- Thread starter
- Banned
- #1

Secondly, I wish to prove this: Let R be an integral domain and let a,b be in R. Then aR=bR iff a=bu for some unit u of R.

- Thread starter Poirot
- Start date

- Thread starter
- Banned
- #1

Secondly, I wish to prove this: Let R be an integral domain and let a,b be in R. Then aR=bR iff a=bu for some unit u of R.

- Jan 29, 2012

- 661

Hint: If $xy$ is a unit, then there exists $z\in R$ such that $(xy)z=1$, that is $x(yz)=1$ so ...let x,y be in a ring such that xy is a unit. Does this imply that both x and y are units?

Hint: If $aR=bR$, then there exists $u, v\in R$ such that $a=a\cdot 1=bu$ and $b=b\cdot 1=av.$ This implies $a(1-vu)=0.$ Now, use that $R$ is an integral domain.Secondly, I wish to prove this: Let R be an integral domain and let a,b be in R. Then aR=bR iff a=bu for some unit u of R.

- Thread starter
- Banned
- #3

Second proof: Suppose that aR=bR. Since a ∈ aR and b ∈ bR, a ∈ bR and b ∈aR. That is, a=bu and b=av for some u and v in R. Hence a=(av)u=a(vu). Now, if a is zero, then aR={0} so b= a=a.1=0 and the theorem is true. So assume a is non-zero so a=a(vu) implies vu=1. Hence u (and v) are units and we are done.

Conversly, suppose a=bu for some unit u of the ring. Then given ar ∈ aR, ar=(bu)r=b(ur) ∈ bR so aR is a subset of bR. Similarly, given br ∈ Br, $br=a(u^{-1}r) ∈aR$ so Br is a subset of aR and the proof is complete.

- Moderator
- #4

- Feb 7, 2012

- 2,771

That argument shows that if $xy$ is a unit then $x$ has a right inverse. In a commutative ring, that is sufficient to show that $x$ is a unit. But in a noncommutative ring it is possible to have two non-units whose product is a unit.Hint: If $xy$ is a unit, then there exists $z\in R$ such that $(xy)z=1$, that is $x(yz)=1$ so ...let x,y be in a ring such that xy is a unit. Does this imply that both x and y are units?

The simplest example that I know of is in the ring of endomorphisms of an infinite sum of abelian groups. Take $G$ to be any nontrivial abelian group, with neutral element 0, and let $H = G\oplus G\oplus G\oplus \ldots.$ Let S be the "right shift" endomorphism on $H$, given by $S(g_1,g_2,g_3,\ldots) = (0,g_1,g_2,\ldots).$ Let T be the "left shift" endomorphism on $H$, given by $T(g_1,g_2,g_3,\ldots) = (g_2,g_3,g_4,\ldots).$ Then $TS$ is the identity map, but $ST$ is not invertible and neither $S$ nor $T$ is a unit.

- Jan 29, 2012

- 661

Well, although I didn't mention it, I considered a commutative ring.In a commutative ring, that is sufficient to show that $x$ is a unit.

- Thread starter
- Banned
- #6

- Moderator
- #7

- Feb 7, 2012

- 2,771

Hmm, that is not part of the usual definition.I consider rings to be commutative by definition.

- Thread starter
- Banned
- #8