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Let $k$ be a field of characteristic $0$. Consider the following $k$-algebra $R$, which is the quotient of a tensor algebra generated by elements $x_i$ in degree $1$ with the relation $x_ix_j=-x_jx_i$ when $i\neq j$.

In other words, $x_i^2$ does not vanish. Has anyone studied this graded ring? Does this graded ring have finite homological dimension? Is there a proof of Hilbert's syzygy theorem that can be doctored to apply to it?

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Presumably the right side of your relation "$x_ix_j=-x_ix_j$" should be "$-x_jx_i$"? – John Palmieri Apr 9 '10 at 19:58
Is your base ring a characteristic zero field? – S. Carnahan Apr 9 '10 at 21:36
Yes, why is the characteristic zero important? – Daniel Pomerleano Apr 9 '10 at 22:01
Edited to reflect the information in the comments and add LaTeX. – Harry Gindi Apr 10 '10 at 13:56

4 Answers 4

I do not know the answer to your first question. As for the next two the answer is positive; one need only slightly modify standard proofs for the usual polynomial ring:

If $R$ is a graded ring then we define the twisted polynomial ring $R[x]_t$ as the graded ring generated by $R$ and $x$ with $\deg x=1$ and the relations $rx=(-1)^{|r|}xr$ where $r$ is homogeneous of degree $|r|$. Your ring is then $k[x_1]_t[x_2]_t\ldots$ (I assume you meant to have a field as base ring) and in general an element of $R[x]_t$ can be uniquely written in the form $\sum_nr_nx^n$. We can now prove that $R[x]_t$ is (left) Noetherian if $R$ is. Let $I\subseteq R[x]_t$ be an ideal and let $I'\subseteq R$ be the ideal of top terms of $I$. Picking generators for $I'$ and representing polynomials for them gives a finite number of element of $I$ such that any element of $I$ can be reduced to a polynomial of fixed degree.

As for finite global dimension by general results it suffices to show that $k$ has a finite resolution. Again we can use induction and assume that $k$ has a finite resolution as $R$-module and then it suffices to show that $R$ has a finite resolution as $R[x]_t$-module. This is done by considering $R[x]_te \to R[x]_t$ with $e$ mapped to $x$. Carrying the induction through gives an explicit resolution which is a sign-twisted version of the Koszul resolution: In homological degree $k$ it has an $R[x]_t$-basis $e _{i_1}e_{i_2}\cdots e_{i_k}$ where $1\leq i_1 < i_2 < \cdots < i_k\leq n$ with $d(e_{i_1}e_{i_2}\cdots e_{i_k})=\sum_r x_{i_r}e_{i_1}\cdots \widehat{e_{i_r}}\cdots e_{i_k}$ (Look Ma no signs!).

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This algebra $R$ is a quadratic Koszul algebra (in fact, it is easily seen to be PBW, that is has a quadratic Groebner basis), from which one can immediately construct a bimodule resolution; the bimodule generators of that resolution come from the Koszul dual algebra $R^!$, which is of finite dimension ($2^n$), hence the resolution is of finite length, and $R$ has finite homological dimension.

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Your ring is what's called a quantum polyniomial ring at $q=-1$, and are quite well studied objects. In particular, Wambst constructed explicit bimodule resolutions for them obtained by "appropriately" modified the usual Koszul complex. The resolution thus obtained is of finite length, and as immediate a corollary of that it turns out the algebra is of finite global (left and right) dimension.

I can give you a precise reference when I get to a place with MathSciNet access.

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I believe you are referring to: MR1252939 (95a:17023) Wambst, Marc Complexes de Koszul quantiques. [Quantum Koszul complexes] Ann. Inst. Fourier (Grenoble) 43 (1993), no. 4, 1089--1156. – Vladimir Dotsenko Apr 10 '10 at 17:12
That's precisely it. Thanks! – Mariano Suárez-Alvarez Apr 10 '10 at 19:06

Hi, if you have n variables $x_i$, then the global dimension is n over any field (since the ground field K has a free graded resolution of length exactly n). The Gelfand-Kirillov dimension is then $n$ + GKdim(F) where F is your field (note that GKdim(F) = tr.deg F). This algebra can be called an anticommutative algebra or, in a broader context graded commutative algebra. It is also a G-algebra, it has Poincare-Birkhoff-Witt basis, is Noetherian domain and so on. Short proofs of dimensions can be found in my PhD thesis (2005), available on the net (e.g. there's a link on my homepage). There you will find the answer for your 3rd question: yes, I prove the global dimension by a generalized Hilbert syzygy theorem (which applies to far more general algebras than this neat one). Moreover, I suspect that the (generalized) Krull dimension of this ring is n as well. Computations with modules over this algebra are constructive with e.g. freely available SINGULAR (

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