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Question: Does $\mathcal{I}_{\mathbb{R}}\mathcal{V}_{\mathbb{R}}(I) = \sqrt{I}$ imply that $\overline{\mathcal{V}_{\mathbb{R}}(I)} = \mathcal{V}_{\mathbb{C}}(I)?$, where $I$ is an ideal of $\mathbb{R}[x_1, \cdots x_n]$ and the closure is taken viewing $\mathcal{V}_{\mathbb{C}}(I) \subset \mathbb{C}^{n}$ obviously with Zariski topology.

I came to this while trying to understand for what ideals does Nullstellensatz holds for real field.

I realised that the failure has to do more with the smallness of the $\mathcal{V}(I)$ over non-closed fields since the behaviour of the radical seems to be similar in all cases. So, I was able to prove that $\overline{\mathcal{V}_{\mathbb{R}}(I)} = \mathcal{V}_{\mathbb{C}}(I)$ does imply that $\mathcal{I}_{\mathbb{R}}\mathcal{V}_{\mathbb{R}}(I) = \sqrt{I}$

Proof:(for general $k$ and $\overline{k}$(algebraic closure))

Then, we have that $\mathcal{I}_{\mathbb{\bar{k}}}\mathcal{V}_{\mathbb{k}}(I) = \sqrt{I^{e}} $........(1) by the Nullstellensatz.

Also note that $\mathcal{I}_{\mathbb{\bar{k}}}\mathcal{V}_{\mathbb{k}}(I) \cap k[x_1, \cdots x_n] = \mathcal{I}_{\mathbb{k}}\mathcal{V}_{\mathbb{k}}(I)$.

But Also $\sqrt{I^{e}} \cap k[x_1, \cdots x_n] = \sqrt{I^{e} \cap k[x_1, \cdots x_n]}$ (inverse image and radical commute).

But also $I^{ec} = I$ for the ring extension $k[x_1, \cdots x_n] \to \bar{k}[x_1, \cdots x_n]$.

Proof of the last claim is elementary. I first sured this by proving it for $\mathbb{R}$ and $\mathbb{C}$.

Because ideals are finitely generated, above boils down to :

Take $f_1 \cdots f_r$ polynomials with real coefficients. if $h_1, \cdots h_r$ are polynomials with complex coefficients such that $f_1h_1 \cdots + f_nh_n \in \mathbb{R}{x_1, \cdots x_n}$ then there exist polynomials $g_1, \cdots g_r$ with real coefficients such that $f_1h_1 \cdots + f_nh_n \in \mathbb{R}{x_1, \cdots x_n} = f_1g_1 \cdots + f_ng_n $ (simply take the real parts).

This also tells us how to prove the fact for any field extension! If $L \subset K$ is a field extension then $I^{ec} = I$ for the inclusion: $L[x_1, \cdots x_n] \to K[x_!, \cdots x_n]$.

THe idea is to simply take a basis of $L$ over $K$ and write every $K$ polynomial uniquely using this basis and $L$ polynomial. Uniqueness is easy to prove.

So, then taking intersection on both sides of (1)

gives

$\mathcal{I}_{\mathbb{k}}\mathcal{V}_{\mathbb{k}}(I) = \sqrt{I}$.

SO, we have one sufficient condition on $V(I)$ for $\mathcal{IV}(I) = \sqrt{I}$ to hold over fields. \

So, now I am thinking about the converse as asked in my original question.

If $\overline{\mathcal{V}_{\mathbb{R}}(I)} \subset \mathcal{V}_{\mathbb{C}}(I)$ then we get that

$\mathcal{I}_{\mathbb{C}}\mathbb{V}_{\mathbb{R}}(I) \supset \sqrt{I^{e}}$ (strictly)

Now I want to show that

$\mathcal{I}_{\mathbb{C}}\mathbb{V}_{\mathbb{R}}(I) \cap \mathbb{R}[x_1, \cdots x_n] \supset \sqrt{I^{e}} \cap \mathbb{R}[x_1, \cdots x_n]$ (strictly) then my question is answered in positive.

