Here is the question I am trying to solve:
What is the factorization of 26 in $\mathbb Z[i]$ into irreducible?
My idea is:
26 = (5-i)(5+i) but it seems like this is an incomplete answer. Could anyone tell me what is missing please? Or what is the correct solution?
Thanks for your question. There is a theorem from 'concrete abstract algebra' by Neils Lauritzen which states (in the ring of Gaussian integers) that
Theorem: [either a number is a prime in $\mathbb{Z}$ (or $i\mathbb{Z}$)] or a number (Gaussian integer) $s$ is prime if and only if $N(s)=p$ where $p$ is prime and $N$ is the norm function. If $z=a+bi$ then $N(z)=a^2+b^2$.
$26=(5+i)(5-i)$ is a good start, but applying the theorem above $5-i$ and $5+i$ are not prime (in this ring the concept of prime and irreducible are the same) elements in $Z[i]$. Given $5+i$ is reducible, if we assume $5+i=(p+qi)(x+yi)$, we see $px-qy=5$ and $py+qx=1$. Hence $26=(p^2x^2+q^2y^2)+(p^2y^2+q^2x^2)$. Hence $26=(p^2+q^2)(x^2+y^2)$. It follows (as $26=2 \cdot 13$) that $p=1$, $q=-1$ and $x=2$, and $y=3$. Hence $5+i=(1-i)(2+3i)$, and $5-i=(1+i)(2-3i)$. The elements $1\pm i$, and $2 \pm 3i$ are irreducible (by using the theorem again). Hence $26=(1-i)(1+i)(2+3i)(2-3i)$ is a factorization [note factorization is unique upto multiplication by units] of 26 into irreducibles. Cheers.
As a similar query, can you write $2+7i$ as a product of irreducibles in $\mathbb{Z}[i]$ ?