By Oscar Zariski

Zariski offers an exceptional advent to this subject in algebra, including his personal insights.

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**Additional info for An Introduction to the Theory of Algebraic Surfaces**

**Sample text**

D ~ r ) = v ( B ( ~ ) J = coefficient of ~ in the divisor r d~r). Since the ~ i are not uniformizing coordinates of is infinite at ~ or ~ is a component of cycles r . Let eihher some ~i (d~ I ... d ~ r )" Thus there are only a finite number of prime divisoria! cycles are not uniformizing coordinates of ~ . , r. , r a~ Denote the right-h~d side of (*) by s(t--). si o sd'-i). Hence each Ai Z o "• , 0 CJ . Let C~ ~ I--1, . . , ~tl--i" (*) l--h~ , We have sho~n ~ t (A•247 k, -~0, and . Trace of a differential.

The system L(S) has no fixed components and is uniquely determined by thim condition and by the given k-module 3. , if L(S) h~permtvfaoes of order S for L(Rq) P~op. 6: Let and S q. If f(y) / O, Proof: Z~ and Fix f(Y) = O. ~ We write L r for L(Rq). q and S t be two finite k-modules of homogeneous q are integral over ~o ~ S, 5o { O. ZI then we have a cycle with the hy~ersurface functions of degree S) f(Yo, ""' Yn ) of is called the linear system cut out on V the intersection cycle of V Lq is the set of forms where S~S k[S], For ~E then S t.

10: is normal if and only if V Proof: Assume C ~ is normal. for some Therefore h. L'q = Zq Assume say V q ~ h. Lq If and q is integrally is complete for large Then we have just seen that is large, Lq Lq k ~ i] R'= R q . h _. Rq h (Yo' "'" Yn )h and so RVq = Rq. is complete. is complete for large q. Therefore R'(Yo, "'" Yn )h c Prep. 9 now shows that V Then R' = Rq q R and so for large is normal. the equation of the hyperplane H Then be k[Va] , V(~ ), the variety determined by ~ ~ and , ~[dh = (I) if and only if is contained in H.