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Reopened, then closed as off topic

Reopened, then closed as off topic, deleted

This question (does the existence of an lcm in a commutative ring imply the existence of the gcd of the two elements) was closed as a duplicate, pointing to this one (which shows inter alia that the existence of an lcm in a domain implies the existence of a gcd). In fact, I had done the same thing yesterday.

However, this is not a real duplicate. The question here is about rings, whereas the alleged duplicate is about domains. The proof there does not easily translate: for example, the first step of the proof in the question pointed to is to take $m=\mathrm{lcm}(a,b)$, let $s$ be such that $ms=ab$, write $m=ar$ and $m=bt$, and then claim that $s=\gcd(a,b)$. The first step is to show $s$ divides $a$, which it does by showing that $(ab/m)*(m/b)=a$. This is equivalent to saying that $st=a$; but I do not see how to conclude this in a ring (in a domain, multiplying through by $b$ and then cancelling will do it, but how do you do it here?).

As I had already closed it, and then re-opened, I can no longer vote to re-open on the page.

Reopened, then closed as off topic

This question (does the existence of an lcm in a commutative ring imply the existence of the gcd of the two elements) was closed as a duplicate, pointing to this one (which shows inter alia that the existence of an lcm in a domain implies the existence of a gcd). In fact, I had done the same thing yesterday.

However, this is not a real duplicate. The question here is about rings, whereas the alleged duplicate is about domains. The proof there does not easily translate: for example, the first step of the proof in the question pointed to is to take $m=\mathrm{lcm}(a,b)$, let $s$ be such that $ms=ab$, write $m=ar$ and $m=bt$, and then claim that $s=\gcd(a,b)$. The first step is to show $s$ divides $a$, which it does by showing that $(ab/m)*(m/b)=a$. This is equivalent to saying that $st=a$; but I do not see how to conclude this in a ring (in a domain, multiplying through by $b$ and then cancelling will do it, but how do you do it here?).

As I had already closed it, and then re-opened, I can no longer vote to re-open on the page.

Reopened, then closed as off topic, deleted

This question (does the existence of an lcm in a commutative ring imply the existence of the gcd of the two elements) was closed as a duplicate, pointing to this one (which shows inter alia that the existence of an lcm in a domain implies the existence of a gcd). In fact, I had done the same thing yesterday.

However, this is not a real duplicate. The question here is about rings, whereas the alleged duplicate is about domains. The proof there does not easily translate: for example, the first step of the proof in the question pointed to is to take $m=\mathrm{lcm}(a,b)$, let $s$ be such that $ms=ab$, write $m=ar$ and $m=bt$, and then claim that $s=\gcd(a,b)$. The first step is to show $s$ divides $a$, which it does by showing that $(ab/m)*(m/b)=a$. This is equivalent to saying that $st=a$; but I do not see how to conclude this in a ring (in a domain, multiplying through by $b$ and then cancelling will do it, but how do you do it here?).

As I had already closed it, and then re-opened, I can no longer vote to re-open on the page.

4 added 26 characters in body
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Reopened, then closed as off topic

This question (does the existence of an lcm in a commutative ring imply the existence of the gcd of the two elements) was closed as a duplicate, pointing to this one (which shows inter alia that the existence of an lcm in a domain implies the existence of a gcd). In fact, I had done the same thing yesterday.

However, this is not a real duplicate. The question here is about rings, whereas the alleged duplicate is about domains. The proof there does not easily translate: for example, the first step of the proof in the question pointed to is to take $m=\mathrm{lcm}(a,b)$, let $s$ be such that $ms=ab$, write $m=ar$ and $m=bt$, and then claim that $s=\gcd(a,b)$. The first step is to show $s$ divides $a$, which it does by showing that $(ab/m)*(m/b)=a$. This is equivalent to saying that $st=a$; but I do not see how to conclude this in a ring (in a domain, multiplying through by $b$ and then cancelling will do it, but how do you do it here?).

As I had already closed it, and then re-opened, I can no longer vote to re-open on the page.

Reopened

This question (does the existence of an lcm in a commutative ring imply the existence of the gcd of the two elements) was closed as a duplicate, pointing to this one (which shows inter alia that the existence of an lcm in a domain implies the existence of a gcd). In fact, I had done the same thing yesterday.

