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Unformatted text preview: INTEGRAL ELEMENTS AND EXTENSIONS PETE L. CLARK Recall that a complex number is said to be an algebraic integer if is the root of a nonconstant monic polynomial with Z coefficients: i.e., if there exists an n and integers a ,...,a n- 1 such that n + a n- 1 n- 1 + ... + a 1 + a = 0 . In order to prove the Quadratic Reciprocity Law, we used the following fact: Proposition 1. Let n be a positive integer and n a primitive n th root of unity. Then every element of the ring R n = Z [ n ] is an algebraic integer. We give two proofs here. The first is a quick one, which however assumes the following fact that one should learn in undergraduate algebra: a subgroup of a finitely generated abelian group is finitely generated. One can deduce this from the structure theory of modules over a PID, although it is in fact easier (if less undergraddy) to use a little bit of the theory of Noetherian rings. Then we will give a second proof, longer but self-contained, of a much more general result. 1. Proof of Proposition 1 Let be any element of R n = Z [ n ], and consider the subring Z [ ] generated by . Note that R n is a finitely generated abelian group: indeed, it is generated by 1 , n ,..., n- 1 n . (With a bit more care, one can verify that R n = Z [ T ] / n ( T ), so that R n = Z ( n ) as an abelian group. But we dont need this.) Instead, we will use the fact that Z [ ], being a subgroup of the finitely gener- ated abelian group R n , is itself finitely generated: that is, there exists a finite set of elements a 1 ,...,a N of Z [ ] such that every element of Z [ ] can be written as a Z-linear combination of the a i s: Z [ ] = = r 1 a 1 + ... + r N a N , r i Z . Now each element a i Z [ ] is, by definition, a polynomial in with integer coef- ficients, say a i = f i ( ). Let t be the maximum degree of these polynomials. We may therefore write = t +1 as t +1 = r 1 f 1 ( ) + ... + r N f N ( ) , which shows that satisfies the monic polynomial...
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