Primary decomposition of an ideal (exercise 7.8 in Reid, Undergraduate Commutative Algebra) [duplicate]

I would like to understand how to use geometry to solve the problem 7.8 from Reid's book Undergraduate Commutative Algebra. The problem is the following

Let $k$ be a field and consider the ideal $I = (xy, x - yz) \subset k[x,y,z]$. Find a primary decomposition of $I$.

So the vanishing locus $V(I)$ is the two axis $Y$ and $Z$, and if $I = q_1 \cap \cdots \cap q_n$ with all $q_i$ being $p_i$-primary, we could guess that we can take $p_1 = (x,z)$ and $p_2 = (x,y)$. Moreover, the vanishing of $xy$ is the $Z$-axis, but with double multiplicity (is this true ?), so we could guess to take $q_2 = (x,y)^2$, but we need $x - yz$ to be in there, so I think that $q_2 = (xy, y^2, x - yz)$ and $q_1 = (x,z)$ would give a decomposition into primary ideals ($x^2$ is automatically in $q_2$, don't need it).

Question 1 : How to make sure that these primes $p_1$ and $p_2$ are the only ones ? We can compute them by $\operatorname{Ass}(k[x,y,z]/(xy,x - yz))$, but wasn't able to do it.

Question 2 : Is $(x,z) \cap (xy, y^2, x - yz)$ a primary decomposition of $I$ ?

Question 3 : What is a better way to get to this answer with geometric intuition and not algebraic brute-force ?

Thanks

Solution 1:

For Question 2, we show that $(xy,y^2,x-yz)$ is a $(x,y)$-primary ideal. Since $xy = y(x-yz) + z y^2$, $(xy,y^2,x-yz) = (y^2,x-yz).$ Write $I_2 = (y^2,x-yz)$. In order to show that $I_2$ is a $(x,y)$-primary ideal, we show that non zero divisors of $k[x,y,z]/I_2$ are nilpotent. Observe that $$ k[x,y,z]/(y^2,x-yz) \cong k[y,z]/(y^2). $$ Hence zero divisors contain $y$ as a factor, and this implies that they are nilpotent.

Let $I_1 = (x,z)$. We show that $I = I_1 \cap I_2$. Since $I \subseteq I_i$ for $i=1,2$, $I \subseteq I_1 \cap I_2$. We show that $I_1 \cap I_2 \subseteq I$. Let $a y^2 + b(x-yz) \in I_1 \cap I_2$ for some $a,b \in k[x,y,z]$. Since $a y^2 + b(x-yz) \in I_1$ and $x-yz \in I_1$, $ay^2 \in I_1 = (x,z)$. This implies that $a \in (x,z)$ since $y$ is a non zerodivisor on $k[x,y,z]/(x,z)$. Write $a = fx + gz$ for some $f,g \in k[x,y,z]$. Then $ay^2 = (fx + gz)y^2 = fxy^2 + gzy^2$. Observe that $zy^2 = -y(x-yz) + xy \in I$. This shows that every element in $I_1 \cap I_2$ is in $I$. Hence we have $I = I_1 \cap I_2$ where both $I_i$ are primary ideals. In particular, it is a primary decomposition of $I$.

An answer to Question 1 follows from this as well.