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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?).

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    $\begingroup$ 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. $\endgroup$ – KReiser Apr 6 '20 at 1:50
  • $\begingroup$ 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! $\endgroup$ – user350031 Apr 6 '20 at 7:27
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Yes, this proof is correct. (Filler text to meet character limit.)

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