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Let $X$ be a variety, $Y \subset \mathbb{A}^n$ an affine variety and $\psi:X \rightarrow Y$ a map such that $x_i \circ \psi$ is a regular function. We want to show that $\psi$ is a morphism of varieties. Since regular functions form a ring, then for any polynomial $f \in k[x_1,\cdots,x_n]$ we have that $f \circ \psi$ is regular and so $\psi$ is continuous. Now we need to show that if $g$ is a regular function on $Y$, then for any open set $V$ of $Y$ the function $g \circ \psi: \psi^{-1}(V) \rightarrow k$ is regular. Hartshorne says that this follows since $g$ is locally a quotient of polynomials.

Here is my "objection": to show that $f \circ \psi$ is regular, we need to show that it is locally a quotient of polynomials, where the polynomials correspond to the "correct" ambient space of $X$. For example, if $X$ is a projective variety, then we need $f \circ \psi$ to be a quotient of homogeneous polynomials of equal degree. In other words, $f \circ \psi$ is locally a quotient of polynomials only after passed through $\psi$... So, what am i missing here?

Manos
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1 Answers1

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The assumption is that $x_i\circ\psi$ is regular for all $i=1,\ldots,n$. Thus, locally in whatever coordinates we have on $X$, we can express $x_i\circ\psi$ as a quotient of polynomials (of equal degree if $X$ is quasi-projective). In other words, we can write the function as $$\psi(p) = (\psi_1(p),\ldots,\psi_n(p))$$ with $p\in X$ and where each $\psi_i(p)$ is a quotient of polynomials in the ambient coordinates on $X$.

Now whatever the expression is for $f$ as a polynomial in the $x_i$'s, we can express $f\circ\psi$ as a quotient of polynomials, namely, we have $$f\circ\psi(p) = f(\psi_1(p),\ldots,\psi_n(p))$$ which is clearly also a quotient of polynomials in the coordinates on $X$, since the $\psi_i(p)$ are and $f$ is a polynomial (and again, things work out in the quasi-projective case).

Andrew
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  • Fantastic answer, +1! – Manos Aug 01 '13 at 16:35
  • @Andrew, please pardon this late question, but I have a lingering concern. Hartshorne writes his regular function on an open set $V\subset Y$ as $g$; you write $f$ here. I will use your notation. We have that $f$ is regular on $V$ and want to show that $f\circ\psi$ is regular on $\psi^{-1}(V)$. You assume $f$ is a polynomial, and in that case it is fine and goes through as you write. But why must $f$ be polynomial instead of a quotient of polynomials? We know that $\psi_i$ is regular at $p$, but do we have an assurance it doesn't vanish there? I am missing this little piece in the argument. – Tomo May 20 '14 at 00:42
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    Dear @Owen, you're right, $g$ is not necessarily a polynomial -- it is only locally a quotient of polynomials. However, I use the same notation as Hartshorne, i.e. $f$ means $f$. If $g = h_1/h_2$ locally near some chosen point $y\in Y$, then $g\circ\psi = (h_1\circ\psi) / (h_2\circ\psi)$ and we can ensure that $h_2\circ\psi$ does not vanish near any chosen point $p$ in the fibre over $y$. – Andrew May 20 '14 at 16:29
  • Dear @Andrew, thank you for your response, and sorry for confusing your notation. This is where I am getting stuck: you say that 'we can ensure that $h_2\circ\psi$ does not vanish near any chosen point $p$ in the fibre over $y$.' But this is what I do not see, since is it not the case that we need $h_2\circ\psi$ to not vanish on all of $\psi^{-1}(y)$,the fibre over $y$, not just on some open subset? That is, as far as I understand, we cannot just throw out $Z(h_2\circ\psi)\cap\psi^{-1}(y)$. So in fact $Z(h_2\circ\psi)\cap\psi^{-1}(y)=\emptyset.$ But this is what I cannot justify to myself. – Tomo May 20 '14 at 18:39
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    Dear @Owen, if $p\in Z(h_2\circ\psi)\cap\psi^{-1}(y)$, then $0 = h_2(\psi(p)) = h_2(y)$. But this cannot happen, since we chose $h_2$ in particular to satisfy $h_2(y)\neq 0$. – Andrew May 20 '14 at 19:26