Why are $a$ and $ab$ associates if $b$ is a unit?












0












$begingroup$



The function $v(x)$ is a Euclidean function on an integral domain, $D$.



Proof :




  • Suppose that $v(a) < v(ab)$. If $b$ were a unit, then $a$ and $ab$
    would be associates.


  • We have $a = abu$ and $ab = au^{-1}$.


  • Then, by Condition 2 of an Euclidean Function, we have $v(ab) leq v(abu) = v(a) leq v(au^{-1}) = v(ab)$. Therefore, $v(a) nless v(ab)$





My main question is, why are $a$ and $ab$ associates if $b$ is a unit in the first place? I'm trying to understand how $amid ab$ = $abmid a$ if $b$ is a unit.



$$a = ab implies b = aa^{-1} implies 1 = aa^{-1} = b?$$



Thank you in advance for the help.






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Clearly, $amid ab$. If $b$ is a unit, $a = abcdot b^{-1}$, so $abmid a$ as well. Hence $a$ and $ab$ are associates.
    $endgroup$
    – Prahlad Vaidyanathan
    Mar 22 '15 at 14:44










  • $begingroup$
    Thanks a lot, wouldn't $ab^{-1} = abb^{-1} implies ab^{-1} = a?$. Since you have to multiply both sides by $b^{-1}$.
    $endgroup$
    – Squires McGee
    Mar 22 '15 at 14:53












  • $begingroup$
    If $a$ is not a unit, there is no $a^{-1}$. In particular, the implication you wrote down ($a = ab implies 1 = b$) is false.
    $endgroup$
    – Michael Albanese
    Jun 26 '15 at 8:15


















0












$begingroup$



The function $v(x)$ is a Euclidean function on an integral domain, $D$.



Proof :




  • Suppose that $v(a) < v(ab)$. If $b$ were a unit, then $a$ and $ab$
    would be associates.


  • We have $a = abu$ and $ab = au^{-1}$.


  • Then, by Condition 2 of an Euclidean Function, we have $v(ab) leq v(abu) = v(a) leq v(au^{-1}) = v(ab)$. Therefore, $v(a) nless v(ab)$





My main question is, why are $a$ and $ab$ associates if $b$ is a unit in the first place? I'm trying to understand how $amid ab$ = $abmid a$ if $b$ is a unit.



$$a = ab implies b = aa^{-1} implies 1 = aa^{-1} = b?$$



Thank you in advance for the help.






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Clearly, $amid ab$. If $b$ is a unit, $a = abcdot b^{-1}$, so $abmid a$ as well. Hence $a$ and $ab$ are associates.
    $endgroup$
    – Prahlad Vaidyanathan
    Mar 22 '15 at 14:44










  • $begingroup$
    Thanks a lot, wouldn't $ab^{-1} = abb^{-1} implies ab^{-1} = a?$. Since you have to multiply both sides by $b^{-1}$.
    $endgroup$
    – Squires McGee
    Mar 22 '15 at 14:53












  • $begingroup$
    If $a$ is not a unit, there is no $a^{-1}$. In particular, the implication you wrote down ($a = ab implies 1 = b$) is false.
    $endgroup$
    – Michael Albanese
    Jun 26 '15 at 8:15
















0












0








0





$begingroup$



The function $v(x)$ is a Euclidean function on an integral domain, $D$.



Proof :




  • Suppose that $v(a) < v(ab)$. If $b$ were a unit, then $a$ and $ab$
    would be associates.


  • We have $a = abu$ and $ab = au^{-1}$.


  • Then, by Condition 2 of an Euclidean Function, we have $v(ab) leq v(abu) = v(a) leq v(au^{-1}) = v(ab)$. Therefore, $v(a) nless v(ab)$





My main question is, why are $a$ and $ab$ associates if $b$ is a unit in the first place? I'm trying to understand how $amid ab$ = $abmid a$ if $b$ is a unit.



$$a = ab implies b = aa^{-1} implies 1 = aa^{-1} = b?$$



Thank you in advance for the help.






abstract-algebra ring-theory







share|cite|improve this question











$endgroup$





The function $v(x)$ is a Euclidean function on an integral domain, $D$.



