(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.
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.