Examples of ring where doesn't exist lcm and gcd












-1












$begingroup$


My question is:



"Examples of ring that doesn't exist lcm and gcd of any elements"



The ring preferely has to be commutative and unitary or olny unitary( as matrix ring). I woul use these two theorems:



Theorem 1



Let $a_1,a_2,a_3,......,a_n$ be nonzero elements of a ring $R$. Then $a_1,a_2,a_3,......,a_n$ have a least common multiple if and only if the ideal $cap (a_i)$ is principal.



Theorem 2
Let $a_1,a_2,a_3,......,a_n$ be nonzero elements of the ring $R$. Then $a_1,a_2,a_3,......,a_n$ have a greatest common divisor $d$, expressible in the form
$$d=r_1a_1+r_2a_2+...+r_na_n quad (r_i in R)$$
if and only if the ideal $(a_1......a_n)$ is principal.






abstract-algebra ring-theory







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  • $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – Paul K
    Dec 7 '18 at 17:18










  • $begingroup$
    Why down vote? .
    $endgroup$
    – Domenico Vuono
    Dec 12 '18 at 20:12
















-1












$begingroup$


My question is:



"Examples of ring that doesn't exist lcm and gcd of any elements"



The ring preferely has to be commutative and unitary or olny unitary( as matrix ring). I woul use these two theorems:



Theorem 1



Let $a_1,a_2,a_3,......,a_n$ be nonzero elements of a ring $R$. Then $a_1,a_2,a_3,......,a_n$ have a least common multiple if and only if the ideal $cap (a_i)$ is principal.



Theorem 2
Let $a_1,a_2,a_3,......,a_n$ be nonzero elements of the ring $R$. Then $a_1,a_2,a_3,......,a_n$ have a greatest common divisor $d$, expressible in the form
$$d=r_1a_1+r_2a_2+...+r_na_n quad (r_i in R)$$
if and only if the ideal $(a_1......a_n)$ is principal.






abstract-algebra ring-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – Paul K
    Dec 7 '18 at 17:18










  • $begingroup$
    Why down vote? .
    $endgroup$
    – Domenico Vuono
    Dec 12 '18 at 20:12














-1












-1








-1





$begingroup$


My question is:



"Examples of ring that doesn't exist lcm and gcd of any elements"



The ring preferely has to be commutative and unitary or olny unitary( as matrix ring). I woul use these two theorems:



Theorem 1



Let $a_1,a_2,a_3,......,a_n$ be nonzero elements of a ring $R$. Then $a_1,a_2,a_3,......,a_n$ have a least common multiple if and only if the ideal $cap (a_i)$ is principal.



Theorem 2
Let $a_1,a_2,a_3,......,a_n$ be nonzero elements of the ring $R$. Then $a_1,a_2,a_3,......,a_n$ have a greatest common divisor $d$, expressible in the form
$$d=r_1a_1+r_2a_2+...+r_na_n quad (r_i in R)$$
if and only if the ideal $(a_1......a_n)$ is principal.






abstract-algebra ring-theory







share|cite|improve this question









$endgroup$




My question is:



"Examples of ring that doesn't exist lcm and gcd of any elements"



The ring preferely has to be commutative and unitary or olny unitary( as matrix ring). I woul use these two theorems:



Theorem 1



Let $a_1,a_2,a_3,......,a_n$ be nonzero elements of a ring $R$. Then $a_1,a_2,a_3,......,a_n$ have a least common multiple if and only if the ideal $cap (a_i)$ is principal.



Theorem 2
Let $a_1,a_2,a_3,......,a_n$ be nonzero elements of the ring $R$. Then $a_1,a_2,a_3,......,a_n$ have a greatest common divisor $d$, expressible in the form
$$d=r_1a_1+r_2a_2+...+r_na_n quad (r_i in R)$$
if and only if the ideal $(a_1......a_n)$ is principal.






abstract-algebra ring-theory




abstract-algebra ring-theory






share|cite|improve this question













share|cite|improve this question











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asked Dec 7 '18 at 16:56









Domenico VuonoDomenico Vuono

2,3161523




2,3161523












  • $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – Paul K
    Dec 7 '18 at 17:18










  • $begingroup$
    Why down vote? .
    $endgroup$
    – Domenico Vuono
    Dec 12 '18 at 20:12


















