Primes in $\mathbb Z$ and $\mathbb Z[i]$ 
Give an example with justification of two primes $p$ and $q$ in $\mathbb Z$ such that $p$ is a prime in $\mathbb Z[i]$ but $q$ is not a prime in $\mathbb Z[i]$. 

I know that $\mathbb Z[i]$ is the set of Gaussian integers which also form a ring under the usual addition and multiplication of complex numbers. But how to define prime number in $\mathbb Z[i]$ since it will involve division operation?
 A: This answer is more for the benefit of other people who come across this question rather than for the original asker. For all we know, the original asker will never come on this site again.
However, I will assume that those who come across this question know about as much as the original asker at the time this question was asked.
Addition of complex numbers is easy. If $a, b, c, d$ are all integers in $\mathbb Z$, it follows that $$(a + bi) + (c + di) = (a + c) + (b + d)i.$$ Subtraction can easily be deduced from the foregoing.
The multiplication formula is easily obtained by applying FOIL: $$(a + bi)(c + di) = ac + adi + bci - bd = (ac - bd) + (ad + bc)i.$$ As an exercise, using a few specific pairs of values of $a$ and $b$, verify that $$(a - bi)(a + bi) = a^2 + abi - abi - (-1)b^2 = a^2 + b^2.$$
This tells us an important fact: the norm of a nonzero complex number is that complex number times its conjugate, and that norm is a positive number.
Furthermore, if the norm is a number that is prime in $\mathbb Z$, that tells us that that number is not prime in $\mathbb Z[i]$. For example, 5 is prime in $\mathbb Z$ but not in $\mathbb Z[i]$ on account of $2 + i$, since now we know that $(2 - i)(2 + i) = 5$ (there's also $1 + 2i$, but that's not a distinct factorization since $(2 + i)i = -1 + 2i$).
With norms, we can bypass actually having to perform division. But... aren't you the least bit curious about how to do division in $\mathbb Z[i]$? The formula is $$\frac{a + bi}{c + di} = \left(\frac{ac + bd}{c^2 + d^2}\right) + \left(\frac{bc - ad}{c^2 + d^2}\right)i.$$ Obviously $c^2 + d^2$ is an integer if $c$ and $d$ are integers. In order for $a + bi$ divided by $c + di$ to be a number in $\mathbb Z[i]$, we need $ac + bd$ and $bc - ad$ to both be multiples of $c^2 + d^2$, which doesn't seem terribly difficult.
However, if $a$ is a positive real integer that is prime in $\mathbb Z$ and we wish to prove it's also prime in $\mathbb Z[i]$, we set $b = 0$ and then we need to demonstrate that $ac$ and $-ad$ (since $bd$ and $bc$ zero out) can never both be multiples of $c^2 + d^2$...
But then it's much easier to use norms. Since $b$ would be 0 and $0^2 = 0$, the norm of a positive real integer $a$ in $\mathbb Z[i]$ is $a^2$. If $a$ is prime in $\mathbb Z$ but not in $\mathbb Z[i]$, its norm is $a^2$ and the norm of a prime in $\mathbb Z[i]$ with nonzero imaginary part that divides $a$ would be $a$ itself.
Then the norm of 5 is 25 and the norm of $2 + i$ is 5. To prove that a prime like 3 or 7 or 11 (or any positive prime of the form $4k - 1$) is not prime in $\mathbb Z[i]$, we just need to show $c^2 + d^2$ has no solution in integers.
A: Division in $\mathbb{Z}[i]$ is the same as in $\mathbb{Z}$ : You say that $a\mid b$ iff $\exists c\in \mathbb{Z}[i]$ such that $b=ac$.
For instance, $(1+i) \mid 2$ since $(1+i)(1-i) = 2$.
Now try to define what it means for an element to be prime.
By what I said above, $2$ is prime in $\mathbb{Z}$ but not in $\mathbb{Z}[i]$.
Try to show that $3$ is irreducible - ie. if you can write $3 = \alpha\beta$ for some $\alpha, \beta \in \mathbb{Z}[i]$, then either $\alpha$ or $\beta$ must belong to the set $\{\pm 1,\pm i\}$ of units of $\mathbb{Z}[i]$.
Check that irreducibility implies that 3 is prime.
A: $p$ is a prime-element in a ring if $p\neq 0$, $p$ is not a unit and if $p\mid uv$ implies that $p\mid u$ or $p\mid v$. Here $p\mid u$ means that $u=pr$ for some $r$.
