# Are sets given in parametric form always algebraic?

If a set is given in parametric form by polynomials, is this set always closed (Zariski topology), i.e algebraic?

For example, take $X=\{(t,t^{2},t^{3}): t \in \mathbb{A}^{1}\}$ and $W=\{(t^{3},t^{4},t^{5}):t \in \mathbb{A}^{1}\}$ one can check these sets are closed, for example $X=V(y-x^{2},z-x^{3})$.

Question: suppose $X \subseteq \mathbb{A}^{n}$ is given by:

$X=\{(f_{1}(t),f_{2}(t),...,f_{i}(t)): t \in \mathbb{A}^{1}\}$ where $f_{i} \in k[t]$ for each $i \in \{1,2,..n\}$. Is it always true that $X$ is closed? why? (by the way the underlying field $k$ is algebraically closed and char(k) is zero)

## 2 Answers

Yes, $X$ is always closed.

Consider the morphism $f: \mathbb A^1_k \to \mathbb P^n_k :t\mapsto [1:f_1(t):\ldots:f_n(t)]$
You can extend it to a morphism $\bar f:\mathbb P^1_k \to \mathbb P^n_k$, because $\mathbb P^1_k$ is smooth of dimension one and on a smooth variety rational maps are defined on an open subset of codimension 2 (for curves this is easy).
The image $\bar X =\bar f(\mathbb P^1_k) \subset \mathbb P^n_k$ of $f$ is closed in $\bar X \subset \mathbb P^n_k$, because $\mathbb P^1_k$ is complete .
Hence $X=\bar X\cap \mathbb A^n_k$ is closed in $\mathbb A^n_k$.

• Going through the proofs, does this also produce polynomials which cut out $X$ (which can be computed by hand in explicit examples)? Jun 22, 2012 at 9:06
• @Martin: I would say yes.The core calculation would be to give equations for the image of the morphism $\mathbb P^1 \to \mathbb P^n$. This is classical elimination theory and the main tool would be resultants. With the advent of computers this has become a tremendous industry, churning out thousands of Ph.D's.The keyword would be implicitization and I needn't tell you that the main tool is Gröbner bases. Jun 22, 2012 at 10:02

Here is a second, more algebraic proof.
Consider the morphism $f:\mathbb A^1_k \to \mathbb A^n_k :t\mapsto (f_1(t),\ldots,f_n(t))$.
It corresponds to the morphism of $k$-algebras $\phi: k[T_1,...,T_n] \to k[T]:T_i\mapsto f_i(T)$.
There are now two cases:
a) If all the polynomials $f_i(T)$ are constant the image of $f$ is a point in $\mathbb A^n_k$, and so obviously a closed set.
b) If some $f_i$ is not constant, then $T$ is integral over $k[f_i(T)]$ and a fortiori over $k[f_1(T),...,f_n(T)]$.
In other words the morphism $\phi: k[T_1,...,T_n] \to k[T]$ is integral (and even finite).

So the morphism $f:\mathbb A^1_k \to \mathbb A^n_k$ is integral and thus closed. In particular $f(\mathbb A^1_k)\subset \mathbb A^n_k$ is closed

Remark
Despite appearances this proof is more elementary than the preceding one. It only uses that integral morphisms are closed, which follows from lying-over.
The simplicity of the first proof is a bit deceptive: it uses as a black box that projective space is complete.
This is often thought trivial because it corresponds to compactness over $\mathbb C$, but the algebraic proof of completeness is not trivial, nor, come to think of it, is the assertion that completeness is equivalent to compactness in the classical case ( a GAGA-type of assertion)

• "Here is a second, more algebraic proof": What was the second proof chronologically has become by decision of the software the first proof on this page because user 10 has accepted it. Aug 27, 2013 at 17:38