- #1
mathmajor2013
- 26
- 0
Let R be an integral domain with elements a,b in R and <a>,<b> the corresponding principal ideals. Prove that <a>=<b> if, and only if, a=bu for some unit u in U(R).
proof
if a=bu, then ar=bur. Since ur in R, call it s. So ar=bs for some s in R. Therefore a times some elements in r is equal to b times a different element in R. Therefore {ra: r in R}={rb: r in R}.
if <a>=<b>, then {ra: r in R}={rb: r in R}. That means ra=r'b for some r,r' in R. However, r'=ru for some unit u in U(R). So ra=rub. Hence, a=ub=bu *commutative ring*.
Therefore <a>=<b> if, and only if, a=bu for some unit u in U(R).
PLEASE HELP I KNOW THIS PROOF HAS SOMETHING WRONG WITH IT.
proof
if a=bu, then ar=bur. Since ur in R, call it s. So ar=bs for some s in R. Therefore a times some elements in r is equal to b times a different element in R. Therefore {ra: r in R}={rb: r in R}.
if <a>=<b>, then {ra: r in R}={rb: r in R}. That means ra=r'b for some r,r' in R. However, r'=ru for some unit u in U(R). So ra=rub. Hence, a=ub=bu *commutative ring*.
Therefore <a>=<b> if, and only if, a=bu for some unit u in U(R).
PLEASE HELP I KNOW THIS PROOF HAS SOMETHING WRONG WITH IT.