1. ## The unity ..

Hi all ,
i want to ask you about the unity .. what do we mean about " this ring has unity and that ring without unity.."

and how i can find the all units in " ring whit unity" for example Z10

sorry for my bad English language..
with my respect to you..

2. Originally Posted by Miral
Hi all ,
i want to ask you about the unity .. what do we mean about " this ring has unity and that ring without unity.."

and how i can find the all units in " ring whit unity" for example Z10

sorry for my bad English language..
with my respect to you..
You of course know that every ring $\displaystyle R$ has two operations, one multiplicative one additive. Since $\displaystyle R$ under addition must be an abelian group there must be some additive identity $\displaystyle 0\in R$ such that $\displaystyle 0+a=a+0=a$ for all $\displaystyle a\in R$. There need not be a multiplicative identity $\displaystyle 1\in R$ such that $\displaystyle a\cdot 1=1\cdot a=a$.

If it has this multiplicative identity it is said to have unity, if it doesn't it is said to be a ring without unity.

The units of $\displaystyle \mathbb{Z}_{10}$ from my limited experience actually refers to those $\displaystyle a\in\mathbb{Z}_{10}$ such that $\displaystyle (a,10)=1$

3. As Drexel 28 said, the "unity" in a ring is the multiplicative identity. A "unit", on the other hand, is a member of the ring that has a multiplicative inverse (which, of course, is only possible in a ring with unity).
In $\displaystyle Z_{10}$, the "unity" is 1 and numbers, n, for which (n, 10) (the least common multiple) is not 1 do not have inverses. Since 10= 2(5), the only possible "least common multiples" are 1, 2, 5, and 10. If (n, 10)= 2, then n= 2a so 5(n)= 5(2a)= 10a= 0 (mod 10). Similarly, if (n, 10)= 5, then n= 5a so 2(n)= 2(5a)= 10a= 0 (mod 10). Of course if (n, 10)= 10, n is itself equivalent to 0 (mod 10).

If mn= 0 with m and n both non-zero, in any ring, then n and m cannot have inverses- if n had an inverse, $\displaystyle n^{-1}$ then we would have $\displaystyle (mn)n^{-1}= m= 0(n^{-1}= 0$, a contradiction.

4. Originally Posted by Drexel28
You of course know that every ring $\displaystyle R$ has two operations, one multiplicative one additive. Since $\displaystyle R$ under addition must be an abelian group there must be some additive identity $\displaystyle 0\in R$ such that $\displaystyle 0+a=a+0=a$ for all $\displaystyle a\in R$. There need not be a multiplicative identity $\displaystyle 1\in R$ such that $\displaystyle a\cdot 1=1\cdot a=a$.

If it has this multiplicative identity it is said to have unity, if it doesn't it is said to be a ring without unity.

The units of $\displaystyle \mathbb{Z}_{10}$ from my limited experience actually refers to those $\displaystyle a\in\mathbb{Z}_{10}$ such that $\displaystyle (a,10)=1$

Originally Posted by HallsofIvy
As Drexel 28 said, the "unity" in a ring is the multiplicative identity. A "unit", on the other hand, is a member of the ring that has a multiplicative inverse (which, of course, is only possible in a ring with unity).
In $\displaystyle Z_{10}$, the "unity" is 1 and numbers, n, for which (n, 10) (the least common multiple) is not 1 do not have inverses. Since 10= 2(5), the only possible "least common multiples" are 1, 2, 5, and 10. If (n, 10)= 2, then n= 2a so 5(n)= 5(2a)= 10a= 0 (mod 10). Similarly, if (n, 10)= 5, then n= 5a so 2(n)= 2(5a)= 10a= 0 (mod 10). Of course if (n, 10)= 10, n is itself equivalent to 0 (mod 10).

If mn= 0 with m and n both non-zero, in any ring, then n and m cannot have inverses- if n had an inverse, $\displaystyle n^{-1}$ then we would have $\displaystyle (mn)n^{-1}= m= 0(n^{-1}= 0$, a contradiction.

thank you Drexel28 and HallsofIvy

the multiplicative identity in the ring = The unity

the units in $\displaystyle {Z}_{10}$ is 1, 3 , 7 , 9
since $\displaystyle {Z}_{10}$ = {1,2,3,4,5,6,7,8,9}
and gcd(1,10)=1 & gcd(3,10)=1 & gcd(7,10)=1 & gcd(9,10)=1

and i have $\displaystyle u$*$\displaystyle u^{-1}$= $\displaystyle u^{-1}$*$\displaystyle u$ = unity

and since the unity in $\displaystyle {Z}_{10}$ =1

Thus

u=1 ----> $\displaystyle u^{-1}$= 1
u=3 ----> $\displaystyle u^{-1}$= 7
u=7 ----> $\displaystyle u^{-1}$= 3
u=9 ----> $\displaystyle u^{-1}$= 9

right?

5. Originally Posted by Miral

u=1 ----> $\displaystyle u^{-1}$= 1
u=3 ----> $\displaystyle u^{-1}$= 7
u=7 ----> $\displaystyle u^{-1}$= 3
u=9 ----> $\displaystyle u^{-1}$= 9

right?
I would agree with that. Notice that in terms of groups the units of $\displaystyle \mathbb{Z}_{n}$ form a group under modular multiplication. This can be seen from above, which is quite a nice Cayley table .

I'm fairly sure that this extends to the general case, but not being a ring theorist, and not (at the moment) being too interested I haven't checked. In particular, if $\displaystyle \left(R,+,\cdot\right)$ is a ring, then $\displaystyle \left(U(R),\cdot\right)$ where $\displaystyle U(R)$ are the units of $\displaystyle R$ form a group that is not necessarily abelian.

6. Originally Posted by Drexel28
I would agree with that. Notice that in terms of groups the units of $\displaystyle \mathbb{Z}_{n}$ form a group under modular multiplication. This can be seen from above, which is quite a nice Cayley table .

Thanks

I'm fairly sure that this extends to the general case, but not being a ring theorist, and not (at the moment) being too interested I haven't checked. In particular, if $\displaystyle \left(R,+,\cdot\right)$ is a ring, then $\displaystyle \left(U(R),\cdot\right)$ where $\displaystyle U(R)$ are the units of $\displaystyle R$ form a group that is not necessarily abelian.

ok i got it ^^"

thank you again ^_*