Ring with unity. Show that if ab units, then a and b are units.

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Suppose that R is a ring with unity, $a, b in R$ and that neither a nor b is a zero divisor. If ab unit, then a and b are units.




I know that a ring with unity means that



$$exists 1_R in R:forall a in R: 1_Ra = a = a1_R.$$



For an element $a in R$ to be a zero divisor we must have $a neq 0_R$ and there must exist an element $hata neq 0_R$ such that $a hata = 0_R$ or $hataa = 0_R$. Therefore, the meaning of not a zero divisor would be that given $a neq 0_R$ we must have for every $hata neq 0_R$ that $ahata neq 0_R$ and $hataa neq 0_R$. Similarly, for b we then have $bhatb neq 0_R$ and $hatbb neq 0_R$.



For $ab$ to be a unit there exists $(ab)^-1 in R$ such that $(ab)^-1ab = 1_R = ab(ab)^-1$.



Now, I have to show that $exists a^-1, b^-1 in R$ such that $a^-1a = 1_R = aa^-1$ and $b^-1b = 1_R = bb^-1$.



I have no starting point thus far so any hint(s) are greatly appreciated.










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    Suppose that R is a ring with unity, $a, b in R$ and that neither a nor b is a zero divisor. If ab unit, then a and b are units.




    I know that a ring with unity means that



    $$exists 1_R in R:forall a in R: 1_Ra = a = a1_R.$$



    For an element $a in R$ to be a zero divisor we must have $a neq 0_R$ and there must exist an element $hata neq 0_R$ such that $a hata = 0_R$ or $hataa = 0_R$. Therefore, the meaning of not a zero divisor would be that given $a neq 0_R$ we must have for every $hata neq 0_R$ that $ahata neq 0_R$ and $hataa neq 0_R$. Similarly, for b we then have $bhatb neq 0_R$ and $hatbb neq 0_R$.



    For $ab$ to be a unit there exists $(ab)^-1 in R$ such that $(ab)^-1ab = 1_R = ab(ab)^-1$.



    Now, I have to show that $exists a^-1, b^-1 in R$ such that $a^-1a = 1_R = aa^-1$ and $b^-1b = 1_R = bb^-1$.



    I have no starting point thus far so any hint(s) are greatly appreciated.










    share|cite|improve this question























      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite












      Suppose that R is a ring with unity, $a, b in R$ and that neither a nor b is a zero divisor. If ab unit, then a and b are units.




      I know that a ring with unity means that



      $$exists 1_R in R:forall a in R: 1_Ra = a = a1_R.$$



      For an element $a in R$ to be a zero divisor we must have $a neq 0_R$ and there must exist an element $hata neq 0_R$ such that $a hata = 0_R$ or $hataa = 0_R$. Therefore, the meaning of not a zero divisor would be that given $a neq 0_R$ we must have for every $hata neq 0_R$ that $ahata neq 0_R$ and $hataa neq 0_R$. Similarly, for b we then have $bhatb neq 0_R$ and $hatbb neq 0_R$.



      For $ab$ to be a unit there exists $(ab)^-1 in R$ such that $(ab)^-1ab = 1_R = ab(ab)^-1$.



      Now, I have to show that $exists a^-1, b^-1 in R$ such that $a^-1a = 1_R = aa^-1$ and $b^-1b = 1_R = bb^-1$.



      I have no starting point thus far so any hint(s) are greatly appreciated.










      share|cite|improve this question














      Suppose that R is a ring with unity, $a, b in R$ and that neither a nor b is a zero divisor. If ab unit, then a and b are units.




      I know that a ring with unity means that



      $$exists 1_R in R:forall a in R: 1_Ra = a = a1_R.$$



      For an element $a in R$ to be a zero divisor we must have $a neq 0_R$ and there must exist an element $hata neq 0_R$ such that $a hata = 0_R$ or $hataa = 0_R$. Therefore, the meaning of not a zero divisor would be that given $a neq 0_R$ we must have for every $hata neq 0_R$ that $ahata neq 0_R$ and $hataa neq 0_R$. Similarly, for b we then have $bhatb neq 0_R$ and $hatbb neq 0_R$.



      For $ab$ to be a unit there exists $(ab)^-1 in R$ such that $(ab)^-1ab = 1_R = ab(ab)^-1$.



      Now, I have to show that $exists a^-1, b^-1 in R$ such that $a^-1a = 1_R = aa^-1$ and $b^-1b = 1_R = bb^-1$.



      I have no starting point thus far so any hint(s) are greatly appreciated.







      abstract-algebra ring-theory






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      salad salad

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          You already have $aunderbraceb(ab)^-1_a^-1? = 1_R$. The question is, can we also show that $b(ab)^-1a = 1_R$ to satisfy the other condition for an inverse?



          This is where the zero divisor condition comes in. Consider the following product:
          $$ab(ab)^-1a = 1_R a = a implies ab(ab)^-1a - a = 0_R.$$
          Then, we have $a(b(ab)^-1a - 1_R) = 0_R$. Since $a$ is not a zero divisor, we may conclude that $b(ab)^-1a - 1_R = 0$, which is what we need. Hence $a^-1 = b(ab)^-1$.



