Let $\mathbb K$ be a field of characteristic $p>0$. Let $f\in\mathbb K[x_1,\dots,x_n]$ be a multivariate polynomial and let $q\in\mathbb K^n$. Is there a computational method to determine the multiplicity of $f$ at $q$ without explicitly computing with Gröbner bases of the powers of the defining ideal of $q$?
$\begingroup$
$\endgroup$
7
-
4$\begingroup$ One way (how good computationally, this is, I do not know) is to change $x_i$'s to $x_i-q_i$ where $q=(q_i)$ and thus assume $q$ is the origin. Then the multiplicity is the least degree of $f$, that is, we can write $f=f_p+f_{p+1}+\cdots$ where $f_k$ is the homogeneous term of degree $k$ and $f_p\neq 0$. The the multiplicity is $p$. $\endgroup$– MohanCommented Aug 20, 2019 at 16:17
-
$\begingroup$ Thank you. I was looking for a linear algebra method, like in characteristic 0 case, where it suffice to take partial derivatives. Is there some analogue in positive characteristic? $\endgroup$– bogCommented Aug 20, 2019 at 18:54
-
$\begingroup$ What is multiplicity? I understand it with respect to a divisor but not with respect to a point... $\endgroup$– Bugs BunnyCommented Aug 20, 2019 at 18:55
-
$\begingroup$ In positive characteristics, derivatives are often ill-behaved. For example, $x^p-y^{p+1}$ has multiplicity $p>0$, but its $p$th derivatives at the origin is zero. $\endgroup$– MohanCommented Aug 21, 2019 at 1:53
-
$\begingroup$ So, is it just the degree of the first non-vanishing homogeneous component, when you write $f$ as a polynomial in $y_k= x_k-q_k$? Why on Earth do you need the Grobner basis in this case? It looks like a straightforward change of variable calculation,,,, $\endgroup$– Bugs BunnyCommented Aug 21, 2019 at 5:37
|
Show 2 more comments