By Oscar Zariski

Zariski offers a superior creation to this subject in algebra, including his personal insights.

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**Example text**

2, so we have |G|C2 = |/c|2 £ C'\ 28 ERIKO HIRONAKA The number of irreducible components in p~l{C) is the index oi He in G. Since the covering is Galois, all the components have the same self intersection. Therefore, \G\C* = for a given C" C p~l(C). C>\ Multiplying both sides of this equation by \Hc\ \IcV\G\ finishes the proof. • If a and /? (C") is nonempty only when they are the same curve. This only happens when aHc equals fiHc, or equivalently, when the intersection aHc H PHc is nonempty. f3(C') = j^\aHcCM3Hc\C2, and we have proven (*).

HI. 10 Examples. The corresponding braid for d = 2 is the generator element cr*. The corresponding braid for d = 3 is i

E r L _ i be the edges in T labelled L so that Px{^i) is the interval between Px(pi) and Px(pi+i)- Let { be any element ofG mapping to