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Can you help me with the follow question? I really don't understand how to solve this one :(

In the ring of integers, find a positive integer a such that

a. $\langle a\rangle = \langle2\rangle + \langle 3\rangle$,

My first thought: $\langle a\rangle = \{1+1, 2+3, 4+9, 6+27\}$, but that doesn't make any sense.

b. $\langle a\rangle = \langle 6\rangle + \langle 8\rangle$,

c. $\langle a\rangle = \langle m \rangle + \langle n\rangle$

Thank you!

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Note that here $\langle a \rangle$ means $\{ak, \ k \in \mathbb{Z} \}$ (the set of multiples of $a$). This is an infinite set (for example $\langle 2 \rangle = \{\ldots, -6, -4,-2, 0, 2, 4 , 6, \ldots\}$ is the set of even numbers). And $\langle m \rangle + \langle n \rangle$ contains all numbers which can be written as a sum of any multiple of $m$ (possibly zero) and any multiple of $n$ (also possibly zero). For example $\langle 2 \rangle + \langle 3 \rangle$, should contain $0 = 2 \times 0 + 3 \times 0$, $1 = 2 \times (-1) + 3 \times 1$, $2 = 2 \times 1 + 3 \times 0$... – Joel Cohen May 11 '12 at 9:50
So the answers should be: a. < a> = <1> b. < a> = <2> c. < a> <m - n> ? – user1255553 May 11 '12 at 9:56
People, you should stop posting your answers as comments. +1! – Jesko Hüttenhain May 11 '12 at 9:57
up vote 1 down vote accepted

Hint $\ $ $\rm\: S = \langle m\rangle + \langle n \rangle\: $ is closed under subtraction, so a $1$-line proof shows that if $\rm\:S \neq \{0\}\:$ then every element of $\rm\:S\:$ is a multiple of the least positive $\rm\: d\in S,\:$ so $\rm\:S = \langle d\rangle,\:$ being a subgroup of $\mathbb Z.\:$ Note $\rm\:d\:|\:m,n,\:$ and $\rm\:d\in S\:\Rightarrow\: d = jm+kn\:$ so $\rm\:c\:|\:m,n\:\Rightarrow\:c\:|\:d.\:$ As a common divisor of $\rm\:m,n\:$ divisible by every other common divisor, $\rm\:d\:$ is the ____ common divisor of $\rm\:m,n.$

I.e. $\rm\: S\subset \mathbb Z\:$ subtraction-closed $\:\Rightarrow\:$ $\rm S$ mod-closed $\:\Rightarrow\:$ $\rm S\:$ gcd-closed $\:\Rightarrow\:$ $\rm\:S = \langle gcd(S)\rangle$

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By definition $\langle a\rangle=\{ak:k\in\Bbb Z\}$, the ideal generated by $a$. The notation $\langle a\rangle+\langle b\rangle$ means $\{ak+b\ell:k,\ell\in\Bbb Z\}$. Thus, $\langle 2\rangle+\langle 3\rangle=\{2k+3\ell:k,\ell\in\Bbb Z\}$; this set contains every multiple of $2$, every multiple of $3$, and every integer that is a sum of a multiple of $2$ and a multiple of $3$. You've probably had already a theorem that says that for any non-zero integers $m$ and $n$ there are integers $k$ and $\ell$ such that $km+\ell n=\gcd(m,n)$. Since $\gcd(2,3)=1$, there are integers $k$ and $\ell$ such that $2k+3\ell=1$. From this you should be able to show easily that every integer can be written in the form $2k+3\ell$ and hence that every integer belongs to $\langle 2\rangle+\langle 3\rangle$. Thus, $\langle 2\rangle+\langle 3\rangle=\langle 1\rangle=\Bbb Z$.

I've given enough hints in the solution of your first problem to give you a good leg up on the remaining problems.

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