By David Mumford

From the reports: "Although a number of textbooks on sleek algebraic geometry were released meanwhile, Mumford's "Volume I" is, including its predecessor the purple booklet of sorts and schemes, now as prior to the most first-class and profound primers of recent algebraic geometry. either books are only precise classics!" Zentralblatt

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**Additional resources for Algebraic Geometry I: Complex Projective Varieties**

**Example text**

G1, ... , gm. Therefore, f belongs to the ideal generated by g1, ... , gm, and we conclude that this set of elements generates Z.

If 0 is a polydisc centered at a boundary point of U, then every component of U n o contains infinitely many of the spheres S3 . Thus, f has zeroes of arbitrarily high total order in each component of o n U. If f could be continued from one of these components to a holomorphic function f on 0, then f and all of its partial derivatives would have to vanish at each point of 0 which is a boundary point of the component. Then f would vanish identically on 0 and, consequently, on every component of U which meets A.

Hence, limsup k-1 log Jfk(z')J < - logc k for all z'EA(a',s'). The functions k-1 log I fk(z')I are bounded and measurable on 0(a', s'), and so, with A = 0(a', s'), Fatou's lemma implies limsupJ k -llogIfkz')IdV(z') < k p limsupk fzA logI k < - JO I log c. We conclude that, if 0 < co < c, then there is a k0 and a 6 > 0 such that L k-l log fk(z')IdV(z') c -JAI logco - S, for all k > k0 Now, since k-1 log I fk(z')I is bounded by M0 for all k and all . z' E 0, we may choose c sufficiently small that, if 0 = 0 (a', s') is replaced by of = 0 (w', s' - 0, where w' E 0 (a', E) and E = (c,.