# Global dimension of the tensor algebra

Let $$R$$ be a semisimple ring with a non-zero $$R$$-bimodule V. Let $$T_R(V):= \bigoplus\limits_{k=0}^{\infty}{V^{\otimes_k}}$$ be the tensor algebra of $$V$$.

Question 1: Is there a simple proof that $$T_R(V)$$ has global dimension equal to one? Note that this contains as a special case that all non-trivial finite quiver algebras have global dimension equal to one. For acyclic quivers this is well known and can be found in most textbooks, but for non-acyclic quivers I have not seen it in a textbook yet.

Note that in case Question 1 of Quadratic algebras and Koszul algebras has a positive answer, a quick proof that the global dimension is equal to one would follow. But I am not sure whether we have that the global dimension of $$T_R(V)$$ is equal to $$\sup \{ i \geq 0 | Ext_{T_R(V)}^i(R,R) \neq 0 \}$$

Question 2: Is there a more general formula for the global dimension (or even the finitistic dimension) of $$T_R(V)$$ when $$R$$ is an arbitrary ring and $$V$$ an arbitrary $$R$$-bimodule? When is the global dimension finite?

• For non-acyclic quivers you can look at Mitchell's rings with several objects to get the Hochschild dimension is 1 which implies global dimension is 1 over a field or semisimple ring. – Benjamin Steinberg Dec 11 '18 at 14:57
• A proof of an arbitrary path algebra having global dimension at most 1 is given in "Multiplicative Bases, Gröbner Bases, and Right Gröbner Bases" by Edward L. Green, Corollary 5.5. – Oeyvind Solberg Dec 12 '18 at 10:33

To answer Question 1, this can be proved in a similar manner to the case of a path algebra of a quiver, where one exhibits a "standard resolution" of a general module over the path algebra.

I will denote $$A = T_R(V)$$. To begin, the kernel of the multiplication map $$\mu \colon A \otimes_R A \to A$$ is known to be isomorphic to $$A \otimes_R V \otimes_R A$$. For instance, see Propositions 2.5 and 2.6 of the paper Algebra extensions and nonsingularity by Cuntz and Quillen. This provides a short exact sequence of $$(A,A)$$-bimodules $$0 \to A \otimes_R V \otimes_R A \overset{j}{\to} A \otimes_R A \overset{\mu}{\to} A \to 0.$$ (I imagine that with enough patience, one can show that $$j$$ is the usual map induced by $$j(1 \otimes v \otimes 1) = v \otimes 1 - 1 \otimes v$$ as in the path algebra case, but this will not be necessary to compute the global dimension.) Note that the surjection $$\mu$$ is split as a morphism of right $$A$$-modules by the map $$a \mapsto 1 \otimes a$$. Thus, when considered as a sequence of right $$A$$-modules, this is a split exact sequence.

Now consider an arbitrary left $$A$$-module $$M$$. Becasuse the sequence is right-split, it remains exact after applying the functor $$- \otimes_A M$$, yielding the short exact sequence of left $$A$$-modules $$0 \to A \otimes_R V \otimes_R M \to A \otimes_R M \to M \to 0.$$ (If $$j$$ above is shown to be the expected map, then this would be the precise analogue of the usual standard resolution for a quiver algebra.) Because every left $$R$$-module is projective (as $$R$$ is semisimple), and because extension of scalars $$A \otimes_R -$$ preserves projectivity of a module, the sequence above is in fact a projective left $$A$$-module resolution of $$M$$. So $$M$$ has projective dimension at most 1.

Thus $$T_R(V) = A$$ has left global dimension at most 1, and similarly for its right global dimension (by the symmetric argument). If $$V = 0$$, then in fact $$T_R(V) = R$$ still has global dimension 0. But if $$V \neq 0$$, then I suppose one can show that the left (or right) module surjection $$T_R(V) \twoheadrightarrow T_R(V)/\left(\bigoplus_{k \geq 1} V^{\otimes k}\right) \cong R$$ is not split, so that the global dimension is in fact equal to 1.

Question 2 seems more difficult to answer, but I would certainly be interested to know any information, even simply partial answers! It seems to me that the sequence for $$M$$ above is still exact for general $$R$$, $$V$$, and $$M$$. (I could be slightly off, as the Cuntz-Quillen results assume that all rings are algebras over a field, but I suspect their proofs are likely to still work.) In this case, I wonder whether in certain situations we may be able to control how the global dimensions behave when passing from $$M$$ (considered as a left $$R$$-module) to the $$A$$-modules $$A \otimes_R M$$ and $$A \otimes_R V \otimes_R M$$.

Concerning Question 2: If $$R$$ is any ring, and $$V$$ is an $$R$$-bimodule, and $$R$$ is a projective left (or right) $$R$$-bimodule. Then the left global dimension of $$T_R(V)$$ is either the left global dimension of $$R$$ or $$l.gl.dim(R)+1$$. Similarly for r.gl.dim.

This is Theorem 2.2.11 of Hazewinkel, Gubareni, Kirichenko, Algebras, Rings and Modules, Vol. 2.

N.B. The Cuntz-Quillen fact mentioned in Manny Reyes' answer is also true in this generality, see Lemma 2.2.10.

• This is a great reference! – Manny Reyes Mar 27 '19 at 18:25
• I just slightly edited my answer: V need not be projective as a bimodule, only as left or right R-module (this is a weaker condition). – nikola karabatic Apr 5 '19 at 13:23