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?
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.
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.
$p$ 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 $r$.
