Primary descomposition of ideals

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I'd appreciate if someone could help me a bit with this problem.

Considering $\mathfrak{p}=(x,y), \mathfrak{q}=(x,z)$ and $\mathfrak{m}=(x,y,z)$ ideals in $k[x,y,z], k$ field.

Is $\mathfrak{p}\mathfrak{q}=\mathfrak{p}\cap \mathfrak{q}\cap\mathfrak{m}^2$ a minimal primary descomposition of $\mathfrak{p}\mathfrak{q}$?

Which component is isolated and which are embbeded?

I tell you what I've thought:

It's known that $k[x,y,z]/\mathfrak{m}\cong k$ and $k$ field, so the ideal $\mathfrak{m}=(x,y,z)$ is maximal $\Rightarrow $ $\mathfrak{m}^k$ are $\mathfrak{m}$-primary, so, particulary $\mathfrak{m}^2$ is $\mathfrak{m}$-primary.

$\mathfrak{p}, \mathfrak{q}$ are prime ideals in $k[x,y,z]$, because it's known that ideals $(x_1,...,x_i), 1\leq i \leq n$ are prime in $k[x_1,...,x_n]$. Hence $\mathfrak{p}, \mathfrak{q}$ are primary ideals.

How could I continue?

Thanks.

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A minimal primary decomposition is one where the radicals of primary ideals are distinct and there is no containment between a primary ideal and the intersection of all the others. It seems to me that this is the case with your decomposition.

The isolated components are $\mathfrak p$ and $\mathfrak q$ since they are minimal among the associated primes of $\mathfrak p\mathfrak q$, that is, $\{\mathfrak p, \mathfrak q, \mathfrak m\}$, while $\mathfrak m$ is an embedded prime.