  • $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – Paul K
    Dec 7 '18 at 17:18










  • $begingroup$
    Why down vote? .
    $endgroup$
    – Domenico Vuono
    Dec 12 '18 at 20:12
















$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Paul K
Dec 7 '18 at 17:18




$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Paul K
Dec 7 '18 at 17:18












$begingroup$
Why down vote? .
$endgroup$
– Domenico Vuono
Dec 12 '18 at 20:12




$begingroup$
Why down vote? .
$endgroup$
– Domenico Vuono
Dec 12 '18 at 20:12










1 Answer
1






active

oldest

votes


















1












$begingroup$

I found on the Web and, also on mathstack, several examples and I decided to write an answer.



Example of ring where the gcd of two elements doesn't exist:



Consider the ring $Bbb Z[sqrt{-d}]={a+bisqrt{-d} : a,bin Bbb Z}$, $dge 3$ ($d$ free-square). Then in this paper D. Khurana has proved that:



Let $a$ be any rational integer such that $aequiv dquad (modquad 2)$ and let $a^2 + d = 2q$. Then the elements $2q$ and $(a+ isqrt{d})q$ do not have a $GCD$.



Two examples of ring where the lcm of two elements doesn’t exist:



I use the following theorem:
Let $D$ be a domain and $a,bin D$.
Then, $text{lcm}(a,b)$ exists if and only if for all $rin Dsetminus{0}$, $gcd(ra,rb)$ exists.



$1)$Consider the ring $Bbb Z[sqrt{-d}]={a+bisqrt{-d} : a,bin Bbb Z}$, $dge 3$ ($d$ free-square); and a rational integer $a$ such that $aequiv d$ $(modquad 2)$, then $lcm(2,a+isqrt{d})$ doesn't exist. Indeed this follows from the previous theorem. Note that $GCD(2,a+isqrt{d})=1$.



$2)$ let $R$ be the subring of $Bbb Z[x]$ consisting of the polynomials $sum_i c_ix^i$ such that $c1$ is an even number. If we consider $p_1(x)=2$ and $p_2(x)=2x$, $lcm(p_1, p_2)$ doesn't exist.






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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    1












    $begingroup$

    I found on the Web and, also on mathstack, several examples and I decided to write an answer.



    Example of ring where the gcd of two elements doesn't exist:



    Consider the ring $Bbb Z[sqrt{-d}]={a+bisqrt{-d} : a,bin Bbb Z}$, $dge 3$ ($d$ free-square). Then in this paper D. Khurana has proved that:



    Let $a$ be any rational integer such that $aequiv dquad (modquad 2)$ and let $a^2 + d = 2q$. Then the elements $2q$ and $(a+ isqrt{d})q$ do not have a $GCD$.



    Two examples of ring where the lcm of two elements doesn’t exist:



    I use the following theorem:
    Let $D$ be a domain and $a,bin D$.
    Then, $text{lcm}(a,b)$ exists if and only if for all $rin Dsetminus{0}$, $gcd(ra,rb)$ exists.



    $1)$Consider the ring $Bbb Z[sqrt{-d}]={a+bisqrt{-d} : a,bin Bbb Z}$, $dge 3$ ($d$ free-square); and a rational integer $a$ such that $aequiv d$ $(modquad 2)$, then $lcm(2,a+isqrt{d})$ doesn't exist. Indeed this follows from the previous theorem. Note that $GCD(2,a+isqrt{d})=1$.



    $2)$ let $R$ be the subring of $Bbb Z[x]$ consisting of the polynomials $sum_i c_ix^i$ such that $c1$ is an even number. If we consider $p_1(x)=2$ and $p_2(x)=2x$, $lcm(p_1, p_2)$ doesn't exist.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      I found on the Web and, also on mathstack, several examples and I decided to write an answer.



      Example of ring where the gcd of two elements doesn't exist:



      Consider the ring $Bbb Z[sqrt{-d}]={a+bisqrt{-d} : a,bin Bbb Z}$, $dge 3$ ($d$ free-square). Then in this paper D. Khurana has proved that:



      Let $a$ be any rational integer such that $aequiv dquad (modquad 2)$ and let $a^2 + d = 2q$. Then the elements $2q$ and $(a+ isqrt{d})q$ do not have a $GCD$.



