# Integral domain and unit of a ring

• Oct 29th 2009, 10:33 AM
knguyen2005
Integral domain and unit of a ring
Lt R be a commutative ring not neccesarily an integral domain. Determine the truth or falsity of the following statements. In each case, give a proof or find a counter-example

Let x, y belongs to R
(i) x.y in R* ---> x in R* , y in R* (R* = R\{0})
(ii) x.y in U(R) ---> x in U(R), y in U(R)
(iii) x.y in R*\ U(R) ---> x in R*\ U(R), y in R*\U(R)

This is my attempt

(i)If x.y in R* then we have x.y not equal to 0 implies that x is nonzero and y is nonzero. So, the statement is true

(ii)I think this statement is false but I dont know how to prove it

(iii)This part I dont know

thank you
• Oct 29th 2009, 11:12 AM
clic-clac
Quote:

(i)If x.y in R* then we have x.y not equal to 0 implies that x is nonzero and y is nonzero. So, the statement is true
Yes

Quote:

(ii)I think this statement is false but I dont know how to prove it
Well $x.y\in U(R)$ means there is $a\in R$ such that $a.(x.y)=1=(x.y).a\ .$ The multiplication in a ring is associative; so what can you deduce from these equalities?

Quote:

(iii)This part I dont know
Try to see what happens if you take $y=1.$
• Oct 29th 2009, 11:29 AM
knguyen2005
(ii)I think this statement is false but I dont know how to prove it

what can you deduce from these equalities?

It implies that xy is the inverse of a, and since xy in U(R) then x in U(R) and y in U(R). Hence the statement is true
• Oct 29th 2009, 11:38 AM
jackie
Quote:

Originally Posted by knguyen2005
(ii)I think this statement is false but I dont know how to prove it

what can you deduce from these equalities?

It implies that xy is the inverse of a, and since xy in U(R) then x in U(R) and y in U(R). Hence the statement is true

I don't think your argument is valid here. I think you're supposed to deduce that $x(ya)=1=(ya)x$ and $y(xa)=1=(xa)y$. You can do this since $R$ is a commutative ring. From those equalities, you get $x,y \in U(R)$
• Oct 29th 2009, 12:05 PM
Swlabr
Quote:

Originally Posted by clic-clac
Try to see what happens if you take $y=1.$

You can't take $y=1$, as this is a unit. We are looking at the non-zero ring element which are not units.
• Oct 29th 2009, 12:21 PM
clic-clac
What we want to prove is the assertion is wrong, i.e. $x.y$ can be non invertible with $x$ or $y$ invertible.

With $y=1,$ it becomes clear that whenever there are non zero non invertible elements in $R,$ then the implication is false.

Edit: Quite close indeed :)
• Oct 29th 2009, 12:21 PM
Swlabr
Quote:

Originally Posted by clic-clac
What we want to prove is the assertion is wrong, i.e. $x.y$ can be non invertible with $x$ or $y$ invertible.

With $y=1,$ it becomes clear that whenever there are non zero non invertible elements in $R,$ then the implication is false.

I know - I'm having a slow evening, and I couldn't get to my laptop in time to change it!

EDIT: Although, by the timing of this post it looks like I was pretty close!