Proving $2\mathbb Z$ is maximal ideal of $\mathbb Z$ I want to prove that $2\mathbb Z$ is a maximal ideal of $\mathbb Z$. That is, suppose there is a ideal $I$ between $\mathbb Z$ and $2\mathbb Z$. When I say between, I say that $I$ has at least one element more than $\mathbb Z$ but is different from $2\mathbb Z$. So, if I suppose that $I$ being between $\mathbb Z$ and $2\mathbb Z$ I conclude that $I = \mathbb Z$, then we proved that any ideal greater than $2\mathbb Z$ is $\mathbb Z$.
So, $2\mathbb Z$ is in the form $2k$ for $k$ integer. Therefore, an Ideal greater than $2\mathbb Z$ must have an element in the form $2k+1$. But in order for this set to be an ideal of $\mathbb Z$, we must have, for $a\in \mathbb Z$ and $b\in 2\mathbb Z$, $ab\in 2\mathbb Z$. So suppose $a = k$, $k$ integer and $b = 2q+1$, $q$ integer. Then $ab = k*(2q+1) = 2(kq)+k$ which is a number in the form $2p+k$ for some integer $p$. I wanted to say that this number is not in the form $2*\mbox{something}$ but sometimes it is. Could somebody help me in proving this, and to generalizing this form $p\mathbb Z$, being $p$ a prime? 
 A: Note that given any ideal $I$ and elements $a,b \in I$ we have $a-b \in I$. Also note that if $1 \in I$ then $I = R$. Using these two facts I'd continue your proof as follows:

So, $2\mathbb Z$ is in the form $2k$ for $k$ integer. Therefore, an
  ideal greater than $2\mathbb Z$ must have an element of the form
  $2k+1$.

Hence $1 = (2k + 1) - 2k \in 2 \mathbb Z$ hence we would have $2 \mathbb Z = \mathbb Z$ but this is false. So $2 \mathbb Z$ does not contain any elements of the form $2k + 1$.
A: If an ideal $I$ contains $2\mathbf Z$ and another element, say $2n+1$, then it contains $2n+1-2n=1$, so it is the whole of $\mathbf Z$.
Generalisation to a prime $p$:
If $I$ contains $p\mathbf Z$ and another element $n$ not divisible by $p$, $n$ and $p$ are coprime. Then you can use Bézout's identity.
A: Let $a\notin p\mathbb{Z}$ thus $<a,p>=1$ since $p$ is prime. Assume now to get contradiction that there exists an ideal $I$ which setisfies $p\mathbb{Z} \lhd I \lhd \mathbb{Z}$ but $I\neq p\mathbb{Z}$ and $I\neq \mathbb{Z}$. Thus we have $a\in I\backslash p\mathbb{Z}$ so we have $<a,p>=1$ as mentioned above which means there exist $\alpha$ and $\beta$ s.t. $\alpha a + \beta p = 1$. That implies $1\in I$ which implies $I=\mathbb{Z}$. $\rightarrow \leftarrow$
