Principal ideals problem.

  • #1
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.
 

Answers and Replies

  • #2
disregardthat
Science Advisor
1,866
34
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}.

You don't use the fact that u is a unit here. Try this: Let u^-1 be an inverse of u, and show that a is in (b) and b is in (a).

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*.

Why does ra = r'b imply that r' = ru for some unit u? Try this: ar = b for some r in R, and a = bs for some s in R since (a) = (b). Substitute, and see what you can say about rs.
 

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