# Find the associated primes of $(x_0) \cdot (x_0, x_1) \cdot \dots \cdot (x_0, \dots, x_r)$ in $k[x_0, \dots x_r]$

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?

Let $R = k[x_0,\dots,x_r]$, $P_n = (x_0,x_1,\dots,x_n)$ and define $I_n = P_0 P_1 \cdots P_n$. Clearly $I_0 = P_0$ is a prime ideal and so the set of associated primes of $R/I_0$ is just $\left\{P_0\right\}$. Now, suppose by induction that the associated primes of $R/I_n$ are precisely $\left\{P_0,P_1\dots,P_n\right\}$. Consider $I_{n+1}$. Since $I_{n+1}R_{x_{n+1}} = I_n R_{x_{n+1}}$, it follows that all $P_0,\dots,P_n$ are associated primes of $R/I_{n+1}$. Moreover, $P_{n+1}$ is precisely $I_{n+1} : x_0 x_1 \dots x_n$, and so $P_{n+1}$ is an associated prime of $R / I_{n+1}$ as well. Now let us show that there are no other associated primes. Towards that end, notice that the associated primes of $I_{n+1}$ are ideals of the form $(x_{j_1},\dots,x_{j_\ell})$, with $j_i \le n+1$, since the smallest ring that contains $I_{n+1}$ is $k[x_0,\dots,x_{n+1}]$. If $Q$ is an associated prime different than $P_0,\dots,P_{n+1}$, then one of its generators has to be $x_{n+1}$ (otherwise localizing at $x_{n+1}$, $Q$ must be one of the $P_0,\dots,P_n$). Since $Q \supset I_{n+1}$, $Q$ must contain one of the $P_0, \dots\, P_{n+1}$. Since $x_{n+1} \in Q$ the only possibility is $Q \supset P_{n+1}$, from which it follows that $Q = P_{n+1}$, contradiction and proof completed.
• Thanks for the help! The only part I'm still confused about is why $Q$ has to be of the form $(x_{j_1}, \dots, x_{j_l})$. Why is that true? Jul 1, 2017 at 19:01
• @user459196: You are welcome. There are more than one ways to see that. One way is that the associated primes of an ideal essentially depend on the smallest ring that contains the ideal. In our case $I_{n+1}$ can be viewed as the extension to $k[x_0,\dots,x_r]$ of an ideal of $k[x_0,\dots,x_{n+1}]$. It is then true that the associated primes of the extended ideal are the extensions of the associated primes; see exercise 4.7 in Atiyah-MacDonald. Another way to see this is by using properties about monomial ideals. Jul 1, 2017 at 19:08
• I see why $j_i \le n + 1$ if $(x_{j_1}, \dots, x_{j_l})$ is an associated prime, but what's stopping something like $(x_1 - x_2)$ from being an associated prime if $n > 1$? (That might have been what you just explained, and I just didn't understand, so sorry if that's the case.) Jul 1, 2017 at 19:21
• @user459196: I see. Here i have used the fact that associated primes of monomial ideals are monomial ideals. $x_1-x_2$ is not a monomial. Or put another way, if $x_1-x_2$ takes the monomial $u$ inside $I_{n+1}$, then so does $x_1$ and $x_2$. Jul 1, 2017 at 19:27