I am trying to prove $(a)$ is a maximal ideal in $\mathbb{Z}$, if and only if $a$ is prime number.
Now I wrote:
assume $a \in\mathbb{Z}$, while it's not prime number
we can write as $a=xy$, for some integers $x$ and $y$.
then $(a)\subset (x)$ and $(a)\subset (y)$
while if $(a)$ is maximal ideal, it's not exist $(k)$ such that $(a)\subset (k)\subset\mathbb{Z}$.
so $(a)$ cannot be maximal ideal.
we can say by contradictory if $a$ is prime, $(a)$ is maximal ideal.
is this correct?
and how to prove the other way.
Best Answer
Hint: can you prove that every ideal is generated by one element in $\mathbb{Z}$? Can you prove that if $ \langle x\rangle \subset \langle y \rangle $, then $ x $ is divisible by $ y$?
And yes, your approach is correct