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4400algebra3

# 4400algebra3 - INTEGRAL ELEMENTS AND EXTENSIONS PETE L...

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INTEGRAL ELEMENTS AND EXTENSIONS PETE L. CLARK Recall that a complex number α is said to be an algebraic integer if α is the root of a nonconstant monic polynomial with Z coefficients: i.e., if there exists an n and integers a 0 , . . . , a n - 1 such that α n + a n - 1 α n - 1 + . . . + a 1 α + a 0 = 0 . In order to prove the Quadratic Reciprocity Law, we used the following fact: Proposition 1. Let n be a positive integer and ζ n a primitive n th root of unity. Then every element of the ring R n = Z [ ζ 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 α be any element of R n = Z [ ζ n ], and consider the subring Z [ α ] generated by α . Note that R n is a finitely generated abelian group: indeed, it is generated by 1 , ζ n , . . . , ζ n - 1 n . (With a bit more care, one can verify that R n = Z [ T ] / Φ n ( T ), so that R n = Z ϕ ( n ) as an abelian group. But we don’t need this.) Instead, we will use the fact that Z [ α ], 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 [ α ] such that every element of Z [ α ] can be written as a Z -linear combination of the a i ’s: β Z [ α ] = β = r 1 a 1 + . . . + r N a N , r i Z . Now each element a i Z [ α ] is, by definition, a polynomial in α with integer coef- ficients, say a i = f i ( α ). Let t be the maximum degree of these polynomials. We may therefore write β = α t +1 as α t +1 = r 1 f 1 ( α ) + . . . + r N f N ( α ) , which shows that α satisfies the monic polynomial T t +1 - r 1 f 1 ( T ) - . . . - r N f N ( t ) and is therefore an algebraic integer.

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