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Every affine variety $V$ has a unique projective closure $\overline{V}$, and there is an injective morphism $\iota : V \rightarrow \overline{V}$ given by something like $\iota(X_1,\ldots, X_n) = (X_1, \ldots, X_n, 1)$. The projective variety $\overline{V}$ can be determined explicitly by homogenizing the polynomials which define $V$.

Now suppose $f: V_1 \rightarrow V_2$ is a morphism of affine varieties. My intuition says that there is a way to "complete" $f$ to give a morphism of projective varieties $\overline{f} : \overline{V_1} \rightarrow \overline{V_2}$ in a way that agrees with $f$ on the affine subvariety $V_1$. In other words, I think there should be some projective morphism $\overline{f}$ that yields a commutative diagram $\iota_2 \circ f = \overline{f} \circ \iota_1$. Since the value of $\overline{f}$ is determined by $f$ on $V_1$, it remains only to define $\overline{f}$ on the points at infinity on $V_1$.

My questions is does such an $\overline{f}$ always exist, is it unique, and how can it be determined?

kless135
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  • A nitpick: the statement "every affine variety has a unique projective closure" can be very false depending on what definitions you take. At the very least you need to specify some embedding in $\Bbb A^n$, otherwise you can have bad things happen. Can you be more specific about what the scenario you're dealing with is? – KReiser Jul 07 '20 at 17:48
  • Mostly I'm interested in curves, and specifically elliptic curves. Does that make the scenario more straight forward? – kless135 Jul 07 '20 at 18:17
  • No. The link in the previous comment gives an example of a curve with two different non-isomorphic projective closures. – KReiser Jul 07 '20 at 18:27
  • In that case, I'm confused. Silverman's Arithmetic of Elliptic Curves textbook defines the projective closure of an affine algebraic set as the projective algebraic set whose homogenous ideal is generated by the homogenizations of the polynomials defining the affine variety. He goes on to remark that each affine variety may be identified with a unique projective variety in this way. – kless135 Jul 07 '20 at 18:47
  • See, this is specific, and now your statement makes sense! You've picked a particular embedding $X\to\Bbb A^n$, and the projective closure of your variety is the closure of the image of the composite morphism $X\to \Bbb A^n\to \Bbb P^n$. Without specifying an embedding, this doesn't make sense. – KReiser Jul 07 '20 at 18:55
  • I'm not understanding how my previous comment was more specific. You mentioned in your original reply about which definitions are being used, so I suspect this is the issue. Doesn't "affine variety" imply the set lives in some particular affine space and has a specific set of polynomials which define it? – kless135 Jul 07 '20 at 19:01
  • There's more than one definition of an affine variety. – KReiser Jul 07 '20 at 19:44

1 Answers1

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It exists, but no uniqueness. Given $f$, take $i:V_1\to W_1$, $j:V_2\to W_2$, some projective closures (highly non-unique). Then, take the closure $Z$ of the graph, $V_1\to W_1\times W_2$, $v\mapsto (i(v), j\circ f(v))$. We have natural maps $V_1\to Z\to W_2$ and I will leave you to check the rest.

Mohan
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