# Why would a field *not* be considered a discrete valuation ring?

There are two theorems in Matsumura (p. 78-9)

Theorem 11.1 Let $R$ be a valuation ring. Then the following conditions are equivalent:

(1) $R$ is a DVR

(2) $R$ is a PID

(3) $R$ is Noetherian

and

Theorem 11.2 Let $R$ be a ring; then the following conditions are equivalent:

(1) $R$ is a DVR

(2) $R$ is a local PID, and not a field

(3) $R$ is a Noetherian local ring, dim $R >0$ and the maximal ideal $\mathfrak{m}_R$ is principal

(4) $R$ is a one-dimensional normal Noetherian local ring.

Question

Suppose we have a field $K$. Then $K$ is trivially a valuation ring. It is also (of course) a PID, then according to Thereorem 11.1 it must be a DVR.

Since $K$ is a DVR, $K$ is not a field according to 11.2. Am I misunderstanding something?

• @user26857 Matsumura himself says that fields are valuation rings. On page 71, he says "The case R=K is the trivial valuation ring". – Artus Dec 30 '14 at 19:04
• I think you're referring to me, not user26857. I made a mistake and misinterpreted his definition as I admitted in the next comment. My apologies. – Weaam Dec 30 '14 at 19:06
• @Weaam I was actually writing to both of you but I can only refer to one. user26857 said he didn't want to argue with Matsumura about whether or not fields were valuation rings...so I just wanted to state Matsumura's opinion. – Artus Dec 30 '14 at 19:08
• I think the author some how assume (without saying clearly) that from now on we will only talk about non-trivial valuation rings. I'm saying this because just before Theorem 11.1 the following definition was given: a valuation ring whose value group is isomorphic to $\mathbb{Z}$ is called a discrete valuation ring (DVR). so the Theorem 11.1 doesn't make sense unless you discard the trivial case (i.e. field). – Krish Dec 30 '14 at 19:10

The definition of DVRs in Matsumura, as valuation rings whose value group is isomorphic to $\mathbb Z$, doesn't allow you to consider the fields as DVRs.