Proof: (⇐) We clearly have a∈(b)=Rb, and since u−1a=u−1(ub)=b, we have b∈(a)=Ra. Therefore, Ra=(a)⊆(b) and Rb=(b)⊆(a) so that (a)=(b). (⇒) If either of a or b are zero, then both are zero, and the case clearly holds, so assume a≠0≠b. We must have a=ub and va=b for some u,v∈R. We can compare the two to obtain va=vub=b, so that vub−b=(vu−1)b=0, and now vu−1=0 implying vu=uv=1, i.e. u is a unit as claimed. ◻
Tuesday, April 30, 2013
Principle Ideals in Integral Domains (7.4.8)
Dummit and Foote Abstract Algebra, section 7.4, exercise 8:
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Let R be an integral domain with a,b∈R. Prove (a)=(b) if and only if a=ub for some unit u.
Proof: (⇐) We clearly have a∈(b)=Rb, and since u−1a=u−1(ub)=b, we have b∈(a)=Ra. Therefore, Ra=(a)⊆(b) and Rb=(b)⊆(a) so that (a)=(b). (⇒) If either of a or b are zero, then both are zero, and the case clearly holds, so assume a≠0≠b. We must have a=ub and va=b for some u,v∈R. We can compare the two to obtain va=vub=b, so that vub−b=(vu−1)b=0, and now vu−1=0 implying vu=uv=1, i.e. u is a unit as claimed. ◻
Proof: (⇐) We clearly have a∈(b)=Rb, and since u−1a=u−1(ub)=b, we have b∈(a)=Ra. Therefore, Ra=(a)⊆(b) and Rb=(b)⊆(a) so that (a)=(b). (⇒) If either of a or b are zero, then both are zero, and the case clearly holds, so assume a≠0≠b. We must have a=ub and va=b for some u,v∈R. We can compare the two to obtain va=vub=b, so that vub−b=(vu−1)b=0, and now vu−1=0 implying vu=uv=1, i.e. u is a unit as claimed. ◻
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