noetherian ring | Arithmetic variety

In this post I will prove a special case of Krull’s intersection theorem, which can be proved without invoking the Artin-Rees lemma. This result is useful in the study of discrete valuation rings. The following proof is from Serre’s Local Fields.

Proposition: Let $R, \frak{m}$ be a noetherian local domain where $\frak{m} = (\pi).$ Then $\bigcap_{n \geq 0} \frak{m}^n = 0$ .

Proof: Suppose $y \in \bigcap_{n \geq 0} \frak{m}^n$ . Then $y = \pi^n x_n$ for all $n \geq 0$ . So for all $n$ $\pi^nx_n = \pi^{n+1}x_{n+1}$ . Since $R$ is a domain, $x_n = \pi x_{n+1}$ .

Therefore consider the ascending chain $(x_0) \subseteq (x_1) \subseteq \cdots$ . This eventually stabilizes for high enough $n$ since $R$ is noetherian, so for some $n$ , $x_{n+1} = cx_n$ . Thus $x_{n+1} = c\pi x_{n+1}$ , so $(1-c\pi)x_{n+1} = 0$ . But $1-c\pi$ is a unit, so $x_{n+1} = 0$ , so $y = 0$ . $\square$

This theorem holds more generally even if $R$ is not assumed to be a domain, but the proof is more complicated (but still among the same lines).

Proposition: Let $R, \frak{m}$ be a noetherian local ring where $\frak{m} = (\pi)$ and $\pi$ is not nilpotent. Then $\bigcap_{n \geq 0} \frak{m}^n = 0$ .

Proof: Let $\frak{u}$ be the ideal of elements that kill some power of $\pi$ . We will use variables $u_1, u_2, \dots$ to refer to elements of $\frak{u}$ . Since $R$ is noetherian, $\frak{u}$ must be finitely generated, so all elements of $\frak{u}$ kill $\pi^N$ for some fixed $N$ .

Now suppose $y \in \bigcap_{n \geq 0} \frak{m}^n$ . $\pi^nx_n = \pi^{n+1}x_{n+1}$ , so $\pi^n(x_n - \pi x_{n+1})$ . Thus $x_n - \pi x_{n+1} \in \frak{u}$ .

Consider the ascending chain $\frak{u} + (x_0) \subseteq \frak{u} + (x_1) \subseteq \cdots$ . Since $R$ is noetherian it must eventually stablize, so for some $n$ , $x_{n+1}$ can be written as $u_1 + cx_n$ . But recall that $x_n = u_2 + \pi x_{n+1}$ . So $x_{n+1} = u_1 + c(u_2 + \pi x_{n+1}) = u_3 + c\pi x_{n+1}$ so $(1-c\pi)x_{n+1} = u_3$ . $1-c\pi$ is a unit since $\frak{m} = (\pi)$ , and $R$ is local, so $x_{n+1} \in \frak{u}$ . If we force $n$ to be large enough to surpass $N$ , then $\pi^{n+1}x_{n+1} = 0$ , so $y = 0$ . $\square$

Here is a proof of Hilbert’s Basis Theorem I thought of last night.

Let $R$ be a noetherian ring. Consider an ideal $J$ in $R[X]$ . Let $I_i$ be the ideal in $R$ generated by the leading coefficients of the polynomials of degree $i$ in $I$ . Notice that $I_i \subseteq I_{i+1}$ , since if $P \in I_i$ , $xP \in I_{i+1}$ , and it has the same leading coefficient. Thus we have an ascending chain $\dots \subseteq I_{i-1} \subseteq I_i \subseteq I_{i+1} \subseteq \dots$ , which must terminate, since $R$ is noetherian. Suppose it terminates at $i = n$ , so $I_n = I_{n+1} = \dots$ .

Now for each $I_i$ choose a finite set of generators (which we can do since $R$ is noetherian). For each generator, choose a representative polynomial in $J$ with that leading coefficient. This gives us a finite collection polynomials: define $J_i$ to be the ideal of $R[x]$ generated by these polynomials.

Let $J' = J_0 + J_1 + \dots + J_n$ . I claim $J = J'$ . Assume for the sake of contradiction that there is a polynomial $P$ of minimal degree (say $i$ ) which is in $J$ but $J'$ . If $i \leq n$ , there is an element of $P' \in J'$ with the same leading coefficient, so $P - P'$ is not in $J'$ but has degree smaller than $i$ : contradiction. If $P$ is of degree $i > n$ , then there is an element of $P'$ of $J_n$ which has the same leading coefficient as $P$ . Thus $P - x^{i-n}P'$ is of degree smaller than $i$ but is not in $J'$ : contradiction.