# On vanishing orders of an ideal via the restriction

Let $Y$ be a submanifold of a complex manifold $X$, and $a$ be an ideal on $X$ which does not vanish along the entire $Y$. Consider a point $\xi$ on $Y$, there are the vanishing order $ord_{\xi}a$ and the vanishing order $ord_{\xi}a\cdot O_{Y}$. Is there any relation between the two numbers?

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Once you know precisely what the definition of the order of vanishing is, it should be easy to see that the order of vanishing of the restriction is at least as large as the original order of vanishing. Furthermore, if Y is a generic submanifold passing through the point, equality is expected. –  Jack Huizenga Apr 2 '12 at 7:32
I guess the equality should be expected in any case... –  Zhengyu Hu Apr 2 '12 at 11:56
I don't think equality should be expected in general. For example, consider the ideal $a = (y) \subseteq k[x,y]$. This vanishes to order $1$ at the origin. Now we restrict this idea to $Y = V(y-x^2)$. In the ring $k[x,y]/(y-x^2) \cong k[x]$, the ideal $(a)O_Y = (y)O_Y = (x^2)O_Y$ vanishes to order 2 at the origin. –  Karl Schwede Apr 2 '12 at 14:09
I figure it out... In fact, the two vanishing orders has no relation.. $f=y+x^{n}$ would give an example... –  Zhengyu Hu Apr 3 '12 at 5:26
Maybe I'm just not understanding what you are saying, but I would say there is a relation: the order of vanishing on $Y$ is at least as big as the order of vanishing on $X$. It is correct to say, however, that there are no further relations: if $X$ is at least 2-dimensional, then for any positive integers $m\leq n$ there exists an ideal $a$ and a submanifold $Y$ such that the order of vanishing in $X$ is $m$ and the order of vanishing in $Y$ is $n$. –  Jack Huizenga Apr 3 '12 at 16:35