          A very similar method works to show that $b^-1 = (ab)^-1a$.






          share|cite|improve this answer



























            up vote
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            $(ab)u=1$ implies that $bu$ is a right unit of $a$, you have $a(bu)a=a$ implies that $a(bua-1)=0$, since $a$ is not a divisor of zero, $bua=1$






            share|cite|improve this answer



























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              Since $ab$ is a unit, there is some $r$ such that $rab=abr=1_R$. Thus $br=a^-1$ (since $abr=1_R$) and $ra=b^-1$ (since $rab=1_R$).






              share|cite|improve this answer
















              • 3




                This only gives you a one-sided inverse, doesn't it? Surely a little more justification is necessary to show $bra = 1_R$?
                – Theo Bendit
                41 mins ago

















              up vote
              0
              down vote













              Hint $ $ Use the special case $,d=1,$ of the following



              Lemma $ $ If $,rmcolor#0a0cancellable$ $,c,$ left-divides $,d,$ then $,c,$ also right-divides $,d,$ if $,color#c00c & d rm commute$



              Proof $, cb = d,oversetlarge times cRightarrow, cbc = color#c00dc=cd,Rightarrow, bc = d,$ by $,c rmcolor#0a0cancellable$






              share|cite|improve this answer






















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                4 Answers
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                active

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                4 Answers
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                active

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                active

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                up vote
                4
                down vote













                You already have $aunderbraceb(ab)^-1_a^-1? = 1_R$. The question is, can we also show that $b(ab)^-1a = 1_R$ to satisfy the other condition for an inverse?



                This is where the zero divisor condition comes in. Consider the following product:
                $$ab(ab)^-1a = 1_R a = a implies ab(ab)^-1a - a = 0_R.$$
                Then, we have $a(b(ab)^-1a - 1_R) = 0_R$. Since $a$ is not a zero divisor, we may conclude that $b(ab)^-1a - 1_R = 0$, which is what we need. Hence $a^-1 = b(ab)^-1$.



                A very similar method works to show that $b^-1 = (ab)^-1a$.






                share|cite|improve this answer
























                  up vote
                  4
                  down vote













                  You already have $aunderbraceb(ab)^-1_a^-1? = 1_R$. The question is, can we also show that $b(ab)^-1a = 1_R$ to satisfy the other condition for an inverse?



                  This is where the zero divisor condition comes in. Consider the following product:
                  $$ab(ab)^-1a = 1_R a = a implies ab(ab)^-1a - a = 0_R.$$
                  Then, we have $a(b(ab)^-1a - 1_R) = 0_R$. Since $a$ is not a zero divisor, we may conclude that $b(ab)^-1a - 1_R = 0$, which is what we need. Hence $a^-1 = b(ab)^-1$.



                  A very similar method works to show that $b^-1 = (ab)^-1a$.






                  share|cite|improve this answer






















                    up vote
                    4
                    down vote










                    up vote
                    4
                    down vote









                    You already have $aunderbraceb(ab)^-1_a^-1? = 1_R$. The question is, can we also show that $b(ab)^-1a = 1_R$ to satisfy the other condition for an inverse?



                    This is where the zero divisor condition comes in. Consider the following product:
                    $$ab(ab)^-1a = 1_R a = a implies ab(ab)^-1a - a = 0_R.$$
                    Then, we have $a(b(ab)^-1a - 1_R) = 0_R$. Since $a$ is not a zero divisor, we may conclude that $b(ab)^-1a - 1_R = 0$, which is what we need. Hence $a^-1 = b(ab)^-1$.



                    A very similar method works to show that $b^-1 = (ab)^-1a$.






                    share|cite|improve this answer












                    You already have $aunderbraceb(ab)^-1_a^-1? = 1_R$. The question is, can we also show that $b(ab)^-1a = 1_R$ to satisfy the other condition for an inverse?



                    This is where the zero divisor condition comes in. Consider the following product:
                    $$ab(ab)^-1a = 1_R a = a implies ab(ab)^-1a - a = 0_R.$$
                    Then, we have $a(b(ab)^-1a - 1_R) = 0_R$. Since $a$ is not a zero divisor, we may conclude that $b(ab)^-1a - 1_R = 0$, which is what we need. Hence $a^-1 = b(ab)^-1$.



