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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.

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  • $\begingroup$ Yes, for the tensor algebra; no for the others. $\endgroup$ – egreg Apr 30 '17 at 23:51
  • $\begingroup$ What is the minimal set for the tensor algebra? And why no for the others? $\endgroup$ – Mike Battaglia Apr 30 '17 at 23:55
  • $\begingroup$ This can't be right - the entire algebra itself would trivially count as such a set, and that has to be more info than you need. $\endgroup$ – Mike Battaglia May 1 '17 at 0:13
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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.

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  • $\begingroup$ Thanks, this is helpful! And I guess that likewise, if you only consider anticommutative algebras, the exterior algebra is free? Is there any nice way to represent an algebra homomorphism at all then? Matrices make it easy for the free case, but what if then it gets more complicated? $\endgroup$ – Mike Battaglia May 1 '17 at 18:46
  • $\begingroup$ @MikeBattaglia Yes, that's the idea. $\endgroup$ – egreg May 1 '17 at 19:39
  • $\begingroup$ One clarification here. Suppose we only care about commutative algebras, and we want to specify some homomorphism from some free commutative algebra onto one that isn't free. For instance, consider the quotient map from $R[x,y] \to R[x,y]/(x^2)$. This would then appear to be completely determined by its image on the set ${x, y, x^2}$. Having both $x$ and $x^2$ in there is redundant and means this isn't a basis for $R[x,y]$, but nonetheless this is sufficient to completely specify the homomorphism. Does something like this always hold when you're going from (but not to) free algebras? $\endgroup$ – Mike Battaglia May 2 '17 at 18:57
  • $\begingroup$ @MikeBattaglia Going from free algebras is easy: just assign arbitrarily the images of the basis elements. $\endgroup$ – egreg May 2 '17 at 19:45
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    $\begingroup$ @MikeBattaglia It seems we don't understand each other. If $B$ is a commutative $R$-algebra, then a homomorphism $R[x,y]\to B$ is completely and uniquely determined by assigning images to $x$ and $y$. You don't care what $B$ is; if it is $R[x,y]/(x^2)$ then there is no a priori connection with $x,y\in R$ and $x,y\in R[x,y]/(x^2)$ (which are actually $x+(x^2)$ and $y+(x^2)$, by the way). $\endgroup$ – egreg May 2 '17 at 20:03
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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 $x \mapsto T$.

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  • $\begingroup$ Thanks - I'm not seeing how a homomorphism on a polynomial ring is completely determined by its image on the generators though. For instance, consider the quotient of $K[x_1,...,x_n]$ by the ideal generated by $(x_1^3)$. Then the image of the associated homomorphism $h$ on ${x_1,...,x_n}$ is the same as the identity - you need to look at $h(x_1^3)$ to see a difference. Perhaps if we extend your statement to all monomials, rather than all generators, there's something there? $\endgroup$ – Mike Battaglia May 1 '17 at 0:45
  • $\begingroup$ I did not understand your example. $\endgroup$ – lhf May 1 '17 at 0:47
  • $\begingroup$ I may have misunderstood what you were saying. How are you defining "free algebra?" The definition here defines it as a non-commutative version of the polynomial algebra, which over the reals is the tensor algebra. But you are talking about the commutative free algebra, so there must be something else to it. $\endgroup$ – Mike Battaglia May 1 '17 at 1:09
  • $\begingroup$ Yes, the commutative free algebras are the free objects in the category of the commutative algebras. $\endgroup$ – lhf May 1 '17 at 2:03
  • $\begingroup$ And so for free algebras, your statement only applies for morphisms from one free commutative algebra to another? i.e. it applies to $R[x,y] \to R[x]$, but not from $R[x,y] \to R[x,y]/(y+1)$, or something like that? $\endgroup$ – Mike Battaglia May 1 '17 at 2:23

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