I'm studying Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry, and I'm having some trouble with the following problem (Exercise 3.9 in the book):
Let $k$ be a field, and let $I$ be the ideal $(x_0) \cdot (x_0, x_1) \cdot \dots \cdot (x_0, \dots, x_r)$ of $k[x_0, \dots x_r]$. Show that the associated primes of $I$ are $(x_0), (x_0, x_1), \dots, (x_0, \dots, x_r)$.
The back of the book contains the hint
Do induction on $r$, inverting $x_r$, and using Theorem 3.1.
Theorem 3.1, among other things, states that the associated primes of $R[U^{-1}]$ correspond to the associated primes of $R$ that are disjoint from $U$. Based on this, it seems like the inductive step ought to involve proving something like this:
Let $R_i$ be the ring $$k[x_0, x_1, \dots, x_r, x_r^{-1}, x_{r-1}^{-1} \dots, x_{i+1}^{-1}],$$ and let $I_i$ be the ideal $$(x_0) \cdot (x_0, x_1) \cdot \dots \cdot (x_0, \dots, x_i)$$ of $R_i$. Assume that the associated primes of $I_i$ are $(x_0), (x_0, x_1), \dots, (x_0, \dots, x_i)$. Then, the associated primes of $I_{i+1}$ are $(x_0), (x_0, x_1), \dots, (x_0, \dots, x_{i+1})$.
By Theorem 3.1, it suffices to show that the only associated prime of $I_{i+1}$ that contains $x_{i+1}$ is $(x_0, \dots, x_{i+1})$. However, I'm not sure how to show this.
Is this what Eisenbud meant with his hint? If so, how do I finish the proof?