                    A very similar method works to show that $b^-1 = (ab)^-1a$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 35 mins ago









                    Theo Bendit

                    15k12045




                    15k12045




















                        up vote
                        2
                        down vote













                        $(ab)u=1$ implies that $bu$ is a right unit of $a$, you have $a(bu)a=a$ implies that $a(bua-1)=0$, since $a$ is not a divisor of zero, $bua=1$






                        share|cite|improve this answer
























                          up vote
                          2
                          down vote













                          $(ab)u=1$ implies that $bu$ is a right unit of $a$, you have $a(bu)a=a$ implies that $a(bua-1)=0$, since $a$ is not a divisor of zero, $bua=1$






                          share|cite|improve this answer






















                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            $(ab)u=1$ implies that $bu$ is a right unit of $a$, you have $a(bu)a=a$ implies that $a(bua-1)=0$, since $a$ is not a divisor of zero, $bua=1$






                            share|cite|improve this answer












                            $(ab)u=1$ implies that $bu$ is a right unit of $a$, you have $a(bu)a=a$ implies that $a(bua-1)=0$, since $a$ is not a divisor of zero, $bua=1$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 35 mins ago









                            Tsemo Aristide

                            53.1k11244




                            53.1k11244




















                                up vote
                                1
                                down vote













                                Since $ab$ is a unit, there is some $r$ such that $rab=abr=1_R$. Thus $br=a^-1$ (since $abr=1_R$) and $ra=b^-1$ (since $rab=1_R$).






                                share|cite|improve this answer
















                                • 3




                                  This only gives you a one-sided inverse, doesn't it? Surely a little more justification is necessary to show $bra = 1_R$?
                                  – Theo Bendit
                                  41 mins ago














                                up vote
                                1
                                down vote













                                Since $ab$ is a unit, there is some $r$ such that $rab=abr=1_R$. Thus $br=a^-1$ (since $abr=1_R$) and $ra=b^-1$ (since $rab=1_R$).






                                share|cite|improve this answer
















                                • 3




                                  This only gives you a one-sided inverse, doesn't it? Surely a little more justification is necessary to show $bra = 1_R$?
                                  – Theo Bendit
                                  41 mins ago












                                up vote
                                1
                                down vote










                                up vote
                                1
                                down vote









                                Since $ab$ is a unit, there is some $r$ such that $rab=abr=1_R$. Thus $br=a^-1$ (since $abr=1_R$) and $ra=b^-1$ (since $rab=1_R$).






                                share|cite|improve this answer












                                Since $ab$ is a unit, there is some $r$ such that $rab=abr=1_R$. Thus $br=a^-1$ (since $abr=1_R$) and $ra=b^-1$ (since $rab=1_R$).







                                share|cite|improve this answer












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                                share|cite|improve this answer










                                answered 43 mins ago









                                Cronus

                                879415




                                879415







                                • 3




                                  This only gives you a one-sided inverse, doesn't it? Surely a little more justification is necessary to show $bra = 1_R$?
                                  – Theo Bendit
                                  41 mins ago












                                • 3




                                  This only gives you a one-sided inverse, doesn't it? Surely a little more justification is necessary to show $bra = 1_R$?
                                  – Theo Bendit
                                  41 mins ago







                                3




                                3




                                This only gives you a one-sided inverse, doesn't it? Surely a little more justification is necessary to show $bra = 1_R$?
                                – Theo Bendit
                                41 mins ago




                                This only gives you a one-sided inverse, doesn't it? Surely a little more justification is necessary to show $bra = 1_R$?
                                – Theo Bendit
                                41 mins ago










                                up vote
                                0
                                down vote













                                Hint $ $ Use the special case $,d=1,$ of the following



                                Lemma $ $ If $,rmcolor#0a0cancellable$ $,c,$ left-divides $,d,$ then $,c,$ also right-divides $,d,$ if $,color#c00c & d rm commute$



                                Proof $, cb = d,oversetlarge times cRightarrow, cbc = color#c00dc=cd,Rightarrow, bc = d,$ by $,c rmcolor#0a0cancellable$






                                share|cite|improve this answer


























                                  up vote
                                  0
                                  down vote













                                  Hint $ $ Use the special case $,d=1,$ of the following



                                  Lemma $ $ If $,rmcolor#0a0cancellable$ $,c,$ left-divides $,d,$ then $,c,$ also right-divides $,d,$ if $,color#c00c & d rm commute$



                                  Proof $, cb = d,oversetlarge times cRightarrow, cbc = color#c00dc=cd,Rightarrow, bc = d,$ by $,c rmcolor#0a0cancellable$






                                  share|cite|improve this answer
























                                    up vote
                                    0
                                    down vote










                                    up vote
                                    0
                                    down vote









                                    Hint $ $ Use the special case $,d=1,$ of the following



                                    Lemma $ $ If $,rmcolor#0a0cancellable$ $,c,$ left-divides $,d,$ then $,c,$ also right-divides $,d,$ if $,color#c00c & d rm commute$



                                    Proof $, cb = d,oversetlarge times cRightarrow, cbc = color#c00dc=cd,Rightarrow, bc = d,$ by $,c rmcolor#0a0cancellable$






                                    share|cite|improve this answer














                                    Hint $ $ Use the special case $,d=1,$ of the following



                                    Lemma $ $ If $,rmcolor#0a0cancellable$ $,c,$ left-divides $,d,$ then $,c,$ also right-divides $,d,$ if $,color#c00c & d rm commute$



                                    Proof $, cb = d,oversetlarge times cRightarrow, cbc = color#c00dc=cd,Rightarrow, bc = d,$ by $,c rmcolor#0a0cancellable$







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited 4 mins ago

























                                    answered 10 mins ago









                                    Bill Dubuque

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