A lemma to identify dvrs

In this post I will prove a commutative algebra lemma, which is proved in page 7 of Serre’s Local Fields. This lemma is useful if you find a ring in the wild and want to know that it’s a discrete valuation ring.

Proposition: Let R, \frak{m} be a noetherian local domain where \frak{m} = (\pi) and \pi is not nilpotent. 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

Hilbert’s Basis Theorem

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.

Thus J = J'. Since J is therefore finitely generated, this proves R[x] is noetherian.