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Let $R$ be a ring (assumed associative and unital) whose additive group is a finitely generated abelian group. As a reduction step in a paper I'm working on, we need to know that $R$ is a quotient of another ring $S$ whose additive group is a finite rank free abelian group. We believe we have a proof, but it is rather involved. Is this something that has appeared already or at least has a reasonably short proof? (The current proof reduces to showing it for finite rings and then proves it by using the basic ring associated to $R$.)

When $R$ is commutative, this has a quick proof: $R$ is generated by $r_1,\dots,r_n$ which are integral over ${\bf Z}$, and hence $R$ is a quotient of ${\bf Z}[x_1,\dots,x_n]/(f_1(x_1),\dots,f_n(x_n))$ where each $f_i$ is a monic univariate polynomial.

So I'm wondering if there is something analogous in the general case.

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    $\begingroup$ It may be good to say at the start that $R$ is "an associative ring" (with identity?) rather than just "a ring". $\endgroup$
    – nfdc23
    Jun 1, 2017 at 15:00

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For a finite ring $R$ you can write it as a quotient of the monoid ring $\mathbb ZR$ (with respect to the multiplicative structure) which has a finitely generated free additive group.

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  • $\begingroup$ For the second part, I write $1=t+f$ where $t$ is torsion and $f$ belongs to the torsionfree part. Are you saying that $t$ is an idempotent? Could you please explain why? $\endgroup$
    – Steven Sam
    Jun 1, 2017 at 17:34
  • $\begingroup$ If you square then since the torsion elements are an ideal you get $f^2=f$. Therefore $t=1-f$ is idempotent. I did forget to check it is central $\endgroup$ Jun 1, 2017 at 17:52
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    $\begingroup$ Either way, doing the finite case was where our proof was more inefficient, so you've already helped trim the proof substantially with the monoid ring observation. $\endgroup$
    – Steven Sam
    Jun 1, 2017 at 18:14
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    $\begingroup$ $t$ can be 0. That is true. But $f$ and $t$ are idempotent. If $f$ Is 0 this is clear. Else $1=1^2=f^2+tf+ft+t^2$ and $tf+ft+t^2$ is torsion. So $f^2$ must be the torsion-free part of $1$. Thus $f^2=f$. $\endgroup$ Jun 1, 2017 at 18:21
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    $\begingroup$ I see that the above argument is slightly off because $f^2$ could have a further torsion part. $\endgroup$ Jun 1, 2017 at 18:32

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