Extension of a morphism $k^*\rightarrow X$ to $\mathbb{P}^1\rightarrow \overline{X}$

Question

EDITED: In the context of defining the Hilbert-Mumford criterion for stability in moduli problems, we need to extend the morphism of varieties $\lambda_x:k^*\rightarrow X$, (where $X$ is a quasi projective variety and $k^*$ is seen as an algebraic group) to a morphism defined in $\mathbb{P}^1$. We identify $a\in k^*$ with $[1:a]\in\mathbb{P}^1$. So we want to extend the morphism to $[0:1]$ and $[1:0]$. It seems that the extension always exist in this context. The answers I'm finding involucrate schemes and the valuation criterion for properness. Is there simpler answer in the context of varieties?

Thank you so much.

I'm working on varieties without knowing so much of shcemes and probably, the answer i was working in is more natural in the context of schemes so here the confusion. I will try to reformulate the question in the context I'm working. Thank you so much for the coment KReiseer. — calitiutiu, May 20 '20 at 10:34
If $X$ is only a quasi projective variety (and not projective), such extensions may not exist. Consider $X=k^*$ and $\lambda$ the identity map. That is where properness of $X$ comes in. — Mohan, May 20 '20 at 12:29
Thanks for the answer @Mohan. I se... And then, if $X$ is projective which is the reason of the existence of such extension? — calitiutiu, May 20 '20 at 17:16
Here is a standard fact. Any rational map from a non-singular curve to a projective variety is in fact a morphism (the valuative criterion you alluded to). — Mohan, May 20 '20 at 17:53
Around $t=0$ the map is $[t:1]\to [\frac{p_0(t)}{q_0(t)} t^{d_0}:\cdots:\frac{p_n(t)}{q_n(t)} t^{d_n}]$ with $p_j(0)\ne 0,q_j(0)\ne 0$ and $d_j\in \Bbb{Z}$, with $d=\min_i d_i$ you get your local definition $[t:1]\to [\frac{p_0(t)}{q_0(t)} t^{d_0-d}:\cdots:\frac{p_n(t)}{q_n(t)} t^{d_n-d}]$ — reuns, May 20 '20 at 20:44

score 1 · Accepted Answer · answered May 20 '20 at 20:31

In your case, there is a not-so-difficult way to do this from your level of technology. If $X$ is quasi-projective, pick some embedding of $X\hookrightarrow \Bbb P^n$. Now a map from $k^*$ to $X$ is the same thing as a map from $k^*$ to $\Bbb P^n$ which lands in $X$. Such a map is given by $[p_0(t)/t^{d_0}:\cdots:p_n(t)/t^{d_n}]$ where the $p_i$ are polynomials, $d_i\geq 0$, and $t$ is a coordinate on $k^*$.

After multiplying through by a sufficiently high power of $t$, we may assume that our map is actually given by polynomials $[p_0(t):\cdots:p_n(t)]$. Now write $t=\frac{u}{v}$ and multiply through by the highest power of $v$ found in a denominator. This gives us that our map is given by $[1:\frac{u}{v}]=[v:u]\mapsto [q_0(u,v):\cdots:q_n(u,v)]$ for polynomials $q_i$. Now we can divide out by the greatest common factor of all of these polynomials to get a map which is globally defined from $\Bbb P^1\to \Bbb P^n$. This lands in $\overline{X}$ by irreducibility of $\Bbb P^1$.

This trick generalizes (though one has to be slightly more careful and work more locally): one can use this idea to show that any rational morphism from a nonsingular projective (=proper) curve to a projective variety is actually defined everywhere: write the map as a collection of rational functions, and then at any point where things aren't defined, clear denominators. Being able to do this relies on the fact that the local ring of a regular point in a curve is a DVR, so we know what to multiply by in order to clear denominators: some power of the uniformizer of this DVR. The valuative criteria is a natural generalization of this procedure - if you keep interested in algebraic geometry, you'll meet this one day and go "oh, yeah, I recognize you".

(Related: Alex Youcis' comments here.)

@calitiutiu After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. — KReiser, May 23 '20 at 18:46

Extension of a morphism $k^*\rightarrow X$ to $\mathbb{P}^1\rightarrow \overline{X}$

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