Answer to Question #147956 in Abstract Algebra for Sourav Mondal

Question

Show that R = {a + b√-3 a, b belongs to Z} i s not a
unique factorization domain by expressing
4 as a product of two irreducible elements in
R in two different ways.

Answer 1

We can clearly see that

$4 = 2\cdot 2 = (1+1\sqrt{-3})\cdot (1-1\sqrt{-3})$

First let's prove that they are irreducible.

We can view elements of $R$ as complex numbers (obviously) and thus we know that if $2=(a+b\sqrt{-3})(c+d\sqrt{-3})$ we have $(a^2+3b^2)(c^2+3d^2)=4$ and as the last expression is an equation in $\mathbb{Z}$ we can easily conclude that either $a^2+3b^2=1,$ or $c^2+3d^2=1$ and thus we have $2= (\pm2) \cdot (\pm1)$ and $2$ is irreducible.

Now let's prove that $1\pm\sqrt{-3}$ are irreducible. By the same argument of absolute values in $\mathbb{C}$ we find that $1+\sqrt{-3} = (\pm1)\cdot (\pm1 \cdot (1+\sqrt{-3}))$ and same for $1-\sqrt{-3}$ . Therefore they are also irreducible.

In addition, while proving this we have studied all unit elements : $\pm1$ and so $1+\sqrt{-3}$ and $2$ are not associated. Thus we conclude that $R$ is not a unique factorization domain.