0

Use the Hilbert Nullstellensatz Theorem to prove the following result:

Given $F_1, F_2, F_3 \in \mathbb{C} [X_1,\dots,X_n]$ polynomials checking the following conditions:

  1. $F_1$ is irreducible;
  2. $F_2$ is not a multiple of $F_1$;
  3. For every element $ x \in \mathbb{C}^n \text{ if } F_1 (x) = 0 \text{ and } F_3 (x) \neq 0,\text{ then } F_2 (x) = 0$.

Show that $F_3$ is a multiple of $F_1$.

rgl4
  • 363
  • 1
    Cuidado: se te colaron algunas palabras en castellano...;) – DonAntonio Mar 17 '13 at 20:36
  • 1
    You still have some words in spanish in there, yet the problem is a little confusing: if we want to prove that $,F_3=kF_1;,;;k,$ a scalar, then condition (3) is fulfilled vacuously... – DonAntonio Mar 17 '13 at 20:43
  • Note: The theorem you're referring to is known in English (!) as the Nullstellensatz. – hmakholm left over Monica Mar 17 '13 at 20:53
  • 2
    Why does it say "Nullstellensatz-Henning Makholm Thorem" @HenningMakholm ? – Pedro Mar 17 '13 at 21:12
  • 1
    @PeterTamaroff: Looks like the OP mistook my signature for part of the name of the theorem I provided. (I wouldn't want my name attached to that theorem -- I can never even remember what it says without looking it up). – hmakholm left over Monica Mar 17 '13 at 21:18
  • I think that this problem is related to the definition of prime element, so, p | ab $\Leftrightarrow$ p | a ó p | b , but I don't know how connect my exercise with this definition – rgl4 Mar 17 '13 at 21:29
  • Condition 3 states that if $F_1(x)=0$ then $F_3(x)=0$ or $F_2(x)=0$. In particular if $F_1(x)=0$ then $F_2(x)F_3(x)=0$. Now apply the Nullstellensatz to get a condition involving the ideal generated by $F_1$ and the ideal generated by $F_2F_3$. – Jeff Mar 17 '13 at 21:38
  • The condition 3 is correct, I can't do anything. Sorry – rgl4 Mar 18 '13 at 18:59
  • 1
    The condition $3$ as it is stated is obviously wrong: how can you pretend to obtain that $F_3$ is a multiple of $F_1$, but in the same time there exists $x\in\mathbb C^n$ such that $F_1 (x) = 0 \text{ and } F_3 (x) \neq 0$? Ah, sure, a false statement can imply a correct one! –  Mar 19 '13 at 00:16

1 Answers1

3

If the condition $3.$ states (as @JeffTolliver suggested in his comment) that $F_1(x)=0$ implies $F_2(x)F_3(x)=0$, then all is easy: this condition means $\mathcal V(F_1)\subset\mathcal V(F_2F_3)$ and then $\mathcal I(\mathcal V(F_2F_3))\subseteq\mathcal I(\mathcal V(F_1))$, that is, $\sqrt{(F_2F_3)}\subseteq\sqrt{(F_1)}$. Since $F_1$ is irreducible the ideal $(F_1)$ is prime, and thus $F_2F_3\in (F_1)$. It follows that $F_1\mid F_2F_3$, and from the condition $2.$ we get $F_1\mid F_3$.