# Show that $\{F/G\in k(X)\mid \deg(G)\geq \deg(F) \}$ is a Discrete Valuation Ring

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.

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

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 $$f\in (1/X)$$.

• 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 $F= O_v[\pi^{-1}]$. – reuns Sep 26 '19 at 1:23
• Yes, what you say is correct. – rr01 Sep 28 '19 at 0:01