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