Factoring into Primes in Quadratic Integers

examplequadratic-integers

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Factor into Primes in $\mathbb{Z}[i]$

$3+4i:N(3+4i)=25=5^2$ Try $1+2i$,

$$\begin{array}{rl}\frac{3+4i}{1+2i}& =\frac{(3+4i)(1-2i)}{(1+2i)(1-2i)},\\ & =\frac{11-2i}{5}\notin \mathbb{Z}[i].\end{array}$$

Try $1-2i$,

$$\begin{array}{rl}\frac{3+4i}{1-2i}& =\frac{(3+4i)(1+2i)}{(1+2i)(1-2i)},\\ & =\frac{-5+10i}{5},\\ & =-1+2i\in \mathbb{Z}[i].\end{array}$$

So,

$$\begin{array}{rl}3+4i& =(1-2i)(-(1-2i)),\\ & =-(1-2i{)}^2,\\ & =i^2(1-2i{)}^2=(i(1-2i){)}^2,\\ & =(2+i{)}^2\end{array}$$

Multiple Prime Factorization in $\mathbb{Z}[\sqrt{2}]$

$$17=(5+2\sqrt{2})(5-2\sqrt{2})=(3\sqrt{2}+1)(3\sqrt{2}-1)=(37+26\sqrt{2})(37-26\sqrt{2})$$

Because we know that unique prime factorization (up to a unit multiple) exists in $\mathbb{Z}[\sqrt{2}]$, let’s try,

$$\frac{5+2\sqrt{2}}{3\sqrt{2}+1}=5+2\sqrt{2}1+3\sqrt{2}=\frac{(5+2\sqrt{2})(1-3\sqrt{2})}{(1+3\sqrt{2})(1-3\sqrt{2})}\notin \text{ Units of }\mathbb{Z}[\sqrt{2}].$$

…some other work… → $3\sqrt{2}-1$ is a unit multiple of $5+2\sqrt{2}$!

This is all just a result of multiplying a factorization by $u\cdot u^{-1}$ where $u$ is a unit and then “attaching” each of them to a different factor.

Lack of Unique Factorization

There are examples of rings in which unique factorization does not hold. For example, in $\mathbb{Z}[\sqrt{5}]$,

$$4=2\cdot 2=(\sqrt{5}+1)(\sqrt{5}-1).$$

Which are all prime in $\mathbb{Z}\sqrt{5}$.