Compute intersection of ideals in a polynomial ring 
Consider the ideals $\mathfrak{p}_1=(x,y)$, $\mathfrak{p}_2=(x,z)$ and $\mathfrak{m}=(x,y,z)$ in $k[x,y,z]$. How to show that $\mathfrak{p}_1\mathfrak{p}_2=\mathfrak{p}_1\cap\mathfrak{p}_2\cap\mathfrak{m}^2$? 

I know how to compute products of ideals, for example $\mathfrak{p}_1\mathfrak{p}_2=(x^2,xy,xz,yz)$. But I don't know how to handle intersections. Also, would it be easy to see that the right hand side equals $(x^2,xy,xz,yz)$ without knowing the answer? Thank you in advance.
 A: All three ideals are monomial, and their intersection is also a monomial ideal generated by the least common multiples of their generators; see here.
A: You can write down how a generic elements in both ideals looks like. The rules are
In an commutative ring $A$ with $1$, the ideal $(a_1,a_2,a_3,...)$ consists of elements of the form $\sum_i s_i x_i$, where $s_i\in A$ and $x_i \in \{a_1,a_2,...\}$, i.e. they are finite linear combinations of the generators. The elements of an ideal $\mathfrak p_1\mathfrak p_2$ are of the form $\sum_i s_i \,x_iy_i$ where $s_i\in A$, $x_i\in \mathfrak p_1$, $y_i \in \mathfrak p_2$. It follows that $(a_1,a_2,...)(b_1,b_2,...)= (a_1b_1, a_1b_2, a_2b_1, a_2b_2, ...)$. (see for example the first chapter of Atiyah's and Macdonald's book).
So... $\mathfrak m^2 = (x^2,y^2,z^2,xy,xz,yz)$. Next show that $\mathfrak p_1 \cap \mathfrak m^2 = (x^2,y^2,xy,xz,yz)$. Use that you know how a generic element of $(x,y)$ looks like, and that you know how a generic element of $(x^2,y^2,z^2,xy,xz,yz)$ looks like, and think about when those overlap. In the same way show that $(x,z) \cap (x^2,y^2,xy,xz,yz)=(x^2, xy, xz, yz)$.
