Principal ideals problem.

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In summary, we have proven that in an integral domain R, <a> = <b> if, and only if, a = bu for some unit u in U(R). This is shown by considering the cases of a = bu and <a> = <b>, and using the commutativity of the ring to show that a = ub = bu. In both cases, we use the fact that a and b are in each other's principal ideals, and in the second case, we also use the fact that r' = ru for some unit u.
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mathmajor2013
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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.
 
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  • #2
mathmajor2013 said:
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).

mathmajor2013 said:
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.
 

What is the principal ideals problem?

The principal ideals problem is a mathematical problem that deals with understanding and analyzing the properties of principal ideals, which are a type of ideal in ring theory. In particular, the problem aims to determine whether all principal ideals in a given ring are generated by a single element.

Why is the principal ideals problem significant?

The principal ideals problem has important implications in the study of algebraic structures, such as rings and fields. It helps to understand the structure of these mathematical objects and has connections to other areas of mathematics, such as algebraic number theory and algebraic geometry.

What is the current status of the principal ideals problem?

The principal ideals problem is a well-known open problem in mathematics. It was first posed in the 1800s and remains unsolved to this day. However, there has been significant progress made towards solving it, with various partial results and approaches proposed by mathematicians.

What are some examples of rings where the principal ideals problem is solved?

The principal ideals problem has been solved for several special cases, such as principal ideal domains (PIDs) and Dedekind domains. Examples of PIDs include the ring of integers and the polynomial ring over a field. However, the general case of the principal ideals problem remains unsolved.

What are some related open problems to the principal ideals problem?

There are several related open problems in mathematics that are connected to the principal ideals problem. These include the inverse Galois problem, the class number problem, and the ideal class group problem. These problems all deal with understanding the structure of ideals in different types of rings and have connections to number theory and algebraic geometry.

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