algebras

Let $K$ be a commutative unital ring (often a field) and $A$ a $K$ -module.Given a bilinear mapping $b:A\\times A\\rightarrow A$ , we say $(K,A,b)$ is a $K$ -algebra. We usually write only $A$ for the tuple $(K,A,b)$ .

Remark 1.

Many authors and applications insist on $K$ as a field, or at least a localring, or a semisimple ring. This enables $A$ to have some notion of dimensionor rank.

This definition is a compact method to encode the property that our multiplication is distributive: the multiplication is additive in both variables translates to

(a+b)c=ac+bc,\\qquad a(b+c)=ab+ac\\qquad a,b,c\\in A.

Furthermore, the assumption that scalars can be passed in and out of the bilinearproduct translates to

(la)b=l(ab)=a(lb),\\qquad a,b\\in A,l\\in K.

Perhaps the most important outcome of these two axioms of an algebra is theopportunity to express polynomial like equations over the algebra. Without the distributive axiom we cannot establish connections between addition and multiplication. Without scalar multiplication we cannot describe coefficients.With these equations we can define certain subalgebras, for example wesee both axioms at work in

Proposition 2.

Given an algebra $A$ , the set

Z_{0}(A)=\\{z\\in A:za=az,a\\in A\\}.

$Z_{0}(A)$ is a submodule of $A$ .

Proof.

For now let elements of $A$ be denoted with $\\hat{a}$ to distinguish themfrom scalars. As a module $0\\hat{a}=\\hat{0}$ for all $a\\in A$ . Then

\\hat{0}\\hat{a}=(0\\hat{a})\\hat{a}=(\\hat{a})(0\\hat{a})=\\hat{a}\\hat{0}.

So $\\hat{0}\\in Z_{0}(A)$ .

Also given $\\hat{z},\\hat{w}\\in Z_{0}(A)$ then for all $a\\in A$ ,

(\\hat{z}+\\hat{w})\\hat{a}=\\hat{z}\\hat{a}+\\hat{w}\\hat{a}=\\hat{a}\\hat{z}+\\hat{a}%\\hat{w}=\\hat{a}(\\hat{z}+\\hat{w}).

So $\\hat{z}+\\hat{w}\\in A$ .

Finally, given $l\\in K$ we have

(l\\hat{z})\\hat{a}=l(\\hat{z}\\hat{a})=l(\\hat{a}\\hat{z})=\\hat{a}(l\\hat{z}).

∎

Although this set $Z(A)$ appears like a reasonable object to define as thecenter of an algebra, it is usually preferable to produce a subalgebra, notsimply a submodule, and for this we need elements that can be regrouped in products associatively, that is, that lie in the nucleus. So the center is commonly defined as

Z(A)=\\{z\\in A:za=az,z(ab)=(za)b,a(zb)=(az)b,(ab)z=a(bz),a,b\\in A\\}.

When the algebra $A$ has an identity (unity) $1$ then we can go further to identify $K$ as a subalgebra of $A$ by $l1$ . Then we see this subalgebra isnecessarily in the center of $A$ . As a converse, given a unital ring $R$ (associativity is necessary), the center of the ring forms a commutative unital subring over which $R$ is an algebra. In this way unital rings and associative unital algebras are often interchanged.