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Let R be a domain that is not a field. R is Noetherian and local, and the maximal ideal is principal. Then R is called a "Discrete Valuation Ring". Further, if $(t)$ is the maximal ideal, then $t$ is the uniformization parameter.

This is from Fulton's Algebraic Curves, question 2.24 (b).

Show that $\{F/G\in k(X)\mid \deg(G)\geq \deg(F) \}$ is a Discrete Valuation Ring, with uniformization parameter $t = 1/X$.

To do this I want to show that it is a Noetherian ring first, but even if it were a Noetherian ring, I don't see how $(1/X)$ is the maximal ideal.

For if $\alpha/\beta$ is non-unital, we can take $\tfrac{X}{(X+1)(X+2)}$ and this will not be expressible as $ut^n$ for unit $u$ and nonnegative $n$, if I am not mistaken.

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For $f\in k[X]$, $\deg(f)$ is the order of the pole at $\infty$

For $h=f/g \in Frac(k[X])$ then $v(h) = \deg(g)-\deg(f)$ is a discrete valuation on $k(X)$

and $O_v=\{ h \in k(X), v(h) \ge 0\}=k[X^{-1}]_{(X^{-1})}$ is a DVR with uniformizer $X^{-1}$ of valuation $1$.

$k(X)=O_v[X]$ means it is a PID with ideals $(X^{-n})$ so it is noetherian

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It turns out both answers were trivial after a bit of thought.

For the first problem, $k(X)$ is a PID, so its subring is a PID.

For the second problem, if $f$ is non-unital, then putting $f=\tfrac{\alpha}{\beta}$, and take factorization of $\beta = (X-\lambda_1)\dots(X-\lambda_1)$. Then $\tfrac{X-\lambda_1}{X}$ is a unit, so multiplying that with $f$, we obtain that $f\in (1/X)$.

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    $\begingroup$ It is of course not true that subrings of fields (ie. all integral domains) are PID. For a discrete valuation on a field $O_v =\{a\in F, v(a)\ge 0\}$ is a DVR : a PID with unique maximal ideal $\pi O_v$ with $\pi$ any element of minimal non-zero valuation, $O_v=O_v^\times\cup \pi O_v$ and $F= O_v[\pi^{-1}]$. $\endgroup$ – reuns Sep 26 '19 at 1:23
  • $\begingroup$ Yes, what you say is correct. $\endgroup$ – rr01 Sep 28 '19 at 0:01

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