Finding a generator of $nZ∩mZ$

Asked Feb 08 '22 at 12:06

Active Feb 08 '22 at 16:48

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I am trying to find a generator of $nZ∩mZ$.

My intuition:

A generator is $c = lcm(m,n)$.

Given some integer $a$ in the intersection $nZ∩mZ$, I know that there exists $x,y$ such that $a = xn$ and $a = ym$.

Now i want to show there exists $k$ such that $c^k = a$, But could no go further.

Any hints will be useful :)

edited Feb 08 '22 at 16:48

user26857

asked Feb 08 '22 at 12:06

aaa bbb

3

It is not $c^k$ but $c k$. We are talking about an additive group. – Gribouillis Feb 08 '22 at 12:09
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@Gribouillis $c^k = ck$ – aaa bbb Feb 08 '22 at 12:10
Perhaps saying with words would work: "any element in $;n\Bbb Z\cap m\Bbb Z;$ is a multiple of $;n;$ and of $;m;$ , thus..." – DonAntonio Feb 08 '22 at 12:10
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Euclidean division shows that a non trivial additive subgroup of ${\mathbb Z}$ has the form $c {\mathbb Z}$ where $c$ is the smallest positive element. (by the way I don't agree that $c^k = c k$. It is ambiguous in this context) – Gribouillis Feb 08 '22 at 12:13
Isn't the definition of $lcm(m,n)$ that it is a generator of the intersection ideal? ;-) – ancient mathematician Feb 08 '22 at 12:18
@aaabbb Don't mix multiplicative and additive notation. – jjagmath Feb 08 '22 at 12:34
@ancient That's one general approach to defining lcm in general domains. The point of the question is to show that general definition is equivalent to the more specific definition in $\Bbb Z$ (or any Euclidean domain), where "least" is measured wrt to the Euclidean size measure ($|n|$ in $\Bbb Z,,$ and $\deg f,$ in $K[x]$), vs. wrt to divisibility order generally. – Bill Dubuque Feb 08 '22 at 16:31
@BillDubuque I did say it was a joke. ;-) – ancient mathematician Feb 08 '22 at 17:30

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