Find minimal primary decompositions of ideals (1) In $R = \mathbb{R}[x,y]/(x^2 + y^2 -1)$, find a minimal primary decomposition of $(\bar{x}^2)$.
(2) In $R = \mathbb{Z}[\sqrt{-5}]$, find a minimal primary decomposition of $(6)$.
For (1), I know that
$$(\bar{x}^2) = (1 - \bar{y}^2)=\left((1-\bar{y})(1+\bar{y})\right)$$
Then
$$((1-\bar{y})(1+\bar{y}))=(1-\bar{y})\cap(1+\bar{y}) $$
and $(1-\bar{y}), (1+\bar{y})$ are prime ideals in $R$,
since
$$R \cong \mathbb{R}[\bar{x}, \bar{y}] \text{ and } R/(1-\bar{y}) \cong \mathbb{R}[\bar{x}, \bar{y}]/(1-\bar{y})\cong \mathbb{R}[\bar{x}]$$ which is an integral domain and similarly for the case of $(1+\bar{y})$.
Then both of these ideals are primary ideals, too.
But for (2), I don’t know how to proceed.
All I’ve done is:
$$(6) = (2\cdot3)=((1+\sqrt{-5})\cdot(1-\sqrt{-5}))$$
And I’ve searched this question but all I’ve found was about ideal factorization.
It is written as a product of ideals, not as an intersection of primary ideals.
Is that factorization same with primary decomposition?
Any comment will be appreciated.
Thank you.
 A: For (1), I'm still working on.
For (2), I think I found a solution.
Here it is:
$Claim: (6) = (2) \cap (3, 1 + \sqrt{-5}) \cap (3 ,1 - \sqrt{-5})$
Consider $(2, 1+ \sqrt{-5}), (3, 1+ \sqrt{-5}$) and $(3, 1-\sqrt{-5})$.
$(2, 1+ \sqrt{-5}) = (2, 1-\sqrt{-5})$,
since $2 - (1+\sqrt{-5}) = 1 - \sqrt{-5}\\$

And $(3, 1+\sqrt{-5})(3, 1-\sqrt{-5}) = (3)$.

We claim that $(2, 1+\sqrt{-5})$ is maximal in $\mathbb{Z}[\sqrt{-5}]$.
Because $\Bbb{Z}[\sqrt{-5}] \simeq \mathbb{Z}[x]/(x^2 + 5)$,

\begin{align}
   \mathbb{Z}[\sqrt{-5}]/(2,1+\sqrt{-5}) &\simeq \mathbb{Z}[x]/(2,1+x,x^2+5) \\
   &\simeq \mathbb{Z}_2[x]/(1+x, 1+x^2) \\
   &\simeq \mathbb{Z}_2[x]/(1+x, (1+x)^2) \\
   &\simeq \mathbb{Z}_2[x]/(1+x) ≃ \mathbb{Z}_2
\end{align}
This shows that $(2, 1+ \sqrt{-5})$ is maximal in $\mathbb{Z}[\sqrt{-5}]$
Similarly, we can show that $(3, 1+\sqrt{-5}),\ (3, 1-\sqrt{-5})$ are both maximal in $\mathbb{Z}[\sqrt{-5}]$.
Since $(2) = (2, 1+\sqrt{-5})^2$, This implies that $(2)$ is $(2,1+\sqrt{-5})$-primary.
And $(3, 1+\sqrt{-5}) + (3, 1-\sqrt{-5}) = (1)$, so that
$$(3, 1+\sqrt{-5}) \cap  (3,1-\sqrt{-5}) = (3,1+\sqrt{-5})(3,1-\sqrt{-5})$$
Then we can prove our claim now.
\begin{align*}
   (6) = (2)\cdot(3) &= (2)\cap(3) \\
   &=(2) \cap \left((3,1+\sqrt{-5})\cdot(3, 1-\sqrt{-5})\right) \\
   &=(2) \cap  (3,1+\sqrt{-5}) \cap (3, 1-\sqrt{-5}) \\
   \end{align*}
$\sqrt{(2)}, \sqrt{(3,1+\sqrt{-5})}, \sqrt{(3, 1-\sqrt{-5})}$ are all distinct.
and none of three primary ideals $(2), (3, 1+\sqrt{-5}), (3, 1-\sqrt{-5})$ contains the intersection of the others.
This shows that this primary decomposition is indeed minimal.
Any comment will be appreciated.