I know that in general that contraction of two different radical ideals under the inclusion $\mathbb{R}[x_1, \cdots x_n] \to \mathbb{C}[x_1, \cdots, x_n] $ is not injective but here the $\mathcal{V}_{\mathbb{R}}(I)$ are special since they are ideals of zero sets which contain all complex conjugates since they are zero sets of real polynomials.

Thank you.

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2 Answers 2

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For the title of the post, the answer is real radical ideals. If $I$ is any ideal defined by real polynomials, then $\mathcal{I}_{\mathbb{R}}(\mathcal{V}_{\mathbb{R}}(I))=\sqrt[\mathbb{R}]{I}$ where

$$\sqrt[\mathbb{R}]{I}=\{f\mid \exists m\in \mathbb{N}, h_1,\ldots, h_k\in\mathbb{R}[x_1,\ldots,x_n], f^{2m}+\sum_i h_i^2 \in I\}$$

Real radical ideals come from the fact that if $h_1,\ldots,h_k$ are real polinomials such that $\sum_ih_i^2$ vanish at a point $p$, then each polynomial $h_1,\ldots,h_k$ also vanishes at $p$. Look for a book in real algebraic geometry.

In the more general case, let $k$ be a perfect field and $\mathbb{K}$ its algebraic closure. If $P$ is a prime ideal in $k[x_1,\ldots,x_n]$, then $P$ is the intersection of prime ideals $Q_1,\ldots,Q_k$ in $\mathbb{K}[x_1,\ldots,x_n]$. All of them with the same height of $P$ (since the extension is integral). Since they have a finite set of generators, it follows that there is a finite Galois extension $k\subseteq k(\alpha)$ such taht all the $Q_i$ are definable by polynomials in $k(\alpha)$. If $\sigma$ is a $k$-automorphism of $k(\alpha)$, then $\sigma$ induces a bijection in the $Q_i$. Geometrically $\mathcal{V}_{\mathbb{K}}(P)$ is the union of irreducible varieties, all of the same dimension, that are conjugate under $k$-automorphisms.

In this situation $\sum_i Q_i$ is an ideal generated by polynomials in $k$. (In the real case, this is just taking the real and complex parts of the generators of a $Q_i$) and

$$\mathcal{I}_{\mathbb{K}}(\overline{\mathcal{V}_{k}(I)})=\sum_{i} Q_i.$$

For the full generality, you can take a look at Zariski, Samuel, Commutative Algebra, Vol II (extension of the ground field).

Edit to answer further clarification:

You can prove the containment without knowing the shape of the real radical.

$\mathcal{V}_{\mathbb{R}}(I)$ is not dense in $\mathcal{V}_{\mathbb{C}}(I)$ if and only if $$ \sqrt{I}=\mathcal{I}_{\mathbb{C}}(\mathcal{V}_{\mathbb{C}}(I))\subsetneq \mathcal{I}_{\mathbb{C}}(\mathcal{V}_{\mathbb{R}}(I)) $$

Assume it is not dense and let $f$ be a polynomial in $\mathcal{I}_{\mathbb{C}}(\mathcal{V}_{\mathbb{R}}(I))$ not in $\sqrt{I}$.

Now, $\overline{f}$ (the conjugate polynomial) also belongs to $\mathcal{I}_{\mathbb{C}}(\mathcal{V}_{\mathbb{R}}(I))$ since it also vanish on the dense set of real points. Then $Re(f)$, $Im(f)\in \mathcal{I}_{\mathbb{C}}(\mathcal{V}_{\mathbb{R}}(I))$. And, since $f\notin \sqrt{I}$, at least one of $Re(f)$, $Im(f)\notin \sqrt{I}$. Assume it is $Re(f)$, this is a a real polynomial that vanishes on $\mathcal{V}_{\mathbb{R}}(I)$, but it cannot belong to $\sqrt{I}\cap \mathbb{R}[x_1,\ldots,x_n]$.

