Let $R= k[x,y,z]$ and $\mathfrak p_1 = (x,y), \mathfrak p_2 = (x,z), \mathfrak m = (x,y,z) \triangleleft R$. Then $$\mathfrak p_1 \mathfrak p_2 = \mathfrak p_1 \cap \mathfrak p_2 \cap \mathfrak m^2$$ is a minimal primary decomposition of $\mathfrak p_1 \mathfrak p_2$. Which components are embedded and which are isolated? My thoughts so far:
By computing quotients one easily sees that $\mathfrak p_1, \mathfrak p_2$ are prime and $\mathfrak m$ is maximal. Hence we know that the right hand side is a decomposition in primary ideals where $\mathfrak p_1$ is $\mathfrak p_1$-primary, $\mathfrak p_2$ is $\mathfrak p_2$-primary and $\mathfrak m^2$ is $\mathfrak m$-primary. Hence the associated primes are $(x,y), (x,z) \subseteq (x,y,z)$, which also shows that $(x,y),(x,y)$ are isolated and $(x,y,z) $ is embedded.
So it remains to show that we actually have equality. We clearly have $$\mathfrak p_1 \mathfrak p_2 \subseteq \mathfrak p_1, \mathfrak p_1 \mathfrak p_2 \subseteq \mathfrak p_2$$ and since $\mathfrak p_1 \subseteq \mathfrak m, \mathfrak p_2 \subseteq \mathfrak m$ also $\mathfrak p_1 \mathfrak p _2 \subseteq \mathfrak m^2$. But I am struggling with the converse inclusion. Is it helpful to explicitly compute $\mathfrak m^2$ or $\mathfrak p_1 \mathfrak p_2$? Are my other thoughts so far correct? Any help appreciated!
$\def\p{\mathfrak{p}} \def\m{\mathfrak{m}} \def\a{\mathfrak{a}} \def\spec{\operatorname{Spec}}$ If $f\in\p_1\cap\p_2$, then $f=gx+hy=\tilde{g}x+\tilde{h}z$ for some $g,h,\tilde{g},\tilde{h}\in k[x,y,z]$. If furthermore $f\in\m^2$, then necessarily $g,h,\tilde{g},\tilde{h}\in\m$. Since $gx\in\a$, we have $f\in\a\Leftrightarrow f-gx\in\a$, so we may replace $f$ by $f-gx$ and $\tilde{g}x$ by $(\tilde{g}-g)x$ and consider an expression $$ \tag{1}\label{1} f=hy=\tilde{g}x+\tilde{h}z, $$ $h,\tilde{g},\tilde{h}\in\m$. Write $h=\alpha y^2+q$, where $\alpha\in k[x,y,z]$ and $q\in\m$ is not a multiple of $y^2$. Since the RHS of \eqref{1} vanishes whenever $x=0$ and $ z=0$, we get $\alpha y^2=h(x=0,y,z=0)=0$; whence $\alpha=0$. Therefore $f=qy\in\a$.
Note that $\spec(k[x,y,z]/\p_1\p_2)$ is not the scheme-theoretic union (inside $\mathbb{A}_k^3$) of the $z$-axis $L_1=\spec(k[x,y,z]/\p_1)$ and the $y$-axis $L_2=\spec(k[x,y,z]/\p_2)$. Indeed, the scheme-theoretic union is $L_1\cup L_2=\spec(k[x,y,z]/\p_1\cap\p_2)$ and $$ \tag{2}\label{2} \p_1\cap\p_2=(x,yz), $$ so that $\p_1\cap\p_2\supsetneq\p_1\p_2$ (thus $\p_1$ and $\p_2$ are not coprime). To see \eqref{2}, note $(\supset)$ is easy. Conversely, if $f\in\p_1\cap\p_2$, then $f=gx+hy=\tilde{g}x+\tilde{h}y$. Thus $h(x=0,y,z=0)=0$, whence $h=\alpha x+\beta z$ and $f=(g+\alpha y)x+\beta yz\in(x,yz)$.