# If a commutative ring is semiprime and its prime ideals are maximal then it is von Neumann regular (absolutely flat).

If a commutative ring is semiprime and its prime ideals are maximal then it is von Neumann regular (absolutely flat).

The converse, although not immediately apparent, can be proven quite easily. But I've been stuck on this one for awhile.

I found that Bourbaki has a question in his treatise on commutative algebra, namely: if all points in Spec(A) are closed (ie. prime ideals are maximal) then A/R (where R is the nilradical) is von Neumann regular (absolutely flat). If this can be proven, the above statement follows immediately since the nilradical of a semiprime ring 0?

Ideas?

Ps: this is my first time posting a question so apologies for any formatting issues.

• You threw me way off with the second line. I thought you were stuck on the converse. So you are actually asking about the title question? – rschwieb Nov 15 '12 at 20:45
• @user49691 Page 96. It is available in googlebooks. Also, it is not J. Lambek's book. It is T.Y. Lam's book. It's my fault for accidentally transposing the words. – rschwieb Nov 16 '12 at 18:46

In a reduced commutative ring with $\dim R=0$ every ideal is radical. (This follows easily from the fact that $R_{\mathfrak{m}}$ is a field for every maximal ideal $\mathfrak{m}$ of $R$.) In particular, for an element $a\in R$ we have that $aR=a^2R$, so $R$ is VNR.

• That's pretty neat, although it looks like it passed the buck to proving "reduced commutative with dimension 0 means all ideals are radical" – rschwieb Nov 15 '12 at 20:54

(Thanks Dr. Lam's wonderful Lectures on Modules and Rings for this proof.)

Suppose $R$ is commutative semiprime and has dimension zero.

Then if $M$ is a maximal ideal, the localization $R_M$ is a local ring with dimension zero also. Since $R$ is reduced, so is $R_M$. But $Nil(R_M)=M=0$, so $R_M$ is a field.

For all maximal ideals $M$, $(aR/a^2R)_M\cong (aR)_M/(a^2R)_M\cong aR_M/a^2R_M$, but since $R_M$ is always a field, we can see that the last term is always zero.

Since then $(aR/a^2R)_M=0$ for all maximal ideals, it follows that $aR/a^2R=0$, that is, $aR=a^2R$.

• Nice, could you give me the page number for the Lambek proof? I'm quite lazy, flipping pages is annoying. – user49691 Nov 16 '12 at 17:27
• @user49691 I put the answer to this in a comment to the original post. Enjoy! – rschwieb Nov 16 '12 at 19:16