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 5 updated status edited Apr 12 at 23:26 Brahadeesh 6,8121010 silver badges1616 bronze badges 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 edited Sep 4 '18 at 14:01 user99914 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 edited Sep 4 '18 at 6:16 Gerry Myerson 154k11 gold badge4040 silver badges6262 bronze badges 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. 2 added 1 character in body edited Sep 4 '18 at 1:05 Arturo Magidin 279k22 gold badges2525 silver badges5454 bronze badges 1 answered Sep 4 '18 at 0:52 Arturo Magidin 279k22 gold badges2525 silver badges5454 bronze badges Post Made Community Wiki by Arturo Magidin occurred Sep 4 '18 at 0:52