This problem is taken from Dummit and M. Foote Books abstract algebra
let R be the quadratics integer ring $\mathbb{Z} [\sqrt -5]$. Define the ideal $I_2 = (2 ,1 + \sqrt -5)$
Is the product of $I_2 \times I_2$ is principal ideal ?
My attempt : i thinks No
$2$ is irreducible in $R$, now take $2 = xy$ hence $x=2$ and $y=1$
$1+ \sqrt {-5} \in (2) \Rightarrow 1+\sqrt {-5}=2(m+n\sqrt {-5}) \Rightarrow m=\frac 12$ which is a contradiction because $m \in \Bbb Z$.
So $I_2$ is not principal ideal
Now $I_2 \times I_2 = (2, 1+ \sqrt{-5}) (2, 1+ \sqrt{-5})= ( 4, 1 +5) = (4,6)$
in the similar way
$4$ is irreducible in $R$, now take $4 = xy$ hence $x=2$ and $y=2$
$1+ \sqrt {-5} \in (4) \Rightarrow 1+\sqrt {-5}=4(m+n\sqrt {-5}) \Rightarrow m=\frac 14$ which is a contradiction because $m \in \Bbb Z$.
So $I_2 \times I_2 = I_2^2$ is not principal ideal
Is its true ?
Best Answer
First, showing that $1 + \sqrt{-5} \notin (2)$ does not show that $(2, 1 + \sqrt{-5})$ is not principal, it shows that $(2, 1 + \sqrt{-5}) \ne (2)$. It is possible that there is some number $a + b\sqrt{-5}$ that is neither $2$, nor $1 + \sqrt{-5}$ such that $(2, 1 + \sqrt{-5}) = (a + b\sqrt{-5})$.
Likewise, the ideal $(4,6)$ is not equal to $(4)$ because $6 \notin (4)$, nor is it equal to $(6)$. However, $(4,6) = (2)$ because $2 = -1 \cdot 4 + 1 \cdot 6 \in (4,6)$, so $(2) \subseteq (4,6)$. And, $4, 6 \in (2)$ so $(4,6) \subseteq (2)$.
Remember that an ideal $(x,y)$ is the set $\{ax + by : a, b \in R\}$ so when you look for a generator, the generator should be in the set, so it should be of the form $ax + by$. But that generator does not need to be $x$ or $y$.
Likewise, when you multiply $(u,v) \cdot (x,y)$ you should be thinking about how each ideal is defined (linear combinations of its generators) and how the product of ideals is defined. So if you work it out: \begin{align} \{s \cdot t : s \in (u,v), t \in (x,y)\} &= \{s \cdot t : s = au + bv, t = cx + dy, a,b,c,d \in R\} \\ &= \{(au + bv)(cx + dy) : a,b,c,d \in R\} \\ &= \{ (ac) ux + (ad) uy + (bc)vx + (bd) vy : a,b,c,d \in R \} \end{align}
So this is contained in the ideal $(ux,uy,vx,vy)$. Without too much effort you can show that $ux, uy, vx, vy \in (u,v) \cdot (x,y)$ so that in fact, $(u,v) \cdot (x,y) = (ux,uy,vx,vy)$.
Once you work out properly what $I_2 \cdot I_2$ is, you can show that $I_2 \cdot I_2 = (2)$ just like how I showed that $(4,6) = (2)$.