Suppose $k$ is a algebraically closed field,$\mathbb{P}_k^n$ is the projective space over $k$,given the Zariski topology.A regular function on $\mathbb{P}_k^n$ is a function $f:\mathbb{P}_k^n \longrightarrow k$ which is locally a quotient of two homogeneous polynomials of the same degree with the denominator not vanishing on that open set.$\mathbb{P}_k^n$ replaced with any other variety $X$, we get a regular function on $X$(with little revision to the definition).Then do there exist such a regular function on $\mathbb{P}_k^n$?How to prove or disprove it?Some hints will be much helpful.
[Math] Do there exist regular functions on the projective space over a field
algebraic-geometry
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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 $$ \mathcal O(\mathbb P^n)=k\subsetneq \text {Rat}(\mathbb P^n)=k(T_1/T_0,\cdots, T_n/T_0)\subsetneq k(T_0,T_1,\cdots, T_n) = \text {Rat}(\mathbb P^n)(T_0) $$
For your first question, if $X$ is affine, then $f$ can be seen globally as a polynomial. I believe Shafarevich proves this.
If $X$ is an irreducible quasiprojective in general, let $f=p/q$ locally at a point $x$, for $p$ and $q$ polynomials. If $y\in X$, let $f=r/s$ at $y$. Define $$Y=\{z\in X:p(z)s(z)-q(z)r(z)=0\}.$$ This is a closed subset of $X$ (since the equation above is homogeneous and can be seen as defining a projective variety which we then intersect with $X$). Moreover, since $f=p/q$ (likewise with $f=r/s$) on an open (and therefore dense) subset of $X$, we see that this closed set contains a dense open set of $X$. Therefore $X=Y$.
Conclusion: You can define $f$ as $p/q$ wherever $q\neq0$. If $q(z)=0$ for some $z$, then you need another representation for $f$ at $z$.
Example: Take $X=\{x^2+y^2-1=0\}\subseteq\mathbb{A}^2$ and $f(x)=x^2/(y^2-1)$. $f$ could possibly have a pole at $y=\pm1$, but when we use the identity $y^2-1=-x^2$, we see that $f(x)$ is actually the constant function $-1$. This is a very boring example but it gets the point across.
For your second question, take a look at Lemma 2 in the section of Quasiprojective Varieties. First assume that $X$ is affine. The equations that Shafarevich finds for the affine variety $X-Z(f)$ are basically the equations for $X$, except that you can replace $T_{n+1}$ by $f$. Try to write the coordinate ring using this, and then see if you can generalize this to an arbitrary quasiprojective variety.
Best Answer
I personally like how Shafarevich answers this question in his book Basic Algebraic Geometry: Varieties in Projective Space. It goes something like this:
First you prove that if you have a regular map $f:X\to Y$, then its graph is closed in $X\times Y$. It's easy to see that it is enough to assume that $Y$ is projective space, so we set $Y=\mathbb{P}^n$. Consider now the regular map $$f\times\mbox{id}:X\times\mathbb{P}^n\to\mathbb{P}^n\times\mathbb{P}^n$$ where $(f\times\mbox{id})(x,y)=(f(x),y)$. We have that the graph of $f$ is then $(f\times\mbox{id})^{-1}(\Gamma)$, where $\Gamma$ is the graph of $\mbox{id}:\mathbb{P}^n\to\mathbb{P}^n$. Since regular maps are continuous with the Zariski topology, we only have to prove that $\Gamma$ is closed. But if $([x_0:\cdots:x_n],[y_0:\cdots:y_n])$ are the coordinates of $\mathbb{P}^n\times\mathbb{P}^n$, we have that $\Gamma$ is given by the equations $x_iy_j=x_jy_i$, and so is closed.
A second theorem you'll need which is a bit more involved (and so I won't prove it here), shows that if $X$ is a projective variety and $Y$ is quasiprojective, then the second projection $p:X\times Y\to Y$ takes closed sets to closed sets.
If you have this, then you can see the following:
Theorem The image of a projective variety under a regular map is closed.
Proof If $f:X\to Y$ is regular and $X$ is projective, then we have that $f$ can factor as $x\mapsto(x,f(x))\mapsto f(x)$. We have that the graph of $f$ is closed, and then the second map is just the second projection. Because of what we said earlier, we have that the image of $f$ is closed.$\Box$
As a corollary to this, we see that if $f:X\to \mathbb{A}^1$ is a regular function and $X$ is projective (and irreducible, Hartshorne's definition includes irreducibility, Shafarevich's does not), then the image of $X$ is closed, and thus is either $\mathbb{A}^1$ or is finite. However we can see $f$ as a regular function to $\mathbb{P}^1$ also, and since its image must be closed in $\mathbb{P}^1$, it's impossible for its image to be $\mathbb{A}^1$. Thus its image must be finite.
We have that $X=\cup_{x\in\mathbb{A}^1}f^{-1}(x)$, and if $X$ is irreducible, we must have that $f$'s image is only one point. Thus $f$ is constant, and the theorem is proved.