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Given two elements $a,b$ in the Euclidean ring $R$ their least common miltiple $c\in R$ is an element in $R$ such that $a\mid c$ and $b\mid c$ and such that whenever $a\mid x$ and $b\mid x$ then $c\mid x$.

a) Prove that any two elements in the Euclidean ring $R$ have a least common multiple in $R$.

b) If the least common multiple of $a$ and $b$ is denoted by $[a,b]$, prove that $[a,b]=ab/(a,b).$

Proof:

(a) Consider the ideals $(a)$ and $(b)$, where $(x)=xR$. Their intersection $(a)\cap (b)$ is also ideal of $R$ since Euclidean ring is the ring of principle ideal then $(a)\cap (b)=(c)$ for some $c\in R$. I proved that this $c$ has all needed properties of least common multiple. Thus the LCM of $a$ and $b$ exists.

(b) We know that $\text{gcd}(a,b)$ exists in Euclidean ring for any $a,b\neq 0 \in R$. But I have no thoughts how to prove that $[a,b]=ab/(a,b)$.

So I would be very thankful answer.

RFZ
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  • Several thoughts are available in pre-existing questions. Indeed, searching around a bit yields complete answers. You might also experiment with the search assist tool https://approach0.xyz – rschwieb May 24 '18 at 16:39
  • It doesn’t have so much to do with euclideanness as with unique factorization in $R$. – Lubin May 24 '18 at 16:42
  • @rschwieb, but the reasoning on the link which you have provided is very short and unclear. That's why I am asking. – RFZ May 24 '18 at 16:50
  • @RFZ If you are unclear about something you read there (or any of the other explanations) you should try commenting on those first, rather than making duplicate content. In fact, at this duplicate Bill Dubuque has crosslinked to no less than three of his other answers on the same topic. – rschwieb May 24 '18 at 17:00

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