Appendix D: Integrality and Algebraic Integers

Appendix D: Integrality and Algebraic Integers

Skript zur Vorlesung: Charaktertheorie SS 2020 69 Example 16 ˆ E.g. the tensor product of two 2 2-matrices is of the form » fi b “ P p q „ ⇢ „ ⇢ — ffi 4 — ffi M K – fl Lemma-Definition›Ñ C.4 (Tensor› productÑ of K-endomorphisms) 1 2 b b If : V V and : W W are two endomorphisms of finite-dimensional K -vector spaces V 1 2 1 2 K and W , then the tensorb productbof and›Ñ is theb K -endomorphism of V W defined by 1 2 bK fiÑ p bK qp b q “ p qb p q : V W V W 1 2 1 2 p b q“ p q p q : 1 2 1 2 Furthermore, Tr Tr Tr b. “ t u “t u 1 2 V Proof : It is straightforward to check that is K -linear. Moreover, choosing ordered bases B 1 W 1 b b “t b | § § § § u and B of V and W respectively,K it is straightforward from the definitions 1 2 V W to check that the matrix of w.r.t. the ordered basis B 1 1 is 1 V 2 W the Kronecker product of the matrices of w.r.t. B and of w.r.t. to B . The trace formula follows. D Integrality and Algebraic Integers Z We recall/introduce here some notions of the Commutative Algebra lecture on integrality of ring ele- ments. However, we are essentially interested in the field of complex numbers and its subring . Definition D.1 (integral element, algebraic integer) Let A be a subring ofP a commutative ring B. P r s ´ p q“ ` ´ ` ` ` (a) An element B is said to be integral over A if is a root of monic 1 polynomial A X , ´ P 1 1 0 that is 0 and is a polynomial of the form X X X with 1 0 A. If all the elements of B are integral over A, then we say that B is integral Z C C Z over“A. “ P (b) If A and B , an element integral over is called an algebraic integer. TheoremÑ D.2 P Let A B be a subring of a commutative ring and let B. TFAE: (a) is integralr s over A; (b) the ring A is finitely generated as an A-module; (c) there exists a subring S of B containing A and which is finitely generated as an A-module. r s Recall that A denotes the subring of B generated by A and . Skript zur Vorlesung: Charaktertheorie SS 2020 70 ´ Proofñ : ´ P ` ´ ` ` ` “ ˚ r s ´ 1 ´ 0 1 1 r s“ 1` `0 ` ´ (a) (b): Let P A`such` that ` “1 • 0 ( ). We1 prove that A is generated as an A-module by 1 1 ,´ i.e. A A A A . Therefore it suffices ¨ “ ˚ “´ ´ ