Specific question on lcm and gcd of rings.
$begingroup$
I can't prove this statement:
Let $a_1,a_2,...,a_n$ and $b_1,b_2,...,b_n$ be non zero elements of an integral domain $R$ such that $a_1b_1=a_2b_2=cdots=a_nb_n=x$
If $gcd(ra_1,ra_2,...,ra_n)$ exists for all $0neq rin R$, then $lcm(b_1,b_2,...,b_n)$ also exists and satisfies $$gcd(a_1,a_2,...,a_n)lcm(b_1b_2,...,b_n)=x$$
Any hints?
I use the following lemma:
Let $a_1,a_2,...,a_n$ and $r$ be nonzero elements of an integral domain $R$.
$1)$ if $lcm(a_1,a_2,..a_n)$ exists, then $lcm(ra_1,ra_2,...,ra_n)$ also exists and $$lcm(ra_1,ra_2,...ra_n)=rlcm(a_1,a_2,...,a_n)$$
$2)$ if $gcd(ra_1,ra_2,...,ra_n)$ exists, then $gcd(a_1,a_2,...a_n)$ also exists and $$gcd(ra_1,ra_2,...,ra_n)=rgcd(a_1,a_2,...,a_n).$$
My attemp:
We suppose that $gcd(ra_1,ra_2,...,ra_n)$ exists, therefore $a:=gcd(a_1,a_2,...,a_n)$ exists for the lemma, and therefore $a| a_i$ for each $i$. Since $x=a_ib_i$ we obtain that $ab_i |x$. We easily can show that $x=lcm(ab_1,ab_2,...,ab_n)$ but i don't know how procede then.
Any hints?
abstract-algebra ring-theory greatest-common-divisor least-common-multiple
asked Dec 10 '18 at 17:19
Domenico VuonoDomenico Vuono
2,3161523
$endgroup$
add a comment |
$begingroup$
I can't prove this statement:
Let $a_1,a_2,...,a_n$ and $b_1,b_2,...,b_n$ be non zero elements of an integral domain $R$ such that $a_1b_1=a_2b_2=cdots=a_nb_n=x$
If $gcd(ra_1,ra_2,...,ra_n)$ exists for all $0neq rin R$, then $lcm(b_1,b_2,...,b_n)$ also exists and satisfies $$gcd(a_1,a_2,...,a_n)lcm(b_1b_2,...,b_n)=x$$
Any hints?
I use the following lemma:
Let $a_1,a_2,...,a_n$ and $r$ be nonzero elements of an integral domain $R$.
$1)$ if $lcm(a_1,a_2,..a_n)$ exists, then $lcm(ra_1,ra_2,...,ra_n)$ also exists and $$lcm(ra_1,ra_2,...ra_n)=rlcm(a_1,a_2,...,a_n)$$
$2)$ if $gcd(ra_1,ra_2,...,ra_n)$ exists, then $gcd(a_1,a_2,...a_n)$ also exists and $$gcd(ra_1,ra_2,...,ra_n)=rgcd(a_1,a_2,...,a_n).$$
My attemp:
We suppose that $gcd(ra_1,ra_2,...,ra_n)$ exists, therefore $a:=gcd(a_1,a_2,...,a_n)$ exists for the lemma, and therefore $a| a_i$ for each $i$. Since $x=a_ib_i$ we obtain that $ab_i |x$. We easily can show that $x=lcm(ab_1,ab_2,...,ab_n)$ but i don't know how procede then.
Any hints?
abstract-algebra ring-theory greatest-common-divisor least-common-multiple
asked Dec 10 '18 at 17:19
Domenico VuonoDomenico Vuono
2,3161523
$endgroup$
add a comment |
$begingroup$
I can't prove this statement:
Let $a_1,a_2,...,a_n$ and $b_1,b_2,...,b_n$ be non zero elements of an integral domain $R$ such that $a_1b_1=a_2b_2=cdots=a_nb_n=x$
If $gcd(ra_1,ra_2,...,ra_n)$ exists for all $0neq rin R$, then $lcm(b_1,b_2,...,b_n)$ also exists and satisfies $$gcd(a_1,a_2,...,a_n)lcm(b_1b_2,...,b_n)=x$$
Any hints?
I use the following lemma:
Let $a_1,a_2,...,a_n$ and $r$ be nonzero elements of an integral domain $R$.
$1)$ if $lcm(a_1,a_2,..a_n)$ exists, then $lcm(ra_1,ra_2,...,ra_n)$ also exists and $$lcm(ra_1,ra_2,...ra_n)=rlcm(a_1,a_2,...,a_n)$$
$2)$ if $gcd(ra_1,ra_2,...,ra_n)$ exists, then $gcd(a_1,a_2,...a_n)$ also exists and $$gcd(ra_1,ra_2,...,ra_n)=rgcd(a_1,a_2,...,a_n).$$
My attemp:
We suppose that $gcd(ra_1,ra_2,...,ra_n)$ exists, therefore $a:=gcd(a_1,a_2,...,a_n)$ exists for the lemma, and therefore $a| a_i$ for each $i$. Since $x=a_ib_i$ we obtain that $ab_i |x$. We easily can show that $x=lcm(ab_1,ab_2,...,ab_n)$ but i don't know how procede then.
Any hints?
abstract-algebra ring-theory greatest-common-divisor least-common-multiple
asked Dec 10 '18 at 17:19
Domenico VuonoDomenico Vuono
2,3161523
$endgroup$
I can't prove this statement:
Let $a_1,a_2,...,a_n$ and $b_1,b_2,...,b_n$ be non zero elements of an integral domain $R$ such that $a_1b_1=a_2b_2=cdots=a_nb_n=x$
If $gcd(ra_1,ra_2,...,ra_n)$ exists for all $0neq rin R$, then $lcm(b_1,b_2,...,b_n)$ also exists and satisfies $$gcd(a_1,a_2,...,a_n)lcm(b_1b_2,...,b_n)=x$$
Any hints?
I use the following lemma:
Let $a_1,a_2,...,a_n$ and $r$ be nonzero elements of an integral domain $R$.
$1)$ if $lcm(a_1,a_2,..a_n)$ exists, then $lcm(ra_1,ra_2,...,ra_n)$ also exists and $$lcm(ra_1,ra_2,...ra_n)=rlcm(a_1,a_2,...,a_n)$$
$2)$ if $gcd(ra_1,ra_2,...,ra_n)$ exists, then $gcd(a_1,a_2,...a_n)$ also exists and $$gcd(ra_1,ra_2,...,ra_n)=rgcd(a_1,a_2,...,a_n).$$
My attemp:
We suppose that $gcd(ra_1,ra_2,...,ra_n)$ exists, therefore $a:=gcd(a_1,a_2,...,a_n)$ exists for the lemma, and therefore $a| a_i$ for each $i$. Since $x=a_ib_i$ we obtain that $ab_i |x$. We easily can show that $x=lcm(ab_1,ab_2,...,ab_n)$ but i don't know how procede then.
Any hints?
abstract-algebra ring-theory greatest-common-divisor least-common-multiple
abstract-algebra ring-theory greatest-common-divisor least-common-multiple
asked Dec 10 '18 at 17:19
Domenico VuonoDomenico Vuono
2,3161523
asked Dec 10 '18 at 17:19
Domenico VuonoDomenico Vuono
2,3161523
edited Dec 10 '18 at 19:36
Domenico Vuono
asked Dec 10 '18 at 17:19
Domenico VuonoDomenico Vuono
2,3161523
asked Dec 10 '18 at 17:19
Domenico VuonoDomenico Vuono
2,3161523
asked Dec 10 '18 at 17:19
Domenico VuonoDomenico Vuono
2,3161523
2,3161523
add a comment |
add a comment |
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