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There is a theorem in Shafarevich's Basic Algebraic Geometry (Theorem 3, pg. 109) which states that if $X$ is a nonsingular variety, and $\varphi\colon X\to\mathbb{P}^N$ a rational map to projective space, then the set of points at which $\varphi$ is not regular as codimension $\geq 2$.

So if $X=\mathbb{P}^1$, this immediately implies that every rational map $\mathbb{P}^1\to\mathbb{P}^N$ is also regular.

Is there a way to see this in a more "down-to-earth" way, I feel like this theorem is a bit overkill for the result. Thanks.

Kaitlyn Schweed
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1 Answers1

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Suppose $f:P^1\to P^n$ is a rational map. On a nonempty open subset of the domain, then, it can be written as $(x:y)\mapsto(f_0(x,y):\cdots:f_n(x,y))$ with the $f_i$ homogeneous of the same degree.

If all the $f_i$ vanish at a point, they all have a factor in common, and you can remove it. That way, you can extend the map to the whole of $P^1$.

Mariano Suárez-Álvarez
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  • This is a consequence of the fact that homogeneous polynomials in two variables $h(x,y)$ are, for lots of intents and purposes, the same thing as univariate polynomials, by doing transformations of the form $g(t)=h(t,1)$. – Mariano Suárez-Álvarez Feb 06 '15 at 21:37
  • This whole business of taking out a common factor will show that a rational map from a smooth curve into projective space is regular too. – Hoot Feb 06 '15 at 21:42