Proof: ($\Leftarrow$) We clearly have $a∈(b)=Rb$, and since $u^{-1}a=u^{-1}(ub)=b$, we have $b∈(a)=Ra$. Therefore, $Ra=(a) \subseteq (b)$ and $Rb=(b) \subseteq (a)$ 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.$~\square$
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: ($\Leftarrow$) We clearly have $a∈(b)=Rb$, and since $u^{-1}a=u^{-1}(ub)=b$, we have $b∈(a)=Ra$. Therefore, $Ra=(a) \subseteq (b)$ and $Rb=(b) \subseteq (a)$ 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.$~\square$
Proof: ($\Leftarrow$) We clearly have $a∈(b)=Rb$, and since $u^{-1}a=u^{-1}(ub)=b$, we have $b∈(a)=Ra$. Therefore, $Ra=(a) \subseteq (b)$ and $Rb=(b) \subseteq (a)$ 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.$~\square$
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