Determining an algebra homomorphism by its image on a subset of vectors

Question

If $T$ is a linear map on a real vector space, it is completely determined by the way it maps any basis of the vector space (by linearity).

This is the general gist of matrix notation - by specifying the image of the linear transformation on $(1,0,0,...), (0,1,0,...), (0,0,1,...), ...$, the linear transformation is uniquely determined.

Is there an analogous concept for (unital, associative) algebras? Something like a minimal set of vectors such that specifying an algebra homomorphism on this set determines it for all others?

I'd be happy enough to answer this just for the tensor algebra, the symmetric algebra, or the exterior algebra, since these would be generally sufficient for my purposes.

Answer 1

Let's limit the analysis to algebras over a field $F$. An algebra $A$ is free if there exists a subset $\mathcal{B}$ of $A$ (a basis) with the property that,

for every algebra $C$ and every map $f\colon\mathcal{B}\to C$, there exists a unique algebra homomorphism $\hat{f}\colon A\to C$ with $\hat{f}(b)=f(b)$, for $b\in\mathcal{B}$.

If you change “algebra” into “vector space” and “algebra homomorphism” into “linear map”, you have the corresponding concept for vector spaces. In this case it turns out that every vector space is free.

For algebras (or groups, rings, and so on) this is false: free algebras have a very special role and most algebras aren't free.

One example is the tensor algebra $T(V)$ over a (finite dimensional, for simplicity) vector space $V$. Here it is true that a basis of the vector space becomes a basis of the tensor algebra. Essentially, no relation is imposed.

It cannot be true for the symmetric algebra $S(V)$, because of the commutative property: $yx=xy$. Suppose $\mathcal{B}$ is a basis and $C$ is an algebra. If $f\colon\mathcal{B}\to C$ is a map and $\hat{f}$ is its extension, we have \begin{align} \hat{f}(xy)&=\hat{f}(x)\hat{f}(y)=f(x)f(y) \\ \hat{f}(yx)&=\hat{f}(y)\hat{f}(x)=f(y)f(x) \end{align} so we must have $f(x)f(y)=f(y)f(x)$. Since the images of elements in $\mathcal{B}$ can be arbitrary, we see that $C$ should be commutative as well, which it may not be according to the definition. (Actually, one symmetric algebra can be free: the one having the empty set as basis, that is, the field $F$.)

Similarly, it cannot be true for the exterior algebra. Indeed, if two algebras are free over bases with the same cardinality, they are isomorphic. Can you sketch a proof?

The “symmetric algebra” example above shows that if we restrict the framework, we can generalize freeness: indeed the symmetric algebra over $V$ is free when we only deal with commutative algebras.

Answer 2

A linear map is completely determined by its action on a basis because a vector space is free module.

It's unlikely that there is a analogous concept for algebras because most interesting algebras are not free algebras.

An analogous result does hold for free algebras.

For instance, it holds for commutative finitely generated free algebras, that is, for polynomial algebras: if $A$ is a commutative algebra, then an algebra homomorphism $K[x_1, \dots, x_n] \to A$ is completely determined by the image of $x_1, \dots, x_n$.

If $A$ is not commutative, then an algebra homomorphism $K[x_1, \dots, x_n] \to A$ needs to map $x_1, \dots, x_n$ to pairwise commuting elements in $A$. This is the only restriction.

So, if $n=1$, you don't need to assume $A$ is commutative. The prime example of this is the map $K[x] \to L(V)$ induced by $x \mapsto T$.