Proof :




  • Suppose that $v(a) < v(ab)$. If $b$ were a unit, then $a$ and $ab$
    would be associates.


  • We have $a = abu$ and $ab = au^{-1}$.


  • Then, by Condition 2 of an Euclidean Function, we have $v(ab) leq v(abu) = v(a) leq v(au^{-1}) = v(ab)$. Therefore, $v(a) nless v(ab)$





My main question is, why are $a$ and $ab$ associates if $b$ is a unit in the first place? I'm trying to understand how $amid ab$ = $abmid a$ if $b$ is a unit.



$$a = ab implies b = aa^{-1} implies 1 = aa^{-1} = b?$$



Thank you in advance for the help.






abstract-algebra ring-theory




abstract-algebra ring-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 26 '15 at 8:14









Michael Albanese

63.5k1599305




63.5k1599305










asked Mar 22 '15 at 14:36









Squires McGeeSquires McGee

1319




1319








  • 5




    $begingroup$
    Clearly, $amid ab$. If $b$ is a unit, $a = abcdot b^{-1}$, so $abmid a$ as well. Hence $a$ and $ab$ are associates.
    $endgroup$
    – Prahlad Vaidyanathan
    Mar 22 '15 at 14:44










  • $begingroup$
    Thanks a lot, wouldn't $ab^{-1} = abb^{-1} implies ab^{-1} = a?$. Since you have to multiply both sides by $b^{-1}$.
    $endgroup$
    – Squires McGee
    Mar 22 '15 at 14:53












  • $begingroup$
    If $a$ is not a unit, there is no $a^{-1}$. In particular, the implication you wrote down ($a = ab implies 1 = b$) is false.
    $endgroup$
    – Michael Albanese
    Jun 26 '15 at 8:15
















  • 5




    $begingroup$
    Clearly, $amid ab$. If $b$ is a unit, $a = abcdot b^{-1}$, so $abmid a$ as well. Hence $a$ and $ab$ are associates.
    $endgroup$
    – Prahlad Vaidyanathan
    Mar 22 '15 at 14:44










  • $begingroup$
    Thanks a lot, wouldn't $ab^{-1} = abb^{-1} implies ab^{-1} = a?$. Since you have to multiply both sides by $b^{-1}$.
    $endgroup$
    – Squires McGee
    Mar 22 '15 at 14:53












  • $begingroup$
    If $a$ is not a unit, there is no $a^{-1}$. In particular, the implication you wrote down ($a = ab implies 1 = b$) is false.
    $endgroup$
    – Michael Albanese
    Jun 26 '15 at 8:15










5




5




$begingroup$
Clearly, $amid ab$. If $b$ is a unit, $a = abcdot b^{-1}$, so $abmid a$ as well. Hence $a$ and $ab$ are associates.
$endgroup$
– Prahlad Vaidyanathan
Mar 22 '15 at 14:44




$begingroup$
Clearly, $amid ab$. If $b$ is a unit, $a = abcdot b^{-1}$, so $abmid a$ as well. Hence $a$ and $ab$ are associates.
$endgroup$
– Prahlad Vaidyanathan
Mar 22 '15 at 14:44












$begingroup$
Thanks a lot, wouldn't $ab^{-1} = abb^{-1} implies ab^{-1} = a?$. Since you have to multiply both sides by $b^{-1}$.
$endgroup$
– Squires McGee
Mar 22 '15 at 14:53






$begingroup$
Thanks a lot, wouldn't $ab^{-1} = abb^{-1} implies ab^{-1} = a?$. Since you have to multiply both sides by $b^{-1}$.
$endgroup$
– Squires McGee
Mar 22 '15 at 14:53














$begingroup$
If $a$ is not a unit, there is no $a^{-1}$. In particular, the implication you wrote down ($a = ab implies 1 = b$) is false.
$endgroup$
– Michael Albanese
Jun 26 '15 at 8:15






$begingroup$
If $a$ is not a unit, there is no $a^{-1}$. In particular, the implication you wrote down ($a = ab implies 1 = b$) is false.
$endgroup$
– Michael Albanese
Jun 26 '15 at 8:15












1 Answer
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$begingroup$

Let $R$ be a commutative ring and let $x, y in R$.



