Understanding $\mathcal V(I)$, $\mathcal I(X)$, and their relationship to each other.

Question

The Details:

Since definitions vary:

A topological space $(X,\tau)$ is a set $\tau$ of subsets of $X$, called closed subsets, such that

• $\varnothing, X\in\tau$,
• The intersection $$\bigcap_{i\in I}X_i$$ of any closed subsets $(X_i)_{i\in I}$ is closed, where $I$ is arbitrary, and
• The union of finitely many closed sets is closed.

Note that $\tau$ is omitted sometimes when the context is clear.

Let $k$ be an algebraically closed field.

We denote by $\mathcal{V}(I)$ the set of all zeros of an ideal $I$ of $S:=k[T_1,\dots, T_n]$. Here, a zero is some $v\in V:=k^n$ such that $f(v)=0$ for all $f\in I$.

For any $X\subseteq V$, let $\mathcal I(X)\subseteq S$ be the ideal of the $f\in S$ with $f(v)=0$ for all $v\in X$.

The topology on $V$ whose closed subsets are $\mathcal{V}(I)$ for ideals $I$ of $S$ is known as the Zariski topology.

We have that the radical $\sqrt{I}$ of an ideal $I$ of $S$ is the ideal of all $f\in S$ with $f^m\in I$ for some $m\in\Bbb N$.

Theorem (Hilbert's Nullstellensatz): If $I$ is an ideal of $S$, then $$\mathcal I(\mathcal V(I))=\sqrt{I}.$$

A proof can be found in any textbook on algebraic geometry. You could also see here.

Lemma: Let $X\subseteq V$. Then $$\mathcal V(\mathcal I(X))=\overline{X},$$ where $\overline{X}$ is the closure of $X$ w.r.t. the Zariski topology.

For a proof, see here.

The Question:

What is the intuition behind $\mathcal V(I)$ and $\mathcal I(X)$, and how does one think about the relationship described by Hilbert's Nullstellensatz and the Lemma above?

Context:

I think I understand some proofs of Hilbert's Nullstellensatz and the Lemma; however, it doesn't feel like my intuition is strong enough yet for my purposes. I am studying for a postgraduate research degree in linear algebraic groups.

Unless I am mistaken, $\mathcal V(I)$ is known as the vanishing set of the ideal $I$. The name makes sense.

---

I am aware of this:

Lemma 2: Let $S$ and $V$ be as above. Then:

• $\mathcal V(\mathcal I(\mathcal V(P)))=\mathcal V(P)$ for any $P\subseteq S$ and
• $\mathcal I(\mathcal V(\mathcal I(X)))=\mathcal I(X)$ for any $X\subseteq V$.

However, I'm not sure I understand these relationships. See here (pdf) for a proof of the first bullet point. The proof of the second is said to be similar.

---

What kind of answer am I looking for?

Something that illuminates $\mathcal V(I)$ and $\mathcal I(X)$. A couple of paragraphs each might be enough. Perhaps some exercises could be suggested.

I'm not looking for a Royal Road; I'm just trying my best to grasp these key concepts.

Do I think I could answer this myself?

No, not any time soon.

---

Please help :)

---

Upon further reading, I have seen & understood proofs of the following:

• For $J\subseteq J'\subseteq S$, we have $$\mathcal V(J')\subseteq \mathcal V(J).$$

• For $X\subseteq X'\subseteq V$, we have $$\mathcal I(X')\subseteq \mathcal I(X).$$

I get that this is basic stuff. However, I need to be thorough as I am building a lot of theory on these concepts.

Answer 1

Forgive me if this is too wishy-washy, but here's how I (knowing just a tiny bit of algebraic geometry) think about these relationships. I should note that my perspective is influenced by my interests, which lie primarily in arithmetic geometry and number theory.

