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How to prove that the field of rational functions on whole of projective $n$-space is constant functions?

By rational function I mean quotients of homogeneous polynomials of same degree defined/regular on whole of projective space.

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  • $\begingroup$ Do you mean regular functions? The field of rational functions on the projective n-space consists of any quotient of homogeneous polynomial of $(n+1)$ variables of the same degree. $\endgroup$
    – Yuchen Liu
    Commented May 9, 2013 at 15:22
  • $\begingroup$ I meant the field of rational functions on the projective n-space, which consists of any quotient of homogeneous polynomial of (n+1) variables of the same degree, but they should be defined on any point of projective space, i.e there should be one representation for which the denominator does not vanish at a point in the projective space, and that should be true for all points. $\endgroup$
    – user4567
    Commented May 9, 2013 at 15:33

1 Answer 1

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A rational function $\phi\in \text {Rat}(\mathbb P^n)$ is defined at $p\in \mathbb P^n$ if there is some representative $P/Q$ of $\phi$ with $P,Q\in k[T_0,\cdots,T_n]$ homogeneous of the same degree such that $Q(p)\neq 0$.
You are asking to characterize those $\phi$'s that are defined at all $p\in \mathbb P^n$.
The answer is that the only such $\phi$ are the elements of $k$, the constants . Here is why:

Any non-constant $\phi\in \text {Rat}(\mathbb P^n)$ has a unique (up to a non-zero constant ) expression of the form $P/Q$ with $P,Q$ homogeneous of the same positive degree and, crucially!, with $P,Q$ relatively prime.
The reason for this pleasant unicity is of course that $k[T_0,\cdots,T_n]$ is a UFD when $k$ is a field.
But now I claim that at any zero $p\in \mathbb P^n$ of $Q$, the rational function $\phi$ is not defined.
Proof of claim
Another representative of $\phi$ will be of the form $P_1/Q_1$ with $P_1,Q_1$ homogeneous of the same degree .
Since $P/Q,P_1/Q_1$ both represent $\phi$, we have $PQ_1=P_1Q$ so that $Q$ divides $PQ_1$ and since $P,Q$ are relatively prime $Q$ must divide $Q_1$.
Ah, but then $Q(p)=0$ implies $Q_1(p)=0$: so try as we can we will never find a representative of $\phi$ whose denominator does not vanish at $p$.
Long story short, $\phi$ is not defined at $p$.

The end of the story is that if we assume $k$ algebraically closed, any homogeneous polynomial of positive degree $Q$ will have some zero $p$ and the reasoning above shows that no non-constant $\phi\in \text {Rat}(\mathbb P^n)\setminus k$ is defined on all of $\mathbb P^n$.

Say it in formulas $$ \mathcal O(\mathbb P^n)=k\subsetneq \text {Rat}(\mathbb P^n)=k(T_1/T_0,\cdots, T_n/T_0)\subsetneq k(T_0,T_1,\cdots, T_n) = \text {Rat}(\mathbb P^n)(T_0) $$

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  • $\begingroup$ Great answer, and great writing style. Very nice. $\endgroup$
    – user4567
    Commented May 10, 2013 at 5:09
  • $\begingroup$ I'm glad you liked the answer, user 4567. $\endgroup$ Commented May 10, 2013 at 6:52

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