Are all parametrizations via polynomials algebraic varieties? - MathOverflow

Are all parametrizations via polynomials algebraic varieties?

2010-08-31T08:24:52Z

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.

Answer by Dan Petersen for Are all parametrizations via polynomials algebraic varieties?

2010-08-31T09:04:51Z

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).

Answer by Sheikraisinrollbank for Are all parametrizations via polynomials algebraic varieties?

2010-08-31T10:06:17Z

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.

Answer by Richard Montgomery for Are all parametrizations via polynomials algebraic varieties?

2010-08-31T15:56:53Z

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.

Answer by JRJ for Are all parametrizations via polynomials algebraic varieties?

2010-09-22T13:56:36Z

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.