We say $x$ divides $y$ (or $y$ is a multiple of $x$) if there is $z in R$ such that $y = xz$; we denote this by $x mid y$.



We say $x$ and $y$ are associates (or associate elements) if $x mid y$ and $y mid x$.




First of all $a mid ab$ by definition. On the other hand, as $b$ is a unit, it has a multiplicative inverse $b^{-1}$, so we have $$a = a1 = a(bb^{-1}) = (ab)b^{-1}.$$ By the first definition, we see that $ab mid a$ (take $x = ab$, $y = a$, and $z = b^{-1}$). As $a mid ab$ and $ab mid a$, we see that $a$ and $ab$ are associates.






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    $begingroup$

    Let $R$ be a commutative ring and let $x, y in R$.



    We say $x$ divides $y$ (or $y$ is a multiple of $x$) if there is $z in R$ such that $y = xz$; we denote this by $x mid y$.



    We say $x$ and $y$ are associates (or associate elements) if $x mid y$ and $y mid x$.




    First of all $a mid ab$ by definition. On the other hand, as $b$ is a unit, it has a multiplicative inverse $b^{-1}$, so we have $$a = a1 = a(bb^{-1}) = (ab)b^{-1}.$$ By the first definition, we see that $ab mid a$ (take $x = ab$, $y = a$, and $z = b^{-1}$). As $a mid ab$ and $ab mid a$, we see that $a$ and $ab$ are associates.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Let $R$ be a commutative ring and let $x, y in R$.



      We say $x$ divides $y$ (or $y$ is a multiple of $x$) if there is $z in R$ such that $y = xz$; we denote this by $x mid y$.



      We say $x$ and $y$ are associates (or associate elements) if $x mid y$ and $y mid x$.




      First of all $a mid ab$ by definition. On the other hand, as $b$ is a unit, it has a multiplicative inverse $b^{-1}$, so we have $$a = a1 = a(bb^{-1}) = (ab)b^{-1}.$$ By the first definition, we see that $ab mid a$ (take $x = ab$, $y = a$, and $z = b^{-1}$). As $a mid ab$ and $ab mid a$, we see that $a$ and $ab$ are associates.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $R$ be a commutative ring and let $x, y in R$.



        We say $x$ divides $y$ (or $y$ is a multiple of $x$) if there is $z in R$ such that $y = xz$; we denote this by $x mid y$.



        We say $x$ and $y$ are associates (or associate elements) if $x mid y$ and $y mid x$.




        First of all $a mid ab$ by definition. On the other hand, as $b$ is a unit, it has a multiplicative inverse $b^{-1}$, so we have $$a = a1 = a(bb^{-1}) = (ab)b^{-1}.$$ By the first definition, we see that $ab mid a$ (take $x = ab$, $y = a$, and $z = b^{-1}$). As $a mid ab$ and $ab mid a$, we see that $a$ and $ab$ are associates.






        share|cite|improve this answer











        $endgroup$



        Let $R$ be a commutative ring and let $x, y in R$.



        We say $x$ divides $y$ (or $y$ is a multiple of $x$) if there is $z in R$ such that $y = xz$; we denote this by $x mid y$.



        We say $x$ and $y$ are associates (or associate elements) if $x mid y$ and $y mid x$.




        First of all $a mid ab$ by definition. On the other hand, as $b$ is a unit, it has a multiplicative inverse $b^{-1}$, so we have $$a = a1 = a(bb^{-1}) = (ab)b^{-1}.$$ By the first definition, we see that $ab mid a$ (take $x = ab$, $y = a$, and $z = b^{-1}$). As $a mid ab$ and $ab mid a$, we see that $a$ and $ab$ are associates.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 3 '18 at 13:27

























        answered Jun 25 '15 at 19:34









        Michael AlbaneseMichael Albanese

        63.5k1599305




        63.5k1599305






























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