# Algebra.....units prove

• Feb 28th 2006, 03:33 PM
suedenation
Algebra.....units prove
Suppose that R and S are rings with identity. Let (r, s) in R × S. Show that
(r, s) is unit in R × S <=> r is a unit in R and s is a unit in S.

Also, if (r, s) is a unit in R×S, show that inverse of (r, s) = (inverse of r, inverse of s)
(Hint: It is efficient to prove the last statement and the implication “<=” in the first statement at the
same time.)

um....can anyone help!!! urgent!!! Thanks very much :)
• Feb 28th 2006, 05:29 PM
ThePerfectHacker
Quote:

Originally Posted by suedenation
Suppose that R and S are rings with identity. Let (r, s) in R × S. Show that
(r, s) is unit in R × S <=> r is a unit in R and s is a unit in S.

Also, if (r, s) is a unit in R×S, show that inverse of (r, s) = (inverse of r, inverse of s)
(Hint: It is efficient to prove the last statement and the implication “<=” in the first statement at the
same time.)

um....can anyone help!!! urgent!!! Thanks very much :)

If $\displaystyle S,R$ are rings with unity, then their direct product is also a ring $\displaystyle S\times R$. Let $\displaystyle (s,r)\in S\times R$. Then, $\displaystyle s\in S$ and $\displaystyle r\in R$. If $\displaystyle (s,r)$ is a unit there exists, $\displaystyle (s^{-1},r^{-1})$ but $\displaystyle s^{-1}\in S$ and $\displaystyle r^{-1}\in R$. Now assume the converse, if $\displaystyle s\in S$ and $\displaystyle r\in R$ are units, then there exists $\displaystyle s^{-1}\in S$ and $\displaystyle r^{-1}\in R$, thus, $\displaystyle (s^{-1},r^{-1})\in R\times S$ is a unit.
Q.E.D.