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Let $P$ a polynomial of two variables, say over the field of real numbers. We define $\partial P$ as $P(\partial_x,\partial_y)$. In this question, it has been shown that if $P_0(x,y)=x^2+y^2$ and $\ker \partial P_O\subset \ker\partial P$ then we can find a polynomial $Q$ such that $P_0Q=P$. Now the questions are

  • What are the polynomials $P_0$ such that if $\ker \partial P_O\subset \ker\partial P$ then one can find a polynomial $Q$ such that $P_0Q=P$?
  • What about a more general case like for example $n$ variables, $n\in\mathbb Z_{\geq 3}$?

For the first question, it works also for constant polynomials, and $P_0(x,y)=x$, $P_0(x,y)=y$.

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+1 very nice question. Do you think that $P(\partial_x,\partial_y)$ can be interpreted as Scalar Product for the Vector Space of Monomial Symmetric Functions? I would be glad if you could have a look. Concerning your question: Can you give an example for $P$ and $Q$? –  draks ... Mar 14 '12 at 18:37

1 Answer 1

up vote 4 down vote accepted

Let's do this in $n$ variables, and over $\mathbb C$. The polynomials $P$ such that $\ker \partial P_0 \subset \ker \partial P$ form an ideal $J$ in ${\mathbb C}[z_1,\ldots,z_n]$ which contains $P_0$. The question is whether this ideal is generated by $P_0$.

For $f(z_1,\ldots,z_n) = \exp(\sum_j a_j z_j)$, where $a_j \in \mathbb C$, we have $(\partial P_0) f = P_0(a_1,\ldots,a_n) f$, so $f \in \ker \partial P_0$ iff $P_0(a_1,\ldots,a_n) = 0$. Let $V_0 = \{(a_1,\ldots,a_n) \in {\mathbb C}^n: P_0(a_1,\ldots,a_n)=0\}$, and let $J(V_0)$ be the ideal of ${\mathbb C}[z_1,\ldots,z_n]$ consisting of polynomials $P$ that are $0$ on $V_0$. This contains $J$. By Hilbert's Nullstellensatz, $J_0$ is actually the radical of the ideal generated by $P_0$. This will be the ideal generated by $P_0$ if $P_0$ is the product of distinct irreducible polynomials. And then $J$ is generated by $P_0$.

EDIT: If $P_0$ has a factor that is the $k$'th power of a polynomial, the situation is less clear to me. I think you have to also consider functions $f$ that are products of a polynomial and an exponential.

EDIT: In fact the statement is true for all polynomials $P_0$.

Suppose $\ker \partial P_0 \subset \ker \partial P$. Let $P_0 = Q^m R$ where $Q$ is irreducible and $\gcd(Q,R) = 1$. I want to show that $P$ is divisible by $Q^m$. In fact let $P = Q^k S$ where $\gcd(Q,S) = 1$, so I need to show $k \ge m$. Take $a = (a_1, \ldots, a_n) \in {\mathbb C}^n$ such that $Q(a_1,\ldots,a_n) = 0$ while $R(a_1,\ldots,a_n) \ne 0$ and $S(a_1,\ldots,a_n) \ne 0$. Note that for any polynomial $p$, $\partial p (g \ e^{a \cdot z}) = e^{a \cdot z} \partial(\tau_a p)(g)$ where $a \cdot z = a_1 z_1 + \ldots + a_n z_n$ and $(\tau_a p)(x_1,\ldots,x_n) = p(x_1 + a_1, \ldots, x_n + a_n)$. $\tau_a$ is an automorphism of the ring ${\mathbb C}[x_1,\ldots,x_n]$, and $\ker \partial P_0 \subset \ker \partial P$ iff $\ker \partial (\tau_a P_0) \subset \ker \partial (\tau_a P)$. Thus without loss of generality we can assume $a = (0,\ldots, 0)$.

Now consider the linear subspace $X_d$ of polynomials in ${\mathbb C}[x_1,\ldots,x_n]$ of degree $\le d$ (for the rest of this paragraph, all operators are considered as acting on this space, which is invariant under all constant-coefficient differential operators). Take $d$ large enough that $\partial Q^m$ is not $0$. Since $Q(0,\ldots,0) = 0$, if $g$ has degree $d \ge 0$ then $\partial Q(g)$ has degree less than $d$, and $\partial Q^{d+1}(g) = 0$. Thus $\partial Q$ is nilpotent. On the other hand, $\partial S$ is of the form $c I + N$, where $c = S(0,\ldots,0) \ne 0$ and $N$ is nilpotent, and therefore $\ker \partial S = \{0\}$. So we have $\ker \partial Q^m \subset \ker \partial Q^k$. Now take $g$ such that $\partial Q^{m-1}(g) \ne 0$, and let $j$ be the least nonnegative integer such that $\partial Q^{m+j}(g) = 0$ (it exists since $Q$ is nilpotent).
Thus $\partial Q^{j}(g)$ is in $\ker \partial Q^m$ but not in $\ker \partial Q^{m-1}$, so $k \ge m$ as required.

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2  
+1. I knew Hilbert's Nullstellensatz, but I didn't make the link. Thanks! Do you think it's still true if we just have $\ker \partial_{P_0}\cap\mathbb C[x_1,\ldots,x_n]\subset\ker \partial_P\cap\mathbb C[x_1,\ldots,x_n]$? –  Davide Giraudo Mar 14 '12 at 19:42
1  
No. $\ker \partial P_0 \cap C[x_1,\ldots,x_n]$ could be $\{0\}$, e.g. for $P_0(x_1,\dots,x_n) = x_1 + 1$. –  Robert Israel Mar 14 '12 at 21:37
    
In fact if $P_0(0,\ldots,0) = a \ne 0$, $\partial P_0$ is one-to-one on ${\mathbb C}[x_1,\ldots,x_n]$. –  Robert Israel Mar 15 '12 at 7:28

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