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so I'm working on a problem in a UFD and in some part of the problem I concluded that $x|1$? what does this tell us about $x$? I was thinking well it means that for some $r$ we have $xr=1$ and that means $x$ is a unit now I have to get back to an equality: $d'=cdx$

I need to get rid of the x? how does $x|1$ help?.

reiji sakamaki
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2 Answers2

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Indeed, $x\mid 1$ means that $xr=1$ for some $r$. Then $x$ is a unit and has a unique inverse $r=x^{-1}$. Then you have $d'x^{-1} = cd$.

Wuestenfux
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  • Only the existence of an inverse is required, not uniqueness. This is an oft-misunderstood point. See here for further discussion. – Bill Dubuque Dec 06 '20 at 10:11
  • If its not unique I cannot write $x^{-1}$. – Wuestenfux Dec 06 '20 at 10:12
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    The proof needs only the existence of $,r,$ with $,xr = 1,,$ not uniqueness. Even if hypothetically there were multiple such $r$ the proof would still work. See the linked thread for details. – Bill Dubuque Dec 06 '20 at 10:19
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For some $\,d\,$ we have $\,da=1,\,$ so $\,a\,$ is a unit. Now I need to get rid of $\,a\,$ in $\,ax = b$

Whenever $\,\color{#c00}{a\mid 1},\,$ scaling by $\,a^{-1}\,$ is a special case of cancelling $\,a,\,$ by $\,c=1\,$ below.

Lemma $\ \ \,\color{#c00}{a\mid c},\ ax = b\!\:\color{#c00}c\ \,\Rightarrow\ x = b(c/a)\ $ if $\,a\,$ is cancellable.

Proof $\ \ \ \,\color{#c00}{da\! =\! c},\ ax = b\color{#c00}{da}\Rightarrow\, x = b\:\!d,\,$ by cancelling $\,a.\ \ \small\bf QED$

So the result is "obvious" once we understand how cancellation works. But inverses are so ubiquitous that it helpful to know this in both forms, i.e. scaling $\,ax = b\,$ by the inverse of $\,a,\,$ or cancelling $\,\color{#c00}a\,$ from its multiple $\color{#c00}1=a\:\!a^{-1}$ in $\,b = b\cdot\color{#c00}1$.

Bill Dubuque
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