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

Question

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?

Answer 1

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.

Answer 2

Following the definitions of the book:

A field is a valuation ring, yet not a DVR, since its valuation group is not ismorphic to the integers, as pointed out in another answer already.

Thus, there is in fact a contradiction there. Indeed, there is a glitch in Theorem 11.1. In the proof, part (2) implies (1), it is assumed (implictly) that the PID is not a field.