samples/texts/102047/page_15.md

(2.9) LEMMA.

$$b(X)\le \dim \mathcal{P}(X)\mathrm{.}b(X)\; \backslash leq\; \backslash dim\; \backslash mathcal\{P\}(X).$$

(2.10) LEMMA.

$$\mathcal{P}(X)\subset I_{\perp }^{\times }\mathrm{.}\backslash mathcal\{P\}(X)\; \backslash subset\; I^\{\backslash times\}\_\{\backslash perp\}.$$

(2.11) LEMMA.

$$\dim I^{\mathcal{X}}\perp \le b(X)\mathrm{.}\backslash dim\; I^\{\backslash mathcal\{X\}\}\; \backslash perp\; \backslash leq\; b(X).$$

Proof [DR1] of (2.9) Lemma. Every polynomial

$$q_B:=\prod _{x\in X\setminus B}(⟨x,\cdot ⟩-{\lambda }_x),q\_B\; :=\; \backslash prod\_\{x\; \backslash in\; X\; \backslash setminus\; B\}\; (\backslash langle\; x,\; \backslash cdot\; \backslash rangle\; -\; \backslash lambda\_x),$$

with $B \in \mathbb{B}(X)$ and $(\lambda_x)$ arbitrary constants, is in $\mathcal{P}(X)$, as follows readily by multiplying out. For each $B \in \mathbb{B}(X)$, there is a unique common zero of the $s$ linear factors $\langle x, \cdot \rangle - \lambda_x$, $x \in B$, which do not occur in the associated $q_B$; call that point $\theta_B$. Choose now, as we may, the constants $(\lambda_x)$ in such a way that $\theta_B \neq \theta_{B'}$ whenever $B \neq B'$. (In fact, almost every choice of the $\lambda_x$ would satisfy this condition.) It then follows that

$$q_B({\theta }_{B^{\mathrm{\prime }}})=0\iff B\mathrm{\ne }{B}^{\mathrm{\prime }},q\_B(\backslash theta\_\{B\text{'}\})\; =\; 0\; \backslash Leftrightarrow\; B\; \backslash neq\; B\text{'},$$

proving the linear independence of the collection $(q_B)_{B \in \mathbb{B}(X)}$ in $\mathcal{P}(X)$. $\square$

Proof of (2.10) Lemma. We have to prove that, for each $h \in \mathbb{H}(X)$, $p_h(D) = (D_{h^\perp})^{#(X\setminus h)}$ annihilates $\mathcal{P}(X)$, i.e., that $p_h(D)p_V = 0$ for every $V \subset X$ for which $X\setminus V$ contains a basis. For this, we note that $D_{h^\perp}p_{V \cap h} = 0$; hence

$$p_h(D)p_V=(p_{V\cap h})p_h(D)p_{V\setminus h}\mathrm{.}p\_h(D)p\_V\; =\; (p\_\{V\; \backslash cap\; h\})\; p\_h(D)\; p\_\{V\; \backslash setminus\; h\}.$$

On the other hand, since $X\setminus V$ contains a basis, $V\setminus h$ cannot co- incide with $X\setminus h$, hence $\deg p_{V\setminus h} < #X\setminus h = \deg p_h$ and therefore $p_h(D)p_{V\setminus h} = 0$. $\square$

Proof of (2.11) Lemma. We prove the lemma by induction on #X. For the case #X = s we observe that I^X is generated by s linearly in- dependent linear homogeneous polynomials, consequently I^X ⊥ con- tains only constants, and so dim I^X ⊥ = 1 = b(X). Assume now that #X > s. We follow the argument used in the proof of [DM3; Thm. 3.1], decompose X as

$$X:=X^{\mathrm{\prime }}\cup X,X\; :=\; X\text{'}\; \backslash cup\; X,$$