Examples of ring where doesn't exist lcm and gcd
$begingroup$
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
asked Dec 7 '18 at 16:56
Domenico VuonoDomenico Vuono
2,3161523
$endgroup$
add a comment |
$begingroup$
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
asked Dec 7 '18 at 16:56
Domenico VuonoDomenico Vuono
2,3161523
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 7 '18 at 16:56
Domenico VuonoDomenico Vuono
2,3161523
$endgroup$
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
asked Dec 7 '18 at 16:56
Domenico VuonoDomenico Vuono
2,3161523
asked Dec 7 '18 at 16:56
Domenico VuonoDomenico Vuono
2,3161523
asked Dec 7 '18 at 16:56
Domenico VuonoDomenico Vuono
2,3161523
asked Dec 7 '18 at 16:56
Domenico VuonoDomenico Vuono
2,3161523
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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oldest
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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.
answered Dec 12 '18 at 16:50
Domenico VuonoDomenico Vuono
2,3161523
$endgroup$
add a comment |
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1 Answer
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$begingroup$
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.
answered Dec 12 '18 at 16:50
Domenico VuonoDomenico Vuono
2,3161523
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 12 '18 at 16:50
Domenico VuonoDomenico Vuono
2,3161523
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 12 '18 at 16:50
Domenico VuonoDomenico Vuono
2,3161523
$endgroup$
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.
answered Dec 12 '18 at 16:50
Domenico VuonoDomenico Vuono
2,3161523
edited Dec 13 '18 at 14:49
answered Dec 12 '18 at 16:50
Domenico VuonoDomenico Vuono
2,3161523
answered Dec 12 '18 at 16:50
Domenico VuonoDomenico Vuono
2,3161523
answered Dec 12 '18 at 16:50
Domenico VuonoDomenico Vuono
2,3161523
2,3161523
add a comment |
add a comment |
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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