# A GAGA question

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.

Elucidation:

(*) 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. )

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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. –  Maharana Dec 30 '09 at 6:30
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. –  Maharana Dec 30 '09 at 6:55

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]