2

Question: Let $S$ be a commutative regular ring with unity. Let $A$ and $B$ be two subrings of $S$ which are also regular. Is the subring $A\cap B$ a regular subring of $S$?

Suppose we take a non-zero non-unit $x\in A\cap B$. Then we need to show that there exists a $k\in A\cap B$ such that $x^2k=x$. Now since $A$ and $B$ are itself regular, we have $x^2y=x$ and $x^2z=x$, (wlog) for some $y\in A\setminus B$ and $z\in B\setminus A$. A simple calculation shows that $xy=xz\in A\cap B$. I am stuck here. I was trying to use the elements $x-xy$,$x-xz$, and regularity of whole $S$, but was unable to conclude anything.

Dots_and_Arrows
  • 1,131

1 Answers1

1

You have to multiply more! We have

$$x^2 (xyz) = x^2 z = x$$

and $xy = xz \in A \cap B$ implies that $xyz = (xy)z = xz^2 = (xz)y = xy^2 \in A \cap B$ also.

This is an elementary calculation but the choice to look at $xyz$ can be motivated, as I mentioned in a recent question on vNr (for von Neumann regular) rings, by looking at the example where $S$ is a product of fields. There is also a less direct argument that proceeds by showing that for commutative vNr rings we can replace the condition $xyx = x$ in the definition with the stronger condition that $xyx = x$ and $yxy = y$, and with this stronger condition $y$ is unique; see e.g. here. Uniqueness immediately solves this problem since with this stronger definition we necessarily have $y = z$.

Qiaochu Yuan
  • 468,795
  • I like the stronger argument. Would you tell me why $xy²$ and $xz²$ belongs to $A\cap B$. I may be missing some subtle observation here. Since y and z may not be in $A\cap B$. – Dots_and_Arrows Aug 03 '24 at 04:17
  • 1
    @Dots: no subtle observation necessary: since $x \in A, y \in A$ we have $xy^2 \in A$. And since $x \in B, z \in B$ we have $xz^2 \in B$. So $xz^2 = xy^2 \in A \cap B$. – Qiaochu Yuan Aug 03 '24 at 04:22