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According to this math.stackexchange.com answer, the following definition of Huybrechts in his book Complex Geometry is nonsensical:

Let $X$ be a complex manifold. Let $Y \subset X$ be a hypersurface and let $x \in Y$. Suppose that $Y$ defines an irreducible germ in $x$. Hence, this germ is the zero set of an irreducible $g \in \mathcal{O}_{X,x}$.

Definition. (D. Huybrechts, Complex Geometry, Definition 2.3.5, page 78) Let $f$ be a meromorphic function in a neighbourhood of $x \in Y$. Then, the order $\mathrm{ord}_{Y,x}(f)$ of $f$ in $x$ with respect to $Y$ is given by the equality $f = g^{\mathrm{ord}(f)}\cdot h$ with $h \in \mathcal{O}^*_{X,x}$.

The definition seems reasonable to me. What's wrong with it?

For example, suppose $X = \mathbb{C}^2$, $Y = \{0\}\times\mathbb{C}$, and $f(x, y) = \frac{1}{x} + y$. Then, $g(x,y) = x$, and we have $f = g^{-1}h$ where $h(x,y)=1+xy \in \mathcal{O}_{X,(0,0)}^*$. So, according to Huybrechts' definition, we have $\mathrm{ord}_{Y,(0,0)}(f)=-1$, which seems correct to me.

Ted Shifrin
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Simon Parker
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  • What happens if $Y$ is singular at $x$ (but still require that the germ of $g$ at $x$ be irreducible)? – Ted Shifrin Jul 22 '20 at 19:00
  • Hmm, I don't think that matters. We doing the order of $f$ along $Y$, not merely at $x$. You should ask Georges to elaborate. – Ted Shifrin Jul 22 '20 at 19:23
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    The condition $f=g^{ord(f)}\cdot h$ with $h\in\mathcal{O}_{X,x}^*$ is very strong, and for most $f$ it seems that an integer $ord(f)$ with this property will not exist (in your example, try $f=y/x$ or $f=1/(x+y)$). – Julian Rosen Jul 22 '20 at 19:46
  • @JulianRosen: Well, your first example is not meromorphic at the origin. In your second example, if we consider $g(x,y) = x+y$, this is a holomorphic function on the plane that does not vanish along $x=0$, so the answer is surely $0$. – Ted Shifrin Jul 22 '20 at 20:06
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    I thought "meromorphic function" meant "ratio of holomorphic functions". In any event, the order of $1/(x+y)$ cannot be $0$ according to Huybrecht's definition since $1/(x+y)$ is not even an element of $\mathcal{O}_{X,(0,0)}$, much less a unit. – Julian Rosen Jul 22 '20 at 22:38
  • Yup, I guess that's right on the second account. No, the ratio of holomorphic functions makes no sense if they both vanish at the same point. (This is like base locus with linear systems.) It's interesting that this mistake seems to be in other books as well, which, frankly, is disturbing. – Ted Shifrin Jul 22 '20 at 22:51
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    @JulianRosen: OK, my commutative algebra is past rusty. Even simpler: What do we do with the holomorphic function $f=y$ with $Y={x=0}$? Surely it doesn't vanish along $Y$; indeed, it gives the local coordinate in $\Bbb C[y]_0$. But it's not a unit. – Ted Shifrin Jul 22 '20 at 23:04
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    That's a nice example, simpler than mine. There is no integer $n$ for which $x^n y$ is a unit in the local ring at $(0,0)$ ($x^ny$ vanishes at the origin if $n\geq 0$, and is not holomorphic in a neighborhood of the origin if $n<0$). So it seems to me that Huybrecht's definition does not define $ord_{{x=0},(0,0)}(y)$. – Julian Rosen Jul 22 '20 at 23:28

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Yes, so Georges is correct. The definition is flawed. The correct definition is that that the order of $f$ is the largest integer $k$ so that you can express $f = g^k h$ in the local ring. There is no reason to expect $h$ to be a unit. (Great question, by the way. I'm embarrassed it took me so long.) See the various examples that @JulianRosen and I offered in the comments.

Ted Shifrin
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  • Sorry for asking a question about this old answer. Let $X = \mathbb C^2$, $Y = V(g)$ and $x = (0,0)$ where $f = 1/(z_1z_2)$, $g = z_1$. Then I don't think $f$ can be written as a form $f = gh$ with $h$ in the local ring $\mathcal O_{X,x}$. – Hydrogen Oct 19 '21 at 00:20
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    @Hydrogen You mean $g^{-1}$, of course. As I commented, this order should be computed at a generic point of $Y$, not at $z_2=0$. – Ted Shifrin Oct 19 '21 at 00:39
  • Hi @Ted Shifrin , the $h$ in the definition is not unit, it's not even to be a holomorphc function but generally a meromorphic function correct? – yi li Oct 25 '22 at 12:01
  • @yili No, incorrect. You then could take $h=f$ and be done. – Ted Shifrin Oct 25 '22 at 16:49
  • Hi @Ted Shifrin , can I asked another silly question , the order of vanishing in the example of the post: $1/x + y$ along the ${0}\times \Bbb{C}$ is $-1$ correct? – yi li Oct 26 '22 at 06:13