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I am trying to reproduce a proof of the Riemann-Roch Theorem. My main references have been Lang's Introduction to Arithmetic and Abelian Functions and Silverman's Arithmetic of Elliptic Curves. There appears to me to be an inconsistency in definitions. We'll let K be an algebraically closed field and K[X] denote the polynomial ring over K in n indeterminates. Here are a few things from Silverman:

A variety V is called affine if its associated ideal $I(V)$ is prime. The quotient $K[X]/I(V)$ is called the co-ordinate ring of V and denoted $K[V]$. Silverman calls the field of fractions $K(V)$ of $K[V]$ the function field of V.

To every point P of a smooth plane algebraic curve C we can associate a discrete valuation ring $K[C]_{P}$ called the local ring of C at P. I am fairly convinced this is given $K[C]_{P}=\{f/g: f,g \in K[C],\text{ }g(P)\neq0\}$, with unique maximal ideal $M_{P}$ the subset with $f(P)=0$.

A property of discrete valuation rings is that any element can be written in the form $t^{s}y$, $t$ a generator for $M_{P}$, $s$ a non-negative integer and $y$ a unit (in this case any element not in $M_{P}$?).

Here is where I am unsure if I've read it correctly. I think Lang takes the quotient field of $K[C]_{P}$ (he calls it K but in keeping with Silverman's notation I'll denote it $K_{Q}[C]_{P}$), so we have a "fraction of a fraction" so to speak. This means we can write an element of $K_{Q}[C]_{P}$ in the form $t^{s}y$ for $s$ any integer.

Now, Lang's definition of the divisor of a function applies to functions in $K_{Q}[C]_{P}$. So let $f \in K_{Q}[C]_{P}$. Then $ord_P(f)$ has absolute value equal to the order of the pole ($ord_P(f)$ is negative) or the order of the zero ($ord_P(f)$ is positive). I think $ord_{P}(f)=s$? We have $div(f)=\sum_{P}ord_P(f)P$. For a divisor $D$, he defines the vector space $L(D)$ to be the set of $f\in K_{Q}[C]_{P}$ with $div(f)\geq -D$.

Silverman's definition of $L(D)$ seems different. He defines $ord_P(f)$ on $K(C)$ similarly as $ord_P(f)=d$ if $f$ has a zero at $P$ of order $d$ and $ord_P(f)=-d$ if $f$ has a pole at $P$ of order $d$. The only real difference is that the zeroes and poles are determined by zeroes of $K[C]_{P}$ for Lang, and of $K[C]$ for Silverman. Lang has the "fraction of a fraction". Silverman just has a "single fraction". Each author defines $ord_{P}$ on a different class of functions. This seems to affect the definition of canonical divisor which comes later in both texts. The theorem is then stated exactly the same, so I believe they are equivalent, perhaps just differently formulated?

Any insight would be greatly appreciated.

jshpmm
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    Relevant: https://math.stackexchange.com/questions/975575/isomorphism-of-ring-localized-twice-atiyah-macdonald-exercise-3-3 – Viktor Vaughn Mar 21 '20 at 06:01
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    You may want to look at how $K[C]_P$ might fail to be a DVR (when $C$ is not smooth) at $P$: $K=\Bbb{C},K[C]=K[t^2,t^3]\cong K[x,y]/(x^3-y^2), P=(t^2,t^3)$, only one discrete valuation above it corresponding to $(t)\subset K[t]$, and $K[C]=K[t^2(t+1), (t+1)^2t], P=(t^2(t+1), (t+1)^2t)$ this time there are two valuations $(t)$ and $(t+1)$. Add the uniformizers: $K[C,t,t+1]$ becomes the coordinate ring of a smooth curve. To the function field $K(t)$ there is only one complete smooth curve $\Bbb{P}^1$ underlying it, this holds for every curves but not in dimension $\ge 2$. – reuns Mar 21 '20 at 06:18
  • @Richard D. James Is this saying that $K_{Q}[C]{P}$ and $K(C)$ are isomorphic so the definitions are equivalent? – jshpmm Mar 21 '20 at 12:32
  • @reuns the smoothness property comes in the statement of the theorem (in fact, I've seen 1 author take as a definition that a curve C is smooth at P when $K[C]{P}$ is a DVR). I'm mainly worried about $K(C)$ vs $K{Q}[C]_{P}$. – jshpmm Mar 21 '20 at 12:41

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Yes, $\DeclareMathOperator{\Frac}{Frac} \Frac(K[C]_P)$ (what you denoted by $K_Q[C]_P$) and $K(C)$ are isomorphic. Using the notation from the linked post, let \begin{align*} S &= \{f \in K[C] : f(P) \neq 0\}\\ T &= K[C] \setminus \{0\} \end{align*} and $U$ be the image of $T$ in $S^{-1} K[C] = K[C]_P$. Then $$ K(C) = T^{-1} K[C] \quad \text{and} \quad \Frac(K[C]_P) = U^{-1}(S^{-1}K[C]) \, , $$ so $\Frac(K[C]_P) \cong K(C)$ by the linked post. (Note that in our case $S \subseteq T$, so $ST = T$.)

The linked post uses universal properties to show this, but you could also work explicitly with your "fractions of fractions" idea.

Viktor Vaughn
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