A GAGA question.

Say I have a ``quasi-projective'' (*) subvariety X over the complex numbers within a smooth complex algebraic variety Z.

True or False: The analytic and algebraic closure of X (within Z) coincide.

I guess the answer must be `True' and is contained somewhere within Serre's GAGA, or some elucidation thereof. If I'm right could someone point me to a precise reference, either within GAGA or elsewhere? If I'm wrong I'd love to hear about it.


(*) By `quasi-projective'' I mean X is defined by a finite number ofalgebraic equations' $f_i = 0$ and inequalities $g_a \ne 0$, As is the case when Z is projective space, the $f_i$ may not be globally defined; same for the $g_a$. In my situation, the Zariski open set defined by intersecting the sets $g_a \ne 0$ is an affine set (in the usual schemy sense) and the $f_i$ are polynomials on this affine set.

My Z is probably projective -- I'm not positive here, just pretty sure. (My Z is obtained by iterating the construction of taking the bundle over a smooth projective variety whose fiber is the Grassmannian of d-planes within said variety's tangent space. )

  • $\begingroup$ X "within" Z is rather confusing. C^* is "in" the affine plane both as a non-closed embedded variety and as a closed embedded variety. A clarification is in order I think. $\endgroup$
    – Maharana
    Dec 30, 2009 at 6:30
  • $\begingroup$ sorry, I got it now. Subvariety already means a closed subset. So your X here has a closed embedding inside Z. Your definition of quasi-projective is still confusing me though! I have never seen the inequalities $g_a\neq{0}$ in any such definition. $\endgroup$
    – Maharana
    Dec 30, 2009 at 6:55

1 Answer 1


Yes, it is true: the analytic and algebraic closures of $X$ in $Z$ coincide (and you don't need at all to assume that $Z$ is smooth).

You may suppose that X is open in $Z$: if it isn't, just replace $Z$ by the Zariski (=algebraic) closure of $X$ in $Z$. Then Serre's Proposition 5, page 11 of his famous GAGA article "Géométrie Algébrique et Géométrie Analytique" says exactly that the analytic closure of $X$ in $Z$ is $Z$, i.e. coincides with its Zariski closure. [ Caution: his "X" takes the place of your "Z" , the big ambient variety]

You can also find the result (in English !) in the fine book by Joseph L. Taylor "Several Complex Variables with Connections to Algebraic Geometry and Lie Groups", published by the AMS in its series Graduate Studies in Mathematics ( Volume 46).There the result you need is Proposition 13.4.6, page 344.


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