prove that maximal ideal in $\mathbb{Z}$ generated by a prime number 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. 
 A: 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
A: 
I am trying to prove $\langle a \rangle$ is a maximal ideal in $\mathbb{Z}$,  if and only if $a$ is prime number.
... assume $a \in \mathbb{Z}$, ... [if] it's not [a] prime number we can write as $a = xy$, for some integers $x$ and $y$. Then $\langle a \rangle \subset \langle x \rangle$ and $\langle a \rangle \subset \langle y \rangle$.

Looks good so far.

while if $\langle a \rangle$ is maximal ideal, [there does not] exist $\langle k \rangle$ such that $\langle a \rangle \subset \langle k \rangle\subset \mathbb{Z}$.

I think the unspoken assumption here is that $k \in \mathbb{Z}$, so that $\langle k \rangle$ is a principal ideal. Then I think you either need to prove that all ideals in $\mathbb Z$ are principal, or refer to a proof of that fact by someone else.
Because without that result, how can you be sure $\langle a \rangle \subset \langle a, k \rangle \subset \mathbb{Z}$ is not a possibility?
Rewrite the previous set comparison as $\langle a \rangle \subseteq \langle a, k \rangle \subseteq \mathbb{Z}$. If $\gcd(a, k) = 1$, then $\langle a, k \rangle = \mathbb{Z}$. But if $\gcd(a, k) = |a|$, then $\langle a \rangle = \langle a, k \rangle$.
But that's just me talking without proving all ideals are principal in this domain.

so $\langle a \rangle$ cannot be maximal ideal.
  we can say by contradictory if $a$ is prime, $\langle a \rangle$ is maximal ideal.
is this correct? 
  and how to prove the other way. 

With the missing assumption filled in, then yes, this is correct. I'm not sure what you mean by "the other way."