Suppose I have some algebraic plane curve defined over some algebraically closed field (e.g., $y^2 = x^3 + x$ defined over $\mathbb{C}$) and I'd like to study various properties of this curve (perhaps I want to know something about the rational or integer points of this curve). A natural thing to try and do is associate to this curve some algebraic structure. Thinking of this curve as a subset $V$ of affine 2-space $k^2$, I realize I can associate to this subset an ideal of $k[x, y]$, namely the ideal generated by $y^2 - x^3 - x$. In general, however, this method of associating ideals to subsets of affine space described by some set of equations is not bijective. This is because I can describe certain subsets of affine space by different sets of equations. For example, the equations $x = 0$ and $x^2 = 0$ describe the same subset of affine space but with my naive method of association I would associate to this subset two different ideals, $(x)$ and $(x^2)$.

If instead I associate to $x = 0$ and $x^2 = 0$ the set of elements of $k[x, y]$ that vanish on the respective subsets $V_1$ and $V_2$ of affine space described by these equations, then I associate to both $x = 0$ and $x^2 = 0$ the ideal $(x)$. This is an improvement to my earlier situation, as I've now associated a single ideal, $(x)$, to the subset of affine space described by both $x = 0$ and $x^2 = 0$. Indeed, the Nullstellensatz tells me that this is the right way to associate ideals to algebraic sets. That is, it says that if we associate ideals to algebraic sets in this way, we obtain a bijection. Furthermore, it specifies the ideal we obtain: it's the radical of the ideal generated by the set of equations we started with. So, two ideals describe the same algebraic set if and only if their radicals are the same.

The lemma $\mathcal{V}(\mathcal{I}(X)) = \overline{X}$ is making a very similar statement. It's telling us that two subsets $U$ and $V$ of affine space describe the same ideal of $k[X_1, \dots, X_n]$ (via the process of forming the ideals consisting of all elements of $k[X_1, \dots, X_n]$ that vanish on $U$ and $V$, respectively) if and only if $\overline{U} = \overline{V}$.

Answer 2

This is adding to user1090793's excellent answer, but is more removed from algebraic geometry. One way to understand this circle of ideas intuitively is to think about the general relationship between a space and "functions" on it. For example, when I think of a space, I always think of a globe $S^2$, a two dimensional sphere (sitting inside three dimensional space). As a way to understand this space, we can look at all real valued functions on it. For any such function $f$, we can look at the (closed) subset of $S^2$ given by the set of points $x$ for which $f(x)=0$. The crazy thing is that this $\mathbb{R}$ algebra $\mathscr{A}:=C(S^2,\mathbb{R})$ of continuous functions, actually determines the topological space $S^2$ completely!

How would one recover the topological space from this algebra? We define the points to be maximal ideals in this algebra $\mathscr{A}$, and the closed sets to be the "vanishing sets" of continous functions in this algebra, where now a function $f$ abstractly is defined to vanish on the point associated to the maximal ideal $\mathfrak{m}$ if $f\in \mathfrak{m}$. To check that this recovers our original space, we then need to do two things, check that every maximal ideal is "functions $f$ that vanish on the north pole" for some choice of fixed point (a north pole, say), and check that every closed set is the vanishing set of a continuous function (this needs some point set topology). Filling in the details/meditating on this example is great for understanding this circle of ideas.

In (abstract) algebraic geometry, one takes the radical approach of interpreting any ring $R$ as functions on a space, called $Spec(R)$. Here it becomes important to consider maximal ideals or all prime ideals, and now this space is strange enough that one can have interesting ideals that aren't defined by just where they vanish, so one has to really shift perspective on what a "space" ought to be (if any ring can be a ring of functions, what "values" do our functions take?). For your case however, we are much closer to the $S^2$ example, we define our space with maximal ideals, and closed sets defined by vanishing sets, and we can recover this algebra from our space. Keeping this general idea in mind is helpful throughout algebraic geometry though, depending on the perspective, one is either building spaces such that our rings are functions on them, or using commutative algebra to study spaces we care about.