However, this is not a real duplicate. The question here is about rings, whereas the alleged duplicate is about domains. The proof there does not easily translate: for example, the first step of the proof in the question pointed to is to take $m=\mathrm{lcm}(a,b)$, let $s$ be such that $ms=ab$, write $m=ar$ and $m=bt$, and then claim that $s=\gcd(a,b)$. The first step is to show $s$ divides $a$, which it does by showing that $(ab/m)*(m/b)=a$. This is equivalent to saying that $st=a$; but I do not see how to conclude this in a ring (in a domain, multiplying through by $b$ and then cancelling will do it, but how do you do it here?).

As I had already closed it, and then re-opened, I can no longer vote to re-open on the page.

Reopened, then closed as off topic

This question (does the existence of an lcm in a commutative ring imply the existence of the gcd of the two elements) was closed as a duplicate, pointing to this one (which shows inter alia that the existence of an lcm in a domain implies the existence of a gcd). In fact, I had done the same thing yesterday.

However, this is not a real duplicate. The question here is about rings, whereas the alleged duplicate is about domains. The proof there does not easily translate: for example, the first step of the proof in the question pointed to is to take $m=\mathrm{lcm}(a,b)$, let $s$ be such that $ms=ab$, write $m=ar$ and $m=bt$, and then claim that $s=\gcd(a,b)$. The first step is to show $s$ divides $a$, which it does by showing that $(ab/m)*(m/b)=a$. This is equivalent to saying that $st=a$; but I do not see how to conclude this in a ring (in a domain, multiplying through by $b$ and then cancelling will do it, but how do you do it here?).

As I had already closed it, and then re-opened, I can no longer vote to re-open on the page.

3 status update
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Reopened

This question (does the existence of an lcm in a commutative ring imply the existence of the gcd of the two elements) was closed as a duplicate, pointing to this one (which shows inter alia that the existence of an lcm in a domain implies the existence of a gcd). In fact, I had done the same thing yesterday.

However, this is not a real duplicate. The question here is about rings, whereas the alleged duplicate is about domains. The proof there does not easily translate: for example, the first step of the proof in the question pointed to is to take $m=\mathrm{lcm}(a,b)$, let $s$ be such that $ms=ab$, write $m=ar$ and $m=bt$, and then claim that $s=\gcd(a,b)$. The first step is to show $s$ divides $a$, which it does by showing that $(ab/m)*(m/b)=a$. This is equivalent to saying that $st=a$; but I do not see how to conclude this in a ring (in a domain, multiplying through by $b$ and then cancelling will do it, but how do you do it here?).

As I had already closed it, and then re-opened, I can no longer vote to re-open on the page.

This question (does the existence of an lcm in a commutative ring imply the existence of the gcd of the two elements) was closed as a duplicate, pointing to this one (which shows inter alia that the existence of an lcm in a domain implies the existence of a gcd). In fact, I had done the same thing yesterday.

However, this is not a real duplicate. The question here is about rings, whereas the alleged duplicate is about domains. The proof there does not easily translate: for example, the first step of the proof in the question pointed to is to take $m=\mathrm{lcm}(a,b)$, let $s$ be such that $ms=ab$, write $m=ar$ and $m=bt$, and then claim that $s=\gcd(a,b)$. The first step is to show $s$ divides $a$, which it does by showing that $(ab/m)*(m/b)=a$. This is equivalent to saying that $st=a$; but I do not see how to conclude this in a ring (in a domain, multiplying through by $b$ and then cancelling will do it, but how do you do it here?).

As I had already closed it, and then re-opened, I can no longer vote to re-open on the page.

Reopened

This question (does the existence of an lcm in a commutative ring imply the existence of the gcd of the two elements) was closed as a duplicate, pointing to this one (which shows inter alia that the existence of an lcm in a domain implies the existence of a gcd). In fact, I had done the same thing yesterday.

However, this is not a real duplicate. The question here is about rings, whereas the alleged duplicate is about domains. The proof there does not easily translate: for example, the first step of the proof in the question pointed to is to take $m=\mathrm{lcm}(a,b)$, let $s$ be such that $ms=ab$, write $m=ar$ and $m=bt$, and then claim that $s=\gcd(a,b)$. The first step is to show $s$ divides $a$, which it does by showing that $(ab/m)*(m/b)=a$. This is equivalent to saying that $st=a$; but I do not see how to conclude this in a ring (in a domain, multiplying through by $b$ and then cancelling will do it, but how do you do it here?).

As I had already closed it, and then re-opened, I can no longer vote to re-open on the page.

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