Terminology:
- Since I want to consider only unital, associative algebras, algebra will be unital and associative unless otherwise stated.
- An ideal will be a two-sided ideal unless otherwise stated.
- Vector spaces will be over an infinite field $\mathbb F$ with characteristic 0.
Background:
Let $V$ be a finite dimensional vector space and let $T(V)$ be its tensor algebra. It is often said that $T(V)$ is the "most general" algebra that contains $V$ as a subspace. This can also be seen by noting that intuitively $T(V)$ is the algebra generated by $1,e_1,…,e_n$ ($e_1,…,e_n$ is a basis of $V$) in a way that $1$ is the unit, the multiplication is associative but from that point on, no additional relations are assumed to hold about the generators.
The tensor algebra also observes the universal property, namely that if $A$ is an algebra, $\phi :V\rightarrow A$ is a linear map, then there exists a unique algebra homomorphism $\phi_\otimes:T(V)\rightarrow A$ that satisfies $\phi=\phi_\otimes\circ i$ where $i:V\rightarrow T(V)$ is the inclusion.
I would like to understand that in what sense the tensor algebra is the most general algebra containing $V$. In particular, this can mean one of several things:
Hypothesis 1: There is no such algebra $A$ that contains $V$ as a subspace and $T(V)$ is a sub algebra of $A$.
Hypothesis 2: There is no such algebra $A$ that contains $V$ as a subspace and it has an ideal $I$ such that $T(V)=A/I$.
Hypothesis 3: If $A$ is an algebra that contains $V$ as a subspace, then there is an ideal $I\vartriangleleft T(V)$ of $T(V)$ such that $A=T(V)/I$.
Hypothesis 3 seems at least believable, since pretty much all famous algebras containing $V$ arise as quotients of the tensor algebra (the exterior algebra, the symmetric algebra, the Clifford and Weyl algebras, the universal enveloping algebra if $V$ is a Lie algebra etc.).
On the other hand, I am not convinced that the universal property stated above actually implies Hypothesis 3.
In particular, assume that $A=T(V)/I$. If $A$ is to contain the entirety of $V$ then we must have $I\cap V=0$. I don't know if $T(V)$ has a nontrivial ideal that intersects $V$ nontrivially or not, but nonetheless let's make this assumption. Then the quotient map $\pi:T(V)\rightarrow T(V)/I$ is surjective. Moreover if $\pi:T(V)\rightarrow A$ is a surjective algebra morphism then $\ker\pi$ is an ideal of $T(V)$ and $A=T(V)/\ker\pi$.
If the universal property is to imply Hypothesis 3, then we must be able to show that if $\phi:V\rightarrow A$ is injective (i.e. the image of $V$ in $A$ is faithful), then $\phi_\otimes :T(V)\rightarrow A$ must be surjective. The universal property does guarantee that the image of $V$ via $\phi_\otimes$ is faithful but to have $A=T(V)/\ker\phi_\otimes$, we must have surjectivity. I don't see it being implied.
Question:
Are any of the hypotheses presented above true? If not, then in what sense is the tensor algebra the most general algebra containing $V$? The "raw" statement of the universal property does not convince me that $T(V)$ is "the most general", what are some additional implications of the universal property that makes $T(V)$ "the most general"?
Best Answer
All three of your hypotheses are false.
The correct statement is that any algebra $A$ equipped with a map $i : V \to A$ (which need not be injective) and which is generated by the image of $V$ must be a quotient of $T(V)$. This follows directly from the universal property.
Loosely speaking, the tensor algebra is the "most general" algebra containing $V$ and nothing "extra." There are ways in which "most general" is a bit misleading, as you've discovered; the way I prefer to describe it is that the tensor algebra is the "laziest" way to build an algebra from $V$, in the sense that you don't impose any conditions on how elements multiply and add except the ones forced on you by the axioms. "Free-est" would also be appropriate, which is why it would also be appropriate to call $T(V)$ the free (associative) $k$-algebra on $V$, and why we call constructions like this free objects in general.
Nothing I've just said requires any hypotheses on the base field $k$, and in fact everything applies without modification to a module $V$ over an arbitrary commutative ring $k$.