Let $R = \mathbb Z[i]$. Show $I \cap \mathbb Z$ is an ideal in $\mathbb Z$, for all $a \in I \cap \mathbb Z$, $10 \mid a^2 = N(a)$. Let $R = \mathbb Z[i]$, $z = 3+i$ and $I = \langle z \rangle$.
I need to show $I \cap \mathbb Z$ is an ideal in $\mathbb Z$, for all $a \in I \cap \mathbb Z$, $10 \mid a^2 = N(a)$ and $10 \mid a$, and $10\mathbb Z = I \cap \mathbb Z$.
I have already proved that $10\mathbb Z \subset I$ since $N(z) = 10$.
 A: Well, if $a\in\Bbb Z$ is such that $10\mid a^2,$ then $2,5\mid a^2,$ so $2,5\mid a$ (why?) and so $10\mid a$ (why?).
Once you've shown that $I\cap\Bbb Z$ is an ideal of $\Bbb Z,$ it follows that $I\cap\Bbb Z=n\Bbb Z$ for some nonnegative $n\in\Bbb Z.$ (Why?) Since $10\Bbb Z\subseteq I\cap\Bbb Z=n\Bbb Z,$ then $n\mid10.$ On the other hand, $n\in n\Bbb Z=I\cap\Bbb Z,$ so $10\mid n$ by the previous work, and so we're done.
A: If, $(a+bi)(3+i)=(3a-b)+(3b+a)i \in I \cap \mathbb Z$ In particular $(a+bi)(3+i)=(3a-b)+(3b+a)i \in \mathcal Z$ implies $3b+a=0$
take an element $(a+bi)(3+i)\in I \cap \mathbb Z $. To show any $m.(a+bi)(3+i) \in I \cap \mathbb Z$ for any $m\in\mathbb Z$. Now $m.(a+bi)(3+i)=m.(3a-b)+m.(3b+a)i $.  
Now, $m.(3b+a)=0$ as $(3b+a)=0$. Therefore,  $m.(a+bi)(3+i)\in \mathbb Z$ $m.(a+bi)(3+i)\in I$ as $I$ is an ideal in $\mathbb Z[i]$.
Therefore, for any $m \in \mathcal Z$, $m.(a+bi)(3+i)\in I \cap\mathbb Z$
Proves that $I\cap \mathbb Z$ is an ideal in $\mathbb Z$
To show $I\cap \mathbb Z$ is a subring is easy. Take $(a+bi)(3+i) =3a-b$  and  $(c+di)(3+i)=3c-d$ belongs to $I \cap \mathbb Z$ 
Then, $(a+bi)(3+i)+(c+di)(3+i)=3a-b+3c-d =3(a+c)-(b+d)$  Observe that   $((a+c)-(b+d)i(3+i)=3(a+c)-(b+d)$ Therefore sum of any two elements of $I \cap \mathbb Z$ belongs to $I \cap \mathbb Z$ . Therefore, it is a subring. 
A simple proof that $I\cap \mathbb Z=10\mathbb Z$
Note that if $(a+bi)(3+i)=(3a−b)+(3b+a)\in I\cap \mathbb Z $ iff $3b+a=0$ or $a=-3b$   i.e., any element of $ I\cap \mathbb Z $ will look like $(3b-ib)(3+i)=b(3-i)(3+i)=10b$ for some b belongs to $\mathbb Z$.
Therefore, $I\cap \mathbb Z =10 \mathbb Z$ 
A: write $\bar z = 3-i$ then the subset $\bar z \mathbb{Z}$ is in bijection with $I \cap \mathbb{Z}$ via multiplication by $z$. thus $z \bar z = 10 \in I \cap \mathbb{Z}$ and any element must be a multiple of 10, all of which are included
