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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?

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4 Answers 4

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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).

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That's a useful criteria. Although I wanted to know an elementary way to see $\mathrm{gl.dim}(A)$, it is good to know the fact. Thanks. –  Michel Aug 7 '12 at 17:21
    
Maybe this is another "big hammer": The global dimension of a graded algebra is the global dimension of $k$ as a module over the algebra. So you need only show that $k$ has a projective resolution of length 1. This isn't too hard to do. –  J. Gaddis Aug 7 '12 at 20:30
    
That's something I couldn't find. –  Michel Aug 11 '12 at 0:33
    
There's a proof in "Some algebras associated to automorphisms of elliptic curves" by Artin, Tate, and Van den Bergh. –  J. Gaddis Aug 11 '12 at 21:27

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.)

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Thanks for the proof ^_^. –  Michel Aug 12 '12 at 6:20
    
@Michel Feel free to thank answers you like with "accept" or at least a vote. People can see your accepted-answer to asked-question ratio, in case you didn't know. –  rschwieb Aug 12 '12 at 11:22

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.

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Yes, in commutative case we have lots of tools and things are much easier. –  Michel Aug 7 '12 at 17:14
    
@Michel Well, I hope you have seen my point that the result you want to prove is false if $n\neq 1$. If you had a proof for (possibly noncommutative) rings having this property, then it would apply to commutative rings. Noncommutatve$\neq$not-commutative! –  rschwieb Aug 7 '12 at 18:17
    
Sorry, but I don't quite understand your comment. This is true as linearfish answers below. –  Michel Aug 11 '12 at 0:36
    
@Michel OK I have to eat my words! While I don't see that linearfish's solution is a complete one, I do see other references on the net saying this is true. I can see now that leaving out commutativity relations might allow this to occur. I'm looking into the problem again. –  rschwieb Aug 11 '12 at 13:12

The Hochschild Cohomological dimension forms an upper bound for the Global dimension of a (unital associative) ring (such as the free $k$-algebra).
Since any polynomial ring is isomorphic to a tensor algebra and the latter must be of Hochschild cohomological dimension $1$ (See Algebra Extensions and Nonsingularity), the result follows (even in the infinitely generated case :) .)

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