On the equivalence relation in a right Ore domain.

First, suppose $D$ is a domain, which isn't assumed to be commutative, such that for any nonzero $a,b$, there is a common nonzero right multiple, that is, $ab_1=ba_1\neq 0$. Consider the set $D\times D^\times$, where $D^\times$ is the set of nonzero elements of $D$. That it is defined that $(a,b)\sim (c,d)$ if for $b_1\neq 0$ and $d_1\neq 0$ such that $bd_1=db_1$ we have $ad_1=cb_1$.

What does it mean that this is independent of the choice of $b_1$? Does it mean that for any other pair $b_2,d_2\neq 0$ such that $bd_2=db_2$, then $ad_2=cb_2$ also? If so, why is this true? I've been trying to prove this, but I'm not sure if it's even the correct assumption.

-

The usual definition for the Ore relation matches yours, except it doesn't say "implies" it says "and": $(a,b)\sim (c,d)$ iff there exists $b',d'\in D$ such that $bb'=dd'\neq 0$ and $ab'=cd'$. (Requiring $b',d'$ to be nonzero turns out to be superfluous.)

So, you can also write the above condition as: $(a,b)\sim(c,d)$ if there exists $b_1,d_1$ such that $(ab_1,bb_1)=(cd_1,dd_1)$.

Maybe something about right Ore domains makes the two conditions equivalent, but I find this one easier to think about.

I'm a little puzzled by the "independence" comment, because it doesn't seem like independence plays any role. Two pairs are related or not, and "related in a way independent of particular choices" seems like a strange thing to tack on.

The next most important thing to do is to determine that addition and multiplication are compatible with this relation.

*Added: * Example verification of addition being well defined 1. (a,b)~(A,B) : (as,bs)=(AS,BS)

1. (c,d)~(C,D) : (ct,dt)=(CT,DT)

2. Addition of (a,b)+(c,d): Choose x,y and form (ax+cy, bx) where bx=dy

3. Addition of (A,B)+(C,D): Choose X,Y and form (AX+CY, BX) where BX=DY

4. SHOW (ax+cy, bx)~(AX+CY, BX)

5. Choose u, U such that bxu=BXU(=dyu=DYU)

6. (ax+cy,bx)~(axu+cyu, bxu)=(axu+cyu, BXU) by #6:

7. Choose v, V such that xuv=sV by (Ore): (axu+cyu, BXU)~(axuv+cyuv, BXUv)=(asV+cyuv, BXUv)

8. (asV+cyuv, BXUv)=(ASV+cyuv, BXUv) by #1.

9. Choose w,W such that yuvw=tW by (Ore): (ASV+cyuv, BXUv)~(ASVw+cyuvw, BXUvw)=(ASVw+ctW, BXUvw)

10. (ASVw+ctW,BXUvw)=(ASVw+CTW, BXUvw) by #2.

11. We proceed to show that (A[SVw]+C[TW], B[XUvw]) proves #5

12. Consider B(SVw-XUvw)=BSVw-BXUvw

=bsVw-BXUvw by #1

=bxuvw-bxuvw by #6 and #8=0

13. Since B is nonzero, we have that SVw=XUvw, and thus AS[Vw]=AX[Uvw]

14. Consider D(TW-YUvw)=DTW-DYUvw

=dtW-DYUvw by #2

=dyuvw-DYUvw by #10

=dyuvw-dyuvw=0 by #6

15. Since D is nonzero, we have TW=YUvw, thus CT[W]=CY[Uvw]

16. Then (ax+cy, bx)~(AX+CY, BX), because

(ax+cy, bx)~(ASVw+CTW, BXUvw) by #12

=(AX[Uvw]+CY[Uvw], BX[Uvw]) by #14 and #16

~(AX+CY, BX)

The thought of working this out for multiplication and distributivity makes me nauseous, and it would also be a bear to work it out for Ore rings which aren't domains!

-
Thanks rschwieb. I'm using your definition of $\sim$, and I see that $\sim$ is an equivalence relation, and I'm still assuming the common right multiple property. I have addition defined as $a/b+c/d=(ad_1+cb_1)/m$ where $m=bd_1=db_1\neq 0$, and $(a/b)(c/d)=ac_1/db_1$ where $cb_1=bc_1$ and $b_1\neq 0$, but I've struggled and don't know how to show these are well-defined. Do you know how to show that they are? – Camilla Vaernes Jun 30 '12 at 6:46
@CamillaVaernes Most authors admit that this verification is quite an exercise in patience :) I'm willing to try one at a time with you... which one do you want to see? – rschwieb Jun 30 '12 at 12:04
@CamillaVaernes Hi here, sorry for the delay. I put a sample computation above. Tangoing with the raw definition really makes one want to use a more technical way to do it! – rschwieb Jul 2 '12 at 19:37
Dear rschwieb sorry for my delay! I hadn't logged in for months. Thank you for writing that out. – Camilla Vaernes Feb 5 '13 at 21:02