-1
$\begingroup$

The Hilbert nullstellsatz tells us that for a complete field $K$, there is a bijective correspondence between K-varieties and finitely generated algegras over $K$.

When the field is not complete, eg $K=R$, what goes wrong here? We still have a map from the set of varieties over $K$ to finitely generated algebras over $K$. So it not to give us a bijection it must fail to be bijective or injective. Which of these happens? Does it fail to be surjective or injective or both?

Improve this question
$\endgroup$
8
  • 4
    $\begingroup$ You certainly mean "algebraically closed" instead of "complete". $\endgroup$
    – Robert Kucharczyk
    Commented Dec 17, 2011 at 21:23
  • 1
    $\begingroup$ Also you have to make precise your use of the term "variety", since this is used differently by different authors. But with any of the usual definitions, your statement of Hilbert's Nullstellensatz is wrong. $\endgroup$
    – Robert Kucharczyk
    Commented Dec 17, 2011 at 21:28
  • 1
    $\begingroup$ In $\mathbb{A}^2_{\mathbb{R}}$ polynomials $x^2+y^2$ and $x^4+y^4$ define the same variety. $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Dec 17, 2011 at 21:31
  • 2
    $\begingroup$ @Mahdi: well, that exactly depends on what you understand by "variety". If you work in scheme theory (as your notation $\mathbb{A}_{\mathbb{R}}^2$ suggests), then they are not the same. Their sets of $\mathbb{R}$-valued points are. $\endgroup$
    – Robert Kucharczyk
    Commented Dec 17, 2011 at 21:34
  • 3
    $\begingroup$ @Mahdi: Well, let's just wait for him to explain what he actually means, because at least for my taste the current formulation can be interpreted in many different ways, yielding many different answers. $\endgroup$
    – Robert Kucharczyk
    Commented Dec 17, 2011 at 21:43

2 Answers 2

Reset to default
5
$\begingroup$

When $K$ is not algebraically closed, one useful choice of replacement for the naive notion of an affine algebraic subset of $K^n$ is an affine algebraic subset of $\bar{K}^n$ which is a union of orbits of the absolute Galois group $\text{Gal}(\bar{K}/K)$. The Nullstellensatz, together with the fundamental theorem of Galois theory, implies that such things correspond bijectively to radical ideals of $K[x_1, ... x_n]$.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Let me try an example to see if I understand this correctly: Take $K=\mathbb{R}$, $n=2$. Then $\overline{K}=\mathbb{C}$. (Do we start with an affine algebraic subset of $\mathbb{R}^2$ and find the union of its orbits under absolute Galois group?) Take $\{(0,0)\}$ as your algebraic subset of $\mathbb{R}^2$ and apply $\mathbf{Gal}(\mathbb{C}/\mathbb{R})$ to $\{(0,0)\}$. We get $\{(0,0)\}$. What radical ideal of $\mathbb{R}[x_1,x_2]$ does it correspond to? (Did I understand what you wrote correctly?) $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Dec 18, 2011 at 20:30
  • $\begingroup$ @Mahdi: an affine algebraic subset of $K^n$ doesn't have an action of the absolute Galois group. You want an affine algebraic subset of $\bar{K}^n$ as I said. In your example the corresponding ideal is just the ideal $(x, y)$ of all real polynomials vanishing at $(0, 0)$. Nothing tricky happening here. $\endgroup$
    – Qiaochu Yuan
    Commented Dec 18, 2011 at 21:08
  • $\begingroup$ Now I see. So a priori such an algebraic set corresponds (bijectively) to a radical ideal of $\overline{K}[x_1,\ldots,x_n]$ but because of Galois Theory the generators are in fact in $K[x_1,\ldots,x_n]$. $\endgroup$
    – Mahdi Majidi-Zolbanin
    Commented Dec 18, 2011 at 21:35
  • $\begingroup$ Would this construction replace usefully using schemes, in some respect? Have you seen it used anywhere? Further ideas or comments will be very welcome. $\endgroup$
    – plm
    Commented Feb 24, 2012 at 19:15
  • 1
    $\begingroup$ @plm: Bjorn Poonen taught it at MIT in fall 2009. You can read his notes here: math.mit.edu/~poonen/782/782notes.pdf . (To my memory he did not give a completely precise definition of an affine variety over a non-algebraically closed field, but he did stress the use of points over finite extensions of $K$.) $\endgroup$
    – Qiaochu Yuan
    Commented Feb 25, 2012 at 19:58
4
$\begingroup$

First, I would like to offer a more rigorous statement of Hilbert's Nullstellensatz from Dummit and Foote's Abstract Algebra:

Let $E$ be an algebraically closed field. Then $\mathcal{I}(\mathcal{Z}(I)) = \mathop{\mathrm{rad}} I$ for every ideal $I$ of $E[x_1, \ldots, x_n].$ Moreover the maps $\mathcal{Z}$ and $\mathcal{I}$ in the correspondence $$\{ \mbox{affine algebraic sets} \} \xleftarrow[\mathcal{Z}]{\xrightarrow{\mathcal{I}}} \{\mbox{radical ideals} \}$$ are bijections of each other.

Now, it should be absolutely obvious why the bijection breaks down when the field considered is $\mathbb{R}$. $\mathbb{R}$ is not algebraically closed (consider $x^2+1 \in \mathbb{R}[x]$, which has a variety that includes $i\not\in \mathbb{R}$) so the Nullstellensatz does not apply and the bijection does not happen.

As for whether or not this is a failure to be injective or surjective, that depends on whether you are talking about $\mathcal{Z}$ or $\mathcal{I}.$

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .