Global dimension of free algebra.

Question

Is there any easy way to see the global dimension of a free algebra $$ A=k\langle x_{1},\dots,x_{n} \rangle $$ is 1?

Answer 1

In "Modules over coproducts of Rings", Bergman proved that the global dimension of a free product is the supremum over the global dimensions of the factors (unless all factors have global dimension 0).

Answer 2

Putting noncommutativity aside for a moment: in the commutative case it's well known that the global dimension of this ring is $n$. You can get references for the ideas in the Wikipedia article on the topic.

Please see my other attempt.

Answer 3

Cognitive error and faulty instincts led to a bad solution, so I have a new one ready.

The free algebra you describe is a free ideal ring, and so all of its right and left ideals are projective. Thus it is left and right hereditary, and hence has global dimension less than or equal to 1. (And in fact equal to 1, since it's not semisimple.)