7
$\begingroup$

Suppose that we have a parametrization via polynomials as follows:

$$t\longrightarrow (f_1(t),\ldots,f_n(t)),$$

where $t$ is a vector in $\mathbb{C}^r$ and $f_i$ are polynomials of arbitrary degree.

Can we always find equations such that the image is an affine algebraic variety?

The question is motivated by Exercise 1.11 in Hartshorne:

Let $Y\subseteq A^3$ be the curve given parametrically by $x = t^3, y= t^4, z = t^5$. Show that $I(Y)$ is a prime ideal of height 2 in $k[x,y,z]$ which cannot be generated by 2 elements.

I am not interested in the exercise in particular. Finding the variety is easy sometimes, for instance $t\rightarrow (t^2,t^3)$ is given by $I(x^3-y^2)$.

I am looking for a result which says that the image is always an affine algebraic variety AND a procedure to find the ideal.

Improve this question
$\endgroup$
1
  • 4
    $\begingroup$ As others note in answers, the image is not necessarily a subvariety if you have more than one source variable---but the image is always constructible. You can calculate the Zariski closure easily by elimination (in theory---may take a long time): for Hartshorne exercise, ask a CAS to eliminate t from the list of defining parametric equations. When you are parameterizing by A^1, the image is always closed, since A^1 --> A^r is a finite map if nontrivial. In Hartshorne exercise, for example, t satisfies the monic polynomial T^3 - x over k[x,y,z]. $\endgroup$
    – user2490
    Commented Aug 31, 2010 at 16:15

4 Answers 4

Reset to default
7
$\begingroup$

I can't comment (b/c I'm not a registered user) but let me add: in case the dimension of the domain is 1 (as in your motivating example) the image is in fact an affine variety. To see this, note that the map can always be extended to a map from the projective line to projective space by homogenizing things (compare with Dan's example---if you tried homogenizing his map, you'd get $[x:y:z] \mapsto [xy:yz:z^2]$ which is not defined if $x=z=0$ or $y=z=0$), and use the fact that the image of a projective variety by a regular map is closed. Finally, observe that the image of the affine line you started with is precisely the intersection of the image of the homogenized map with the affine coordinate patch determined by your homogenization. Therefore the image of the map you started with is an affine variety. So Hartshorne's example is not an accident.

Improve this answer
$\endgroup$
14
$\begingroup$

Consider $f \colon {\mathbb C}^2 \to {\mathbb C}^2$ given by $(x,y) \mapsto (xy,y)$. The image consists of ${\mathbb C}^2$ minus the subset $y = 0, x \neq 0$. Since the image is not closed, it is not a variety.

The notion of a constructible subset was invented to deal with questions like this. A constructible subset is one which can be constructed from subvarieties using "boolean operations". Equivalently it is a subset defined by polynomial equations and inequations. It is true that the image of a constructible subset is again constructible (Chevalley's theorem).

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Just to clarify: It's not exactly equivalent to just using equations and inequations; without additional boolean operations, this just gives you the affine locally closed sets, but a general constructible set is a union of locally closed sets. $\endgroup$
    – Joshua Grochow
    Commented Feb 2, 2016 at 21:49
4
$\begingroup$

If I change the question slightly the answer is `yes'. Changed question: `` Is the Zariski closure of the image of a polynomial map always an (affine) algebraic variety''.

Explicitly finding this variety (i.e the ideal defining it) is the subject of `elimination theory''. See ch. 4 of the bookIntroduction to Algebraic Geometry' by Brendan Hasset.

Improve this answer
$\endgroup$
1
  • 9
    $\begingroup$ While what you write is correct, it's not what you intend, I think. The Zariski closure is an affine variey by definition. But the stronger fact, mentioned by Dan Petersen above, is the constructibility of the image. So the image is Zariski open inside its Zariski closure. This implies, in particular, that the 'usual' closure (with respect to the complex topology) is an affine variety. Meanwhile, all these closures are indeed computable using elimination theory. $\endgroup$
    – Minhyong Kim
    Commented Aug 31, 2010 at 16:07
1
$\begingroup$

I think this is ok,. Suppose you have $t\longrightarrow (f_1(t),\ldots,f_n(t)).$ Let's call Y the image of that. Then to see if $Y$ is a variety, you can see that $I(Y)$ is prime. But let $\phi: k[X_1, \ldots, X_n] \longrightarrow k[T]$ be given by $X_i \mapsto f_i(T)$. Then $$Ker \phi = \{f: \phi(f) = 0\} = \{f: f(f_1(T), \ldots, f_n(T)) = 0\} = I(Y).$$ So $I(Y)$ is prime because it is the kernel of a map whose image is an integral ring, and then $Y$ is a variety.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ How do you know that the zero set of $I(Y)$ is the image of the map (and not something bigger)? $\endgroup$
    – Sheikraisinrollbank
    Commented Sep 22, 2010 at 14:18
  • 2
    $\begingroup$ I think you have just proved that the Zariski closure of the image is irreducible, which is certainly true if the source is irreducible. But this is not what was asked; e.g. $(x,y) \mapsto (x,xy)$ does not have Zariski closed, or even locally closed, image. $\endgroup$
    – Emerton
    Commented Sep 22, 2010 at 15:37

You must log in to answer this question.

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