3
$\begingroup$

I'm reading Algebraic Geometry of Hartshorne. I have a question about a proof. First of all I'll write the important definitions that Hartshorne uses.

Let $Y\subset \mathbb{A}^n$ be a quasi-affine variety.

  • A function $f:Y\to \mathbb{k}$ is said to be regular at $P\in Y$ if is an open neighborhood $U$ of $P$ such that $U\subset Y$ and polynomials $g,h\in A=\mathbb{k}[x_1,...,x_n]$ such that $h$ is nowhere zero on $U$ and $f=g/h$ on $U$. We say that $f$ is regular at $Y$ if is regular at each $P\in Y$.

  • We denote by $O(Y)$ the ring of all regular functions on $Y$. If $P\in Y$, we define the local ring of $P$ on $Y$, $O_P$ to be the ring of germs of regular functions on $Y$ near $P$. In other words, an element of $O_P$ is a pair $\left<U,f\right>$ where $U$ is an open neighborhood of $P$ and $f$ is regular on $U$. And where we identify two pairs $\left<U,f \right>$ and $\left<V,g \right>$ if $f=g$ on $U\cap V$.

  • Finally we define the function field $K(Y)$ as the set of equivalence relations, given by elements of the form $\left<U,f \right>$ for some $U$ open subset of $Y$ (not necessary containing $P$).

The operations that turns $O_p$ and $K(Y)$ into a rings an actually $K(Y)$ into a field are the usual for germs.

I'm reading the proof of Theorem $3.2$, I don't understand the proof of part $d)$. It uses the following observation:

My question: Prove that for each $p \in Y$ the quotient field of $O_p$ is isomorphic to the field $K(Y)$.

An element $z \in Frac(O_P)$ is writen in the form $z=f/g$ where $f,g\in O_P$ and $g$ is not the zero on $O_P$. Recall that $f,g$ are equivalence classes of the form $\left<U,f \right>$ and $\left<V,g \right>$, without lost of generality we can assume that $U=V$ otherwise we intersect the open sets and restrict the functions. I think that the most natural map is given by $\phi: Frac(O_p) \to K(Y)$ defined by:

$$ \left<U,f \right> / \left<U,g \right> \mapsto \left<U,f/g \right>$$

I proved that $\phi$ is an injective ring homomorphism. I think that this is the map that gives the isomorphism, but I'm not sure how to prove that it's surjective (or it's not but I don't think so).

Please help me

$\endgroup$
0

1 Answer 1

2
$\begingroup$

Fix $[(U,f)] \in K(Y)$ an equivalence class of regular functions. Let $V \subseteq U$ be any open subset such that there exist polynomials $g$ and $h$ with $Z(h) \cap V = \emptyset$ and $(U,f) \sim (V,g/h)$. We then have both $[(Y,g)]$ and $[(Y,h)]$ elements of $\mathscr{O}_P$. Since $h$ is not the zero polynomial, it is not the zero element of $\mathscr{O}_P$, and also $Z(h) \neq Y$. So $\phi$ sends the element $[(Y,g)]/[(Y,h)]$ in the field of fractions of $\mathscr{O}_P$ to $[(Y\setminus Z(h),g/h)] \in K(Y)$, and this is precisely $[(U,f)]$.

By the way, you have made a slight error in your definition of $\phi$. You will need to set $$ \phi\left(\frac{[(U,f)]}{[(V,g)]}\right) = [(W,f/g)], $$ where $W = (U \cap V) \setminus Z(g)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.