3), or 16 - 15 (k = 8, m = -5).

the set {2,3} is not Z-linearly independent: we have 2(3) + 3(-2) = 0, without 3 = -2 = 0. in this example if we choose N = Z as well, with f:X-->N as f(2) = f(3) = 1, we get (trying to use a similar F):

F(1) = F(2k + 3m) = kf(2) + mf(3) = -f(2) + f(3) = -1 + 1 = 0, using k = -1, m = 1
F(1) = F(2k + 3m) = kf(2) + mf(3) = -4f(2) + 3f(3) = -4 + 3 = -1, using k = -4, m= -3
F(1) = F(2k + 3m) = kf(2) + mf(3) = 8f(2) - 5f(3) = 8 - 5 = 3, using k = 8, m= -5, showing our definition of F is not "well-defined" (we get different values for F depending on how we pick k and m).

*********************************

now, in general an R-module need not HAVE a basis, and even if it does, two bases need not be isomorphic. but for all commutative rings R, if we HAVE a basis, the cardinality of the basis is an invariant. if R = k, a field, then not only is the size of any basis of a k-module invariant, but every k-module also HAS a basis (the proof of this for arbitrary k-modules (vector spaces) involves the axiom of choice, and is a bit involved).

my apologies for the length of the exposition above. suffice to say, modules are "almost like vector spaces" (except the scalars are a ring), and when the ring is a field, they ARE vector spaces. so if you start with a basis set B of a vector space V, and create the free k-module over B, you wind up with something isomorphic to V, which being a subspace of V, IS V. in purely linear algebraic terms, if B is a basis, span(B) = V (this sort of "hides" the fact that B is LI).

********************************

regarding "compatibility", the product in a k-algebra A is required to be BILINEAR (over k). this means two things:

1) (u)(v + w) = uv + uw

(u + v)(w) = uw + vw (these are, of course, just the distributive laws for a ring) for all u,v,w in A.

2) (cu)(v) = (u)(cv) = c(uv) (the scalar multiplication of A as a vector space respects the multiplication of A as a ring), for all c in k, and u,v in A.

note that if we defined A in terms of a field monomorphism η:k-->Z(A), then:

c.(uv) = η(c)(uv) = (η(c)u)(v) = (c.u)(v) by associativity of multiplication in A, while:

c.(uv) = η(c)(uv) = (η(c)u)(v) = (uη(c))(v) = (u)(η(c)v) = (u)(c.v), by associativity, and the fact that η(c), being in Z(A), commutes with u.

(the "dot" is normally omitted in the scalar product, because of these rules).

Thank you again - extremely helpful

WIll now work through the post carefully!

Peter