By Florentin Smarandache

A set of definitions, questions, and theorems edited by way of M. L. Perez, corresponding to Smarandache sort conjectures, difficulties, numerical bases, T-numbers, progressions, sequence, capabilities, Non-Euclidean geometries, paradoxes (such as Smarandache Sorites Paradox that our seen international consists through a totality of invisible particles), linguistic tautologies, Smarandache speculation that there's no pace barrier within the universe - which has been prolonged to SRM-theory.

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**Extra info for Definitions, Solved and Unsolved Problems, Conjectures, and Theorems in Number Theory and Geometry**

**Example text**

R1 ) are the real embeddings and {σi , σi+r2 } (i = r1 + 1, . . r1 + r2 ) the pairs of complex conjugate embeddings. Viewed as a Z-module, OK is free of rank d. Taking any Z-basis {ω1 , . . , ωd } of OK , we define the discriminant of K by 2 DK := DK/Q (ω1 , . . ,d . This is a non-zero rational integer which is independent of the choice of the basis. Let M ⊃ L ⊃ Q be a tower of algebraic number fields with [M : L] = n. , the ideal of OL generated by all numbers DM/L (α1 , . . , αn ) with α1 , .

8). The completion of a field K with discrete valuation v is the completion of K with respect to the absolute value D−v for any D > 1. The discrete valuation v can be extended uniquely to a discrete valuation on this completion. 3 Absolute Values and Places on Number Fields 39 Let K be a field. A place of K is an equivalence class of non-trivial absolute values of K. As mentioned above, two equivalent absolute values of K give rise to the same completion. So we can speak about the completion of K at a particular place v, which we denote by Kv .

After multiplying P, Q with suitable elements of K ∗ , we may assume that v(P) = v(Q) = 0. Then the reductions P, Q of P, Q modulo pv are non-zero polynomials in kv [X1 , . . , Xr ]. Hence P · Q 0, which implies v(PQ) = 0. 1 implies that v can be extended to a discrete valuation on K(X1 , . . , Xr ), also denoted by v, given by v(R) = v(P) − v(Q) for R = P/Q with P, Q ∈ K[X1 , . . , Xr ], Q 0. Let A be a Dedekind domain with quotient field K. For a polynomial P ∈ K[X1 , . . , Xr ], we denote by (P) the fractional ideal of A generated by the coefficients of P.