Let $K$ be a field. A surjective transformation: $v: K \to \mathbb{Z}\cup\{\infty\}$ is defined as a discrete valuation, if for any $a, b \in K$, the following statements hold true:

  1. $v(ab) = v(a) + v(b)$
  2. $v(a) = \infty \iff a = 0$
  3. $v(a+b) \geq \min\{v(a), v(b)\}$

Using this definition, I want to show that $A := \{x \in K|v(x) ≥ 0\}$ is an euclidean subring of $K$ (called the discrete valuation ring). I also want to show that $A$ has exactly one maximal ideal $(p)$ with p being a prime element.

  • $\begingroup$ $\mathfrak{m} = \{x:v(x) > 0\}$ is the maximal ideal and $\mathfrak{m} = (x)$ for some $x$ .It's easy to prove that if $v(a) = 0 \implies \exists a^{-1}$ . U can prove that $a \in A \implies a = x^{v(a)}*b$ where $b$ is invertible. $\endgroup$ – qwenty May 1 '15 at 16:05

What did you try ? To prove that $A$ is a subring of $K$, you need to prove that:
(i) $1 \in A$ (note: it is easy to forget this one, so I put it first!), which means that $v(1) \geq 0$. We could easily prove this one using fact 1. of your question: namely, $v(1 \times 1) = v(1) + v(1)$, what could $v(1)$ be ?
(ii) for $x, y \in A$, $x+y$ and $xy \in A$. These two should be trivial consequences of your points 1. and 3.

Your second point, however, is not true as you wrote it. You must replace “p is a prime number” by “p is a prime element of $A$” (and it is more common to call it $\pi$ instead of $p$, precisely because of this; we usually restrict the name $p$ for prime numbers). Counter-example: $K = \mathbb{Q}((X))$, $A = \mathbb{Q}[[X]]$, ${\frak m} = (X)$.

If $A$ has only one maximal ideal $\mathfrak m$ then it is a local ring, which means that $A^\times = A \setminus \mathfrak m$ is the set of invertible elements of $A$. You may easily prove that an element $x \in A$ is invertible iff $v(x) = 0$. So a candidate for $\mathfrak m$ is the set $$ \mathfrak m = \left\{ x \in A, v(x) > 0 \right\}.$$ Proving this should now be as easy as proving that $A$ is a subring of $K$.

The only remaining point to prove is that $\frak m$ is principal; namely, any maximal ideal is always prime, and if it is principal, then its generator must be a prime element. Since multiplying elements add their valuations, any generator must have valuation $1$. Why does such an element always exist ?

Oh, and by the way, condition 2. in your question is redundant. Since $K$ is a field, for any $a \neq 0$, we have $v(1) = v(a \times 1/a) = v(a) + v(1/a)$, therefore $v(a) = \infty \Rightarrow a = 0$; conversely, for $a$ such that $v(a) = 1$, $v(0) = v(0 a)$ proves that $v(0) = v(0) + 1$, which is possible only for $v(0) = \infty$.


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