This question is related to How to factor ideals in a quadratic number field?
In Algebraic Number Theory by W. Stein he makes a remark about the factorization of $65537$ in $\mathbb Z[i]$. I checked this in Sage and the result is different.
What is an explanation of this difference?
Things are made easier by the fact that $\mathbb{Z}[i]$ is a principal ideal domain with unique factorization. This means all ideals are principal ideals, so any prime ideals correspond to principal ideals even if presented to not look like principal ideals, and all prime ideals are generated by prime numbers. So if $\langle n \rangle = \langle p \rangle \langle q \rangle$, then $n = upq$, where $p$ and $q$ are primes and $u$ is some unit.
As you know, 65537 is a prime number in $\mathbb{Z}$. So to factor $\langle 65537 \rangle$ in $\mathbb{Z}[i]$, we need to show that either it is also prime in $\mathbb{Z}[i]$ or that it can be expressed as $(a - bi)(a + bi)$, and that answer can be found by solving $n = a^2 + b^2$. A well-known result tells us an odd prime in $\mathbb{Z}$ can be expressed as a sum of two squares if and only if is of the form $4k + 1$, and 65537 abundantly qualifies. Furthermore, since $65536 = 2^{16}$, from basic facts of exponentiation we can readily determine that $\sqrt{2^{16}} = 2^8$. Therefore $65537 = (2^8 - i)(2^8 + i)$ (there are three different answers which are technically the same because of multiplication by units). Both $2^8 - i$ and $2^8 + i$ are prime, as a standard issue examination of norms will show.
So $\langle 2^8 - i \rangle$ is a principal ideal and a prime ideal in $\mathbb{Z}[i]$, and the same goes for $\langle 2^8 + i \rangle$ and therefore $\langle 65537 \rangle = \langle 2^8 - i \rangle \langle 2^8 + i \rangle$.
But what about this $\langle 65537, 2^8 - i \rangle$ business, what's that all about? It doesn't look like a prinicipal ideal, but since $\mathbb{Z}[i]$ is a principal ideal domain, it must be a principal ideal, even if a trivial one, e.g., $\langle 1 \rangle$. $\langle 65537, 2^8 - i \rangle$ consists of numbers of the form $65537r + (2^8 - i)s$, with $\{r, s\} \in \mathbb{Z}[i]$. If we set $r = 0$, we see that $\langle 65537, 2^8 - i \rangle$ contains all numbers in $\langle 2^8 - i \rangle$. But if we set $s = 0$, we obtain all the same numbers we'd obtain if instead we set $r = 0$ and $s = (2^8 + 1)t$, with $t \in \mathbb{Z}[i]$. So $\langle 65537, 2^8 - i \rangle$ is just a confusing way of saying $\langle 2^8 - i \rangle$.
I know next to nothing about Sage, so I could be wrong about this, but it seems to be saying that $\langle 65537 \rangle = \langle 256 + i \rangle \langle 256 - i \rangle$. That's how I'm understanding that stuff about
Fractional ideal (w + 256). It looks to me like the book and the computer are in complete agreement.