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Suppose we have a quasi-projective morphism $F: V \rightarrow W$ for $V, W$ affine varieties (quasi-projective varieties isomorphic to some Zariski closed subset of affine space). I was wondering if $F$ is not only locally polynomial, but 'globally', i.e. the homogeneous polynomials that define the map on open subsets of $V$ is the same for each open subset.

I feel that this may not be the case...however, the polynomial maps defined locally have to agree on the intersections, so there is some restriction to defining any such map.

How do I show that from the definition of a quasi-projective morphism that such a morphism between affine varieties is described as a 'global' polynomial?

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No, this is impossible: any map of affine varieties $X\to Y$ is equivalent to a map of their coordinate algebras $k[Y]\to k[X]$, which provides you the "global" polynomial description you seek to avoid.

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  • $\begingroup$ How do I see this? (Let z.c. variety denote a Zariski closed subset of affine space) I know that a morphism of z.c. varieties gives rise to a homomorphism of coordinate algebras, but why does a quasi-projective morphism behave the same way? The morphism of z.c. varieties is already defined to be "globally" polynomial so this is clear, but the definition of q.p. morphisms seems to be involved in only local behavior so I can't see it in this case. $\endgroup$
    – green frog
    Jan 14, 2020 at 23:08
  • $\begingroup$ Any good definition of affine varieties establishes that the coordinate algebra is an intrinsic invariant of the variety - this means that it does not depend on how we embed the affine variety in any other space. From here, any morphism of affine varieties induces a morphism of coordinate algebras by pullback: if we have $f:X\to Y$, a regular function $g$ on $Y$ pulls back to a regular function $g(f(-))$ on $X$. This morphism of coordinate algebras will then give you the global polynomial description you want. $\endgroup$
    – KReiser
    Jan 14, 2020 at 23:13
  • $\begingroup$ Does the following argument make sense: Let $f$ be a q.p. morphism $f: X \rightarrow Y.$ The projection onto the $i^{th}$ coordinate of points in $Y$ is a regular function. Hence, the composition with $f$ is a regular function. That is, the mapping of $f$ onto each coordinate is defined by a rational function, which is in fact a polynomial on $X$. Hence, $f$ is a global polynomial? So in fact, we don't really have to differentiate z.c. morphism and q.p. morphism? $\endgroup$
    – green frog
    Jan 14, 2020 at 23:25
  • $\begingroup$ I don't understand what you're getting at here. Any morphism of affine varieties gives you a map on coordinate algebras - this is the point of the definition of a morphism of varieties! If you're having specific issues with specific definitions, perhaps you should edit your post to make these clear. $\endgroup$
    – KReiser
    Jan 14, 2020 at 23:32
  • $\begingroup$ It seems like I'm supposed to deduce that $f$ is a z.c. morphism from the fact that $g(f(-))$ is a regular function on $X$? Isn't that what I did in my comment? I understand a z.c. morphism of a.a. varieties gives a map on coordinate algebras, but I'm trying to see that a q.p. morphism does the same. To my knowledge, these morphisms aren't defined the same... $\endgroup$
    – green frog
    Jan 14, 2020 at 23:38

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