Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Sorry, but I have to ask a dumb question:

Algebraically, a hyperbola has only one irreducible component (given by an irreducible polynomial).

Why, then, does the real image of a hyperbola show two components?

Better yet: How should I interpret, visually, the components of an algebraic variety?

share|cite|improve this question
Short and rather dumb answer: because the Zariski topology is much coarser than the Hausdorff topology. (every polynomial vanishing on one Hausdorff component must already vanish entirely on both). However, each of the Hausdorff components is what is sometimes called semi-algebraic (that is, if you allow polynomial inequalities, then you can $x + y \geq 0$. – t.b. May 27 '11 at 15:21
The real points are the intersection of the complex points in $\mathbb{C}^2$ with the real plane, and there's no reason for an intersection to have the same number of components as the original thing. – Qiaochu Yuan May 27 '11 at 15:35

As an algebraic variety, the hyperbola is isomorphic to $\mathbb A^1\setminus \{0\}$ (via the map $(x,y) \mapsto x$), so you should consider this (more elemental) case: why does $\mathbb R\setminus \{0\}$ have two components, although $\mathbb A^1 \setminus \{0\}$ is irreducible?

You can imagine various possible answers --- here is one: there is no polynomial map that vanishes identically on the negative real numbers but not the positive real numbers (so we can't detect the decomposition of $\mathbb R\setminus \{0\}$ into two components using the basic investigative tools of algebraic geometry, which are simply polynomial maps).

One way to see this is as follows: if there were such a polynomial, it would extend over $\mathbb R$, but a polynomial cannot vanish on infinitely many points of $\mathbb R$ (e.g. on all negative real numbers) without vanishing identically.

Another way: If we could use polynomial maps to disconnect $\mathbb R\setminus \{0\},$ they would also give a disconnection of $\mathbb C\setminus \{0\},$ but this latter space is connected.

An important point here is that $\mathbb C\setminus \{0\}$ is topologically connected (i.e. not only can it not be disconnected by polynomial maps, it cannot be disconnected at all!). This is because $\mathbb C$ is algebraically closed, so the topology of complex algebraic varieties is reflected by their Zariski topology (e.g. connected in the Zariski topology is equivalent to being connected in the usual topology).

Because $\mathbb R$ is not algebraically closed, there is a less precise link between the topology of real algebraic varieties and their Zariski topology.

share|cite|improve this answer

You should look in projective space. The hyperbola becomes closed. For other equations, such as $y^2=x^3-x$, you may also need to look in complex projective space.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.