Thus if $$\overline{\mathcal{V}_{\mathbb{R}}(I)}\subsetneq \mathcal{V}_{\mathbb{C}}(I)$$ then $$ \sqrt{I^e}\cap \mathbb{R}[x_1,\ldots,x_n] \subsetneq \mathcal{I}_{\mathbb{R}}(\mathcal{V}_{\mathbb{R}})(I)$$

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  • $\begingroup$ Thank you but can you also give me a hint of how to use this to answer my question? My question now becomes: If $sqrt{\mathbb{R}}{I} = \sqrt{I}$ then is $\mathcal{V(I)}$ dense in $\mathcal{V}_{\mathbb{C}}(I)$..given the description of the real radical I dont see how to proceed $\endgroup$ May 14, 2022 at 17:57
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In addition to the good answer of lftabera, let me add some pointers to the literature and some more $\mathbb{R}$-oriented explanations.

Let $R$ be a real closed field (for instance, $R = \mathbb{R}$). We say that an ideal $I \subset R[x_1, \dots ,x_n]$ is real if, for any $f_i, \dots ,f_p \in R[x_1, \dots ,x_n]$, $f_1^2+ \dots +f_p^2 \in I \Rightarrow f_i \in I$ for all $i$. For any ideal $I \subset R[x_1, \dots ,x_n]$, the real radical of $I$ is the smallest real ideal containing $I$.

As a consequence of the Real Nullstellensatz (see Bochnak, Coste, Roy, Real Algebraic Geometry, th. 4.1.4 or Marshall, Positive Polynomials and Sums of Squares, th. 12.5.3) we have the following:

Corollary (see for instance Marshall, Positive Polynomials and Sums of Squares, cor. 12.5.4(2)). An ideal $I$ is in the image of the map $V \mapsto \mathcal{I}_{R}(V)$ if and only if $I = \sqrt[R]{I}$ is a real ideal.

Now, if $\mathcal{I}_R(\mathcal{V}_R(I)) = \sqrt{I}$ then $\sqrt{I} \subset R[x_1,\dots,x_n]$ is a real ideal, and $\mathcal{V}_R(I) = \mathcal{V}_R(\sqrt{I})$. Moreover the radical of $I$ in $R[x_1, \dots ,x_n]$ is clearly contained in the radical of $I$ in $C[x_1, \dots x_n]$ (where $C$ is the algebraic closure of $R$), so that $\mathcal{V}_C(I) = \mathcal{V}_C(\sqrt{I})$. Then it is enough to prove the following:

Claim. If $I$ is a real ideal, $\overline{\mathcal{V}_R(I)} = \mathcal {V}_C (I)$.

This is well known: let me repeat the argument of lftabera for completeness. Obviously, we have $\mathcal{V}_R(I) \subset \mathcal {V}_C (I)$, and from the definition of Zariski topology $\overline{\mathcal{V}_R(I)} \subset \mathcal {V}_C (I)$. From the Nullstellensatz $\sqrt{I}_C = \mathcal{I}_C(\mathcal{V}_C(I)) \subset \mathcal{I}_C(\mathcal{V}_R(I)) = J_C$, where $\sqrt{I}_C$ denotes the radical in $C[x_1, \dots, x_n]$. We want to prove the converse inclusion. Pick then $f = h +i g \in J_C$, where $h, g$ are real polynomials. Then, for all $x \in \mathcal{V}_R(I)$: $$ 0 = f(x) = h(x) + i g(x) \Rightarrow 0 = \overline{f(x)}= \overline{f}(x) = h(x)- ig(x). $$ Then we can easily show that $h, g \in J_C \cap R[x_1, \dots , x_n] = \mathcal{I}_R(\mathcal{V}_R(I)) = \sqrt[R]{I} = I \subset \sqrt{I}_C$. Therefore $f \in \sqrt{I}_C$, proving the reverse inclusion and $\overline{\mathcal{V}_R(I)} = \mathcal {V}_C (I)$.

Another way to prove the claim is the so-called Simple Point Criterion, see for instance Marshall, th. 12.6.1 or BCR, prop. 3.3.16. The proof uses the fact that the complex and real variety have the same dimension, since locally at the simple point, the real variety (resp. the complex variety) is a real (resp. complex) manifold of the dimension equal to the dimension of the variety (using the rank condition for the Jacobian at the simple point). See also Mangolte, Real Algebraic Varieties, ex. 2.2.6 and th. 2.2.9. For prime real ideals $I=(f)$ defined by a single real polynomial $f$, another closely related criterion is the Sign Changing Criterion, see Marshall, th. 12.7.1. In this case, the real and complex varieties are hypersurfaces.

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