By Zariski O.

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Structure and Representation of Jordan Algebras

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Additional info for On the linear connection index of the algebraic surfaces z^n=f(x,y)

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Continuing in our optimism: Now we hope that we can adjust the solution 10 mod 72 by adding some multiple of 72 to it in order to get a solution mod 73. That is, we hope to nd y so that 10 + 72 y2 2 mod 73 Multiplying out and simplifying, this is Dividing through by 72 gives 294y ,98 mod 73 6y ,2 mod 7 Again, the inverse of 6 mod 7 is just 6 again, so this is y 6,2 2 mod 7 Therefore, satis es 10 + 72  2 = 108 1082 2 mod 73 This was considerably faster than brute-force hunting for a square root of 2 mod 73 directly.

The next integer after 2 on the list which hasn't been marked is 3. Starting with 3 + 3, mark every 3rd integer counting those already marked. This marks all multiples of 3 bigger than 3 itself. The next integer after 3 on the list which hasn't been marked is 5. Starting with 5 + 5, mark every 5th integer counting those already marked.  ... Take the next integer n on the list which has not yet been crossed-o . This n is prime. Starting with n + n, cross o every nth integer counting those already marked.

And, funnily enough, the procedure to do so is exactly parallel to Newton's method for numerical approximation to roots, from calculus. In particular, we will use a purely algebraic form of Taylor expansions to prove the result. First we'll do a numerical example to illustrate the idea of the process to which Hensel's Lemma refers. Suppose we want to nd x so that x2 2 mod 73. Noting that a solution mod 73 certainly must give a solution mod 7, we'll start by nding a solution mod 7. This is much easier, since there are only 7 things in Z=7, and by a very quick trial and error hunt we see that 32 = 9 = 2 mod 7.