# Definition of the order of a meromorphic function

Let $X$ be a complex manifold and $Y \subset X$ a hypersurface. Let $x \in Y$ and $f$ a meromorphic function on $X$ near $x$. In Huybrecht's Complex Geometry the order of $f$ along $Y$ at $x$ is defined as $$f = g^{ord_{Y,x}(f)} \cdot h$$ where $h \in \mathcal O_{X,x}^*$, the sheaf of germs of nonvanishing sections on $x$, and $g\in \mathcal O_{X,x}$ is irreducible and defines $Y$ near $x$.

I'm having trouble coming to grips with this definition. Consider the case $X = \mathbb C^2$ with coordinates $(z_1,z_2)$ and $Y = \{z_1 = 0\}$. Then what is $ord_{Y,(0,0)} z_2$? It seems impossible to write $z_2 = z_1^d h$ for some $h \in \mathcal O_{X,x}^*$. Further, it is stated that the order at any point on an irreducible hypersurface is independent of the point. But clearly $ord_{Y,x} z_2 = 0$ if $x \ne (0,0)$. So if the order is 0 at (0,0) as well, then this is saying that $z_2 \in \mathcal O_{X,(0,0)}^*$. What am I missing?

Thanks.

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In your example: isn't $z_2$ the uniformizer in the local ring of $Y$ at the origin? What do you mean by "defines" $Y$ near $x$: do you mean is the function giving the hypersurface (in your example $z_1$) or the function giving the coordinate on $Y$ (in your example $z_2$)? – tkr Oct 27 '11 at 23:21
@tkr: i mean that around $x$, $Y$ is given as the zero locus of $g$. – Eric O. Korman Oct 27 '11 at 23:25
No wonder you're having trouble: Huybrechts's definition is incorrect. See the Edit to my answer. – Georges Elencwajg Nov 2 '11 at 22:39

Everything becomes crystal-clear once you recall:

$$\large \mathcal O_{X,x} \text {is a UFD}$$
Then if the germ of $Y$ at $x$ is defined by the irreducible element $g\in \mathcal O_{X,x}$, you can write-as in every UFD- any non-zero $f\in \mathcal M_x=Frac(O_{X,x})$ uniquely as: $$f=g^n\Pi h^{n_h}=g^n (stuff) \quad (*)$$ where
- the $h$'s are pairwise non associated and run through the irreducibles of $\mathcal O_{X,x}$ not associated to $g$,
- $n,n_h\in \mathbb Z$ are almost all zero ( "unicity" is up to order of factors and up to invertible elements, as usual in a UFD) .

The order of $f$ along $Y$ at $x$ is then, very naturally, the exponent $n$ of $g$ in $(*)$.

In your example $z_1$ and $z_2$ are two non-asociated irreducibles so that if $Y$is defined locally by $z_1=0$, you write $f=z_2=(z^1)^0.( stuff)$ and you get that the order of vanishing of $z_2$ along $Y$ is $0$
[of course $(\text stuff)=z_2$, but that's irrelevant!]

Edit QiL's crisp and definitive comment made me want to check Huybrechts's definition.
His Definition 2.3.5.on page 78 states:

"Let $f$ be a meromorphic function in a neighbourhood of $x\in Y$. Then the order $ord_{Y,x}(f)$ of $f$ in $x$ with respect to$Y$ is given by the equality $f=g^{\text {ord}(f)}.h$ with $h\in \mathcal O^*_{X,x}$."

This is a completely wrong definition since it is impossible to find such a factorization of $f$ in general: it would imply in particular that any germ at $x$ of a holomorphic function on $X$ would have a zero set coinciding with $Y$ at $x$, an obviously preposterous conclusion.

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My issue is how can we guarantee that $(stuff)$ is in $\mathcal O_{X,x}^*$, the germs of nonzero functions at $x$. – Eric O. Korman Oct 29 '11 at 1:25
@Eric: in general, you can't. – user18119 Nov 2 '11 at 19:56

I think there is actually a problem with this definition. The best one can do is to have $h$ meromorphic and belonging to $\mathcal O_{X,y}^*$ for some $y\in Y$ near $x$.

The idea is the order of $f$ at $(Y,x)$ is an integer $r$ such that the divisor of $f/g^r$ is not supported in $Y$ in a neighborhood of $x$.

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The order function measures the extent to which a meromorphic function $f$ on a complex manifold $X$ vanishes along a codimension one subvariety $Y$. It appears that your notation for that is $ord_{X,Y}(f)$.

So in my mind, there are two sensible interpretations of your example. The less likely is that you want $ord_{X,Y}(z_2)$, which is $0$ because $z_2$ is holomorphic on $Y$ and does not vanish identically. More likely, you want $ord_{Y,(0,0)}(z_2)$, which is $1$ because $Y \cong \mathbb{C}$, with coordinate $z_2$, and of course, $z_2$ vanishes to order $1$. Said more verbosely, $z_2 = g^1 h$, where $g = z_2$ generates the maximal ideal of germs of functions at $(0,0)$ that vanish there (more colloquially, "$z_2$ defines $(0,0)$ in $Y = 0 \times \mathbb{C}$") and $h = 1$ is an invertible germ (i.e. $h(0,0) \neq 0$).

Please note that if $x \in X$, where $X$ is a complex manifold, then the defintion $ord_{X,x}(f)$ only makes sense if $\dim X = 1$.

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