What is a primary decomposition of the ideal $I = \langle xy, x - yz \rangle$? 
Given the ring $k[x,y,z]$, where $k$ is a field, and an ideal $I=(xy,x-yz)$, find a primary decomposition of $I$. 

I tried to draw the graph of the variety of $I$ and get a decomposition of $(x,y)\cap(x,z)$ (the two prime ideals corresponds to the irreducible components of the variety), but apparently these two ideals are not the primary decomposition of $I$, and I don't know how to fix this.
I would really appreciate it if someone can help me. thx
 A: If you are planning to work more on ring theory, I recommend installing Macaulay2 on your computer. It is a computer algebra software for computing with rings and ideals and modules.
For example, to compute the primary decomposition of your ideal, one would type the following commands:
R = QQ[x,y,z]
I = ideal (x*y, x-y*z)
primaryDecomposition I

{ideal (z, x), ideal (y*z - x, y^2 , x*y)}

This actually disagrees with Jim Belk's answer above. 
A: You can only use geometry to find the primary decomposition if the given ideal is actually an intersection of prime ideals.  In your case, it is true that
$$
V(xy,x-yz) \;=\; V(x,y) \cup V(x,z)
$$
but
$$
(xy,x-yz) \;\subsetneq\; (x,y) \cap (x,z).
$$
For example, $x\in (x,y)\cap (x,z)$, but $x\notin (xy,x-yz)$.
Instead, the given ideal is the intersection of two primary ideals
$$
(xy,x-yz) \;=\; Q_1 \cap Q_2
$$
where the radicals of $Q_1$ and $Q_2$ are the prime ideals that you have found:
$$
\sqrt{Q_1}=(x,y) \qquad\text{and}\qquad \sqrt{Q_2} = (x,z).
$$
To find $Q_1$ and $Q_2$, observe that
$$
(xy,x-yz) \;=\; (y^2z,x-yz),
$$
since $xy - y^2 z = y(x-yz)$.  We can now factor the $y^2z$:
$$
(y^2z,x-yz) \;=\; (y^2,x-yz) \cap (z,x-yz)
$$
To prove this equation, observe that all three ideals contain $(x-yz)$.  The quotient $k[x,y,z]/(x-yz)$ is isomorphic to $k[y,z]$, and obviously $(y^2z)=(y^2)\cap (z)$ in $k[y,z]$, so lifting back to $k[x,y,z]$ gives the desired equation.
It is easy to see that $(y^2,x-yz)$ is primary, and $(z,x-yz) = (x,z)$ is actually prime.  We conclude that
$$
Q_1 = (y^2,x-yz) \qquad\text{and}\qquad Q_2 = (x,z)
$$