      Two examples of ring where the lcm of two elements doesn’t exist:



      I use the following theorem:
      Let $D$ be a domain and $a,bin D$.
      Then, $text{lcm}(a,b)$ exists if and only if for all $rin Dsetminus{0}$, $gcd(ra,rb)$ exists.



      $1)$Consider the ring $Bbb Z[sqrt{-d}]={a+bisqrt{-d} : a,bin Bbb Z}$, $dge 3$ ($d$ free-square); and a rational integer $a$ such that $aequiv d$ $(modquad 2)$, then $lcm(2,a+isqrt{d})$ doesn't exist. Indeed this follows from the previous theorem. Note that $GCD(2,a+isqrt{d})=1$.



      $2)$ let $R$ be the subring of $Bbb Z[x]$ consisting of the polynomials $sum_i c_ix^i$ such that $c1$ is an even number. If we consider $p_1(x)=2$ and $p_2(x)=2x$, $lcm(p_1, p_2)$ doesn't exist.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        I found on the Web and, also on mathstack, several examples and I decided to write an answer.



        Example of ring where the gcd of two elements doesn't exist:



        Consider the ring $Bbb Z[sqrt{-d}]={a+bisqrt{-d} : a,bin Bbb Z}$, $dge 3$ ($d$ free-square). Then in this paper D. Khurana has proved that:



        Let $a$ be any rational integer such that $aequiv dquad (modquad 2)$ and let $a^2 + d = 2q$. Then the elements $2q$ and $(a+ isqrt{d})q$ do not have a $GCD$.



        Two examples of ring where the lcm of two elements doesn’t exist:



        I use the following theorem:
        Let $D$ be a domain and $a,bin D$.
        Then, $text{lcm}(a,b)$ exists if and only if for all $rin Dsetminus{0}$, $gcd(ra,rb)$ exists.



        $1)$Consider the ring $Bbb Z[sqrt{-d}]={a+bisqrt{-d} : a,bin Bbb Z}$, $dge 3$ ($d$ free-square); and a rational integer $a$ such that $aequiv d$ $(modquad 2)$, then $lcm(2,a+isqrt{d})$ doesn't exist. Indeed this follows from the previous theorem. Note that $GCD(2,a+isqrt{d})=1$.



        $2)$ let $R$ be the subring of $Bbb Z[x]$ consisting of the polynomials $sum_i c_ix^i$ such that $c1$ is an even number. If we consider $p_1(x)=2$ and $p_2(x)=2x$, $lcm(p_1, p_2)$ doesn't exist.






        share|cite|improve this answer











        $endgroup$



        I found on the Web and, also on mathstack, several examples and I decided to write an answer.



        Example of ring where the gcd of two elements doesn't exist:



        Consider the ring $Bbb Z[sqrt{-d}]={a+bisqrt{-d} : a,bin Bbb Z}$, $dge 3$ ($d$ free-square). Then in this paper D. Khurana has proved that:



        Let $a$ be any rational integer such that $aequiv dquad (modquad 2)$ and let $a^2 + d = 2q$. Then the elements $2q$ and $(a+ isqrt{d})q$ do not have a $GCD$.



        Two examples of ring where the lcm of two elements doesn’t exist:



        I use the following theorem:
        Let $D$ be a domain and $a,bin D$.
        Then, $text{lcm}(a,b)$ exists if and only if for all $rin Dsetminus{0}$, $gcd(ra,rb)$ exists.



        $1)$Consider the ring $Bbb Z[sqrt{-d}]={a+bisqrt{-d} : a,bin Bbb Z}$, $dge 3$ ($d$ free-square); and a rational integer $a$ such that $aequiv d$ $(modquad 2)$, then $lcm(2,a+isqrt{d})$ doesn't exist. Indeed this follows from the previous theorem. Note that $GCD(2,a+isqrt{d})=1$.



        $2)$ let $R$ be the subring of $Bbb Z[x]$ consisting of the polynomials $sum_i c_ix^i$ such that $c1$ is an even number. If we consider $p_1(x)=2$ and $p_2(x)=2x$, $lcm(p_1, p_2)$ doesn't exist.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 13 '18 at 14:49

























        answered Dec 12 '18 at 16:50









        Domenico VuonoDomenico Vuono

        2,3161523




        2,3161523






























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