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Let $n$ be a natural number; I am studying (commutative) rings $R$ and $R$-algebras $A$ such that $A\cong R^n$ as $R$-modules. There is a universal such algebra: a ring $R_0$ and a free, rank-$n$ $R_0$-algebra $A_0$, such that if $(R, A)$ is any other such pair, there is a ring homomorphism $R_0\to R$ making $A\cong R\otimes_{R_0} A_0$.

The construction is as follows: Let $R_0$ be the quotient of the polynomial ring

$\mathbb{Z}[\eta^i,\alpha_{ij}^k:i,j,k\in\{1,\ldots,n\}]$

by the ideal generated by elements of the form

$\alpha_{ij}^k-\alpha_{ji}^k$,
$\displaystyle\sum_{m=1}^n \alpha_{ij}^m \alpha_{km}^\ell - \sum_{m=1}^n\alpha_{ik}^m \alpha_{jm}^\ell$, and
$\displaystyle\sum_{m=1}^n\alpha_{im}^j\eta^m - \delta_i^j$, as $i,j,k,$ and $\ell$ range from $1$ to $n$.

Then $A_0$ is the $R_0$-algebra $R_0[x_1,\ldots,x_n]/(1-\sum_m\eta^mx_m, x_ix_j - \sum_m\alpha_{ij}^mx_m)$; the relations ensure that $\{x_1,\ldots,x_n\}$ is a free $R_0$-basis for $A_0$.

My question is:

Can 2 be a zerodivisor in $R_0$, for some value of $n$?

For technical reasons, this would turn an ugly, intractable calculation for general rank-$n$ algebras into a nice, tidy one. Any other information about $R_0$ (is it a domain? how would I tell?) would be appreciated.

share|improve this question
    
Presumably, you are looking at commutative algebras, judging from the relation $\alpha_{ij}^k = \alpha_{ji}^k$. –  David Speyer Aug 29 '12 at 18:05
    
You're right - I work with commutative rings so much I forget there is another kind... I'll edit to make that explicit. :) –  Owen Biesel Aug 29 '12 at 18:08
    
Could you please make the question in body and title match. (Once it ask zero div once non-zero div. Sure a detail and likel this will not cause major confusion, but it could still be odd if two people answered yes and no, resp, both wanting to say the same thing.) –  quid Aug 29 '12 at 18:08
    
@quid: Good point! Thanks for catching that. –  Owen Biesel Aug 29 '12 at 18:11
    
I also think you want to require $n\neq 1$. –  darij grinberg Aug 29 '12 at 19:29
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