I am reading Laurent's notes Sums of squares, moment matrices and optimization over polynomials and, since I am not really experienced in algebraic geometry, I have some questions. This one concerns the following theorem (Theorem 2.6, p.15):
Theorem. (i) An ideal $I\subseteq\mathbb{R}[\mathbf{x}]$ is zero-dimensional (i.e. $|V_{\mathbb{C}}(I)|<+\infty$) if and only if the vector space $\mathbb{R}[\mathbf{x}]/I$ is finite dimensional.
(ii) (Assuming that $V_{\mathbb{C}}(I)$ is finite) $|V_{\mathbb{C}}(I)|\leq \dim(\mathbb{R}[\mathbf{x}]/I)$, with equality if and only if the ideal $I$ is radical.
Some preliminary background, before I set the main question:
For an ideal $I$ of $\mathbb{R}[\mathbf{x}]$, where $\mathbf{x}$ denotes, in general, a multidimensional variable, define the varieties
$$V_{\mathbb{C}}(I)=\{ v\in\mathbb{C}^n: \ f(v)=0, \ \forall f\in I \} \ \ \text{and } \ V_{\mathbb{R}}(I)=V_{\mathbb{C}}(I)\cap\mathbb{R}.$$
The radical ideal of $I$, $\sqrt{I}$, is defined to be $$\sqrt{I}=\{ f\in\mathbb{R}[\mathbf{x}]: \ f^k\in I, \ \text{for some } k\in\mathbb{N} \},$$ and the real radical ideal of $I$, $\sqrt[\mathbb{R}]{I}$, $$\sqrt[\mathbb{R}]{I}=\{ f\in\mathbb{R}[\mathbf{x}]: f^{2k}+\sum_{j=1}^{m}p_j^2\in I, \ \text{for some } k,m\in\mathbb{N} \ \text{and } p_j\in\mathbb{R}[\mathbf{x}], \ \forall j=1,\dots,m \}.$$ Finally, the vanishing ideal of a set $V\subseteq\mathbb{C}^n$, $\mathcal{I}(V)$, is defined to be the set
$$\mathcal{I}(V)=\{ f\in\mathbb{R}[\mathbf{x}]: \ f(v)=0, \ \forall v\in V \}.$$
Then the Hilbert's Nullstellensatz and the Real Nullstellensatz imply that $\mathcal{I}(V_{\mathbb{C}}(I))=\sqrt{I}$ and $\mathcal{I}(V_{\mathbb{R}}(I))=\sqrt[\mathbb{R}]{I}$, respectively.
In order to prove the Theorem, we need a Lemma. Before we state the Lemma (Lemma 2.5), we need to define the Lagrange interpolation polynomials, $\{p_v\}_{v\in V}\subseteq\mathbb{C}[\mathbf{x}]$, for a finite subset $V$ of $\mathbb{C}^n$. More precisely, the polynomials $p_v$ are defined so that
$$p_v(u)=\delta_{v,u}, \ \forall v,u\in V, \ \ \text{and } \ p_{\bar{v}}(x)=\bar{p}_v(x), \ \forall x\in\mathbb{R}^n.$$
Lemma. Let $I$ be a zero-dimensional ideal of $\mathbb{R}[\mathbf{x}]$. Partition $V_{\mathbb{C}}(I)$ into $V_{\mathbb{C}}(I)=S\cup V_{+}\cup\bar{V}_{+}$, where $S=V_{\mathbb{R}}(I)$ and $V_+=\{v\in V_{\mathbb{C}}(I): \ \Im(v)>0\}$. If $\{p_v\}_{v\in V_{\mathbb{C}}(I)}$ are the interpolation polynomials that were defined above, then the set $\mathcal{L}=\{p_v\}_{v\in S}\cup\{\Re(p_v),\Im(p_v)\}_{v\in V_+}$ is linearly independent in $\mathbb{R}[\mathbf{x}]/I$ and generates $\mathbb{R}[\mathbf{x}]/\mathcal{I}(V_{\mathbb{C}}(I))$.
My first question is about the proof of this lemma. More precisely, in order to prove the linear independence, M. Laurent considers a selection of scalars $\{\lambda\}_{v\in S}$, $\{\lambda'_v\}_{v\in V_+}$ and $\{\lambda''_v\}_{v\in V_+}$ such that
$$\sum_{v\in S}\lambda_v p_v+\sum_{v\in V_+}(\lambda'_v\Re(p_v)+\lambda''_v\Im(p_v))\in I.$$
Then, she states that "evaluating this polynomial at $v\in V_{\mathbb{C}}(I)$ yields that all scalars are $0$". I think I would agree with that for $\lambda_v$ and $\lambda'_v$, but why does this hold true for $\lambda''_v$? I mean $\Im(p_v(u))=0$, for all $u\in V_{\mathbb{C}}(I)$, right? And in that case, the scalars $\lambda''_v$ do not even involve.
But if there is a glitch indeed, then the actual statement of the Lemma probably should be slightly amended into: $\mathcal{L}=\{p_v\}_{v\in V_{\mathbb{R}}(I)}$ is linearly independent in $\mathbb{R}[\mathbf{x}]/I$ and it generates $\mathbb{R}[\mathbf{x}]/\sqrt[\mathbb{R}]{I}$ (having used the Real Nullstellensatz here).
But such an alternation would also affect the second part of the Theorem. More precisely, I would expect something like $|V_{\mathbb{R}}(I)|\leq\dim(\mathbb{R}[\mathbf{x}]/I)$, with equality if and only if the ideal $I$ is real radical. I have omitted the proof of part (ii), but I would be willing to add this, if someone thinks it's necessary.
As I mentioned, I am not totally familiarised with algebraic geometry so, apologies if I missed something obvious...
