# The function field of an affine part of a projective variety

Let $$\phi:\mathbb{A}^n \to U_0\subseteq \mathbb{P}^n$$ be given by $$\phi(a_1,\ldots,a_n)=(1:a_1:\ldots:a_n).$$

Let $$X\subseteq \mathbb{P}^n$$ be an irreducible Zariski-closed subspace (I call this a projective variety) such that $$X\cap U_0\neq \emptyset.$$

Then $$Y:=\phi^{-1}(X\cap U_0)\subseteq \mathbb{A}^n$$ is an irreducible Zariski-closed subspace (I call this an affine variety).

Let $$\theta:k[y_1,\ldots,y_n] \to k(X)$$ be the $$k$$-algebra homomorphism such that $$\theta(y_i)=x_i/x_0$$ for $$i=1,\ldots,n.$$

PROBLEM. I'm struggling to show (in an algebraic fashion) that $$\ker\theta$$ is the vanishing ideal of $$Y.$$

Any hints or help greatly appreciated!

ATTEMPT. Recall that $$k(X)$$ consists of formal fractions $$g/h$$ where

1. $$g,h \in k[x_0,\ldots,x_n]$$ are homogeneous of the same degree,

2. $$h$$ does not vanish on $$X$$ i.e. $$h\notin I(X),$$

3. we identify two fractions $$g/h$$ and $$g'/h'$$ if and only if $$gh'-g'h \in I(X).$$

Note that, for any $$f \in k[y_1,\ldots,y_n],$$ we have $$\theta(f)=\frac{F(x_0,x_1,\ldots,x_n)}{x_0^{\deg f}}$$ where $$F$$ is the homogenisation of $$f$$ at $$x_0.$$

It follows that $$f \in \ker \theta$$ if and only if $$F \in I(X).$$

Clearly, if $$F \in I(X),$$ then $$f=F(1,y_1,\ldots,y_n) \in I(Y).$$

Conversely, if $$f \in I(Y),$$ then $$F \in I(X\cap U_0).$$

Hence, since $$X\cap U_0$$ is dense in $$X,$$ it follows that $$F \in I(X).$$

This proves the claim (right?).

• One slight nitpick: the kernel should be the vanishing ideal of $Y$. Anyways, this ought to be fairly direct from the definition of $k(X)$. Please add your definition and any attempts you've made to your post. – KReiser Apr 6 '20 at 1:50
• Oh yes, sorry, I've changed to "vanishing ideal of Y". I think I've managed to give the proof now - does it look right to you? Thanks! – user350031 Apr 6 '20 at 7:27