Theorem: If $\nu:F\to\mathbb R\cup\{\infty\}$ is a valuation of a functional field, then the set $$\mathfrak O_{\nu}=\{x\in F: \nu(x)\geq 0\}$$ is a local ring with maximal ideal $$\mathfrak M_{\nu}=\{x\in F: \nu(x) > 0\}$$

and a quotient field $F$.

If the valuation is discrete then $\mathfrak O_{\nu}$ is a discrete valuation ring and an arbitrary parameter $t$ of $\mathfrak O_{\nu}$ generates the maximal ideal $\mathfrak M_{\nu}$.

Proof: Assume that $\mathfrak O_{\nu}$ is a local ring with maximal ideal $\mathfrak M_{\nu}$ and a quotient field $F$. I have no doubts.

"...If $\nu:F\to\mathbb R\cup\{\infty\}$ is a discrete valuation then $\nu[F^*]=(d\mathbb Z,+)$ for some $d\in\mathbb R^+$. Assume $t\in F^*$ is such that $\nu(t)=d$. We will show that $t$ is a local parameter of $\mathfrak O_{\nu}$..."

Then the proof goes on to show the representation $z=ut^m$ for any nonzero $z\in\mathfrak O_{\nu}$, and some $u\in\mathfrak O_{\nu}^*$.

Now "...If $xy\in\langle t\rangle$ for some $x=ut^m$, $y=vt^n\in\mathfrak O_{\nu}$ and $u,v\in\mathfrak O_{\nu}^*$ ... then $x\in\langle t\rangle$ or $y\in\langle t\rangle$. This shows that $\mathfrak O_{\nu}$ is a discrete valuation ring with local parameter $t$.

Definition: The commutative domain $R$ with a multiplicative identity is a discrete valuation ring if an indecomposable element $t\in R\setminus R^*$ exists such for any $z\in R\setminus \{0\}$, $z=ut^m$ for some $u\in R^*$ and $m\in \mathbb N$. The element $t$ we call a local parameter.

Problem: I do not understand how from $0\not=xy\in\langle t\rangle$ and $xy\not\in\mathfrak O_{\nu}^*\Rightarrow x\in\langle t\rangle$ or $y\in\langle t\rangle$ follows that $t$ is indecomposable. In more detail if $t=t_1t_2$ then $t_1\in\langle t\rangle$ or $t_2\in\langle t\rangle$, but how come then $t_1\in \mathfrak O_{\nu}^*$ or $t_2\in \mathfrak O_{\nu}^*$.

Any help would be appreciated.

  • $\begingroup$ A functional field is any transcendental field extension. We call the base field a field of constants. $\endgroup$ – superAnnoyingUser Nov 2 '13 at 17:24
  • $\begingroup$ Now I see that I don't. $\endgroup$ – superAnnoyingUser Nov 2 '13 at 18:54

Your result comes from the following:

Exercise: In a domain $R$, a principal ideal $(a)$ being prime implies $a$ is irreducible.

Proof: Suppose $(a)$ is prime. Suppose there are $b,c$ so that $a = bc$. Then without loss of generality we get $b = aa'$ for some $a' \in A$. Hence $a = aa'c$ and so $a'c = 1$ since $R$ is a domain. In other words, $a'$ is a unit and so $a = aa'c$ from which we get $a'c = 1$. In other words, $c$ is a unit and so $a$ is irreducible.

  • $\begingroup$ Thank you, sir. You are not by any chance Alexander Grothendieck, who learned from Laurent Schwartz himself, are you? $\endgroup$ – superAnnoyingUser Nov 2 '13 at 18:01
  • $\begingroup$ No I'm not Grothendieck. I just admire him. $\endgroup$ – user38268 Nov 3 '13 at 2:32
  • $\begingroup$ Actually we have more: In a domain $R$, a principal ideal $(a)$ being prime implies $a$ is prime. $\endgroup$ – user89712 Nov 3 '13 at 8:24

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