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!

## Best Answer

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 $\:\Rightarrow\:$ $\rm\:S = \langle gcd(S)\rangle$