Addition in ring of integers 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!
 A: 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.
A: 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 $\:\Rightarrow\:$ $\rm\:S = \langle gcd(S)\rangle$
