Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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
add comment

2 Answers 2

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?

share|improve this answer
add comment

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:

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

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.