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Let $k$ be a field and $R$ be the exterior ring over $k^d $, that is, $k$-algebra generated by elements $$x_1,\ldots,x_d,$$ where $$\ x_ix_j= - x_jx_i?$$

Is $R$ Artinian?

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Hint: is $R$ finite-dimensional over $k$? – Pete L. Clark Feb 28 '11 at 8:14
could you please explain more? – saba Feb 28 '11 at 8:17

To expand Pete's ḧint:

The key observation is that your $R$ is a finite dimensional $k$-vector space, and each of its left ideals is a vector subspace.

Now consider a decreasing chain of left ideals in $R$... It is in particular a decreasing chain of subspaces in a finite dimensional vector space: can it not stop?

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Actually, I am not sure the question is true as stated if $1 = -1$ in $k$!

If $2$ is invertible, then you get $x_i^2=0$ for all $i$. As suggested in the comments and Mariano's answer, the key point is $R$ has finite dimension as a $k$-vector space. Since the monomials span $R$, you want a lot of them to vanish. For example:

$$x_1x_2x_1 = -x_1x_1x_2= -x_1^2x_2 = 0$$

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Well, the OP says first that she wants $R$ to be the exterior ring of $k^d$ and second that "that is the $k$-algebra generated by..." As you say, if $k$ has characteristic $2$, then that is not the exterior algebra! So probably she means that $k$ does not have characteristic $2$. – Pete L. Clark Feb 28 '11 at 15:51
@Professor Clark, I do not know the definition of exterior algebra. I was using the OP's definition, hence the words "as stated". – curious Feb 28 '11 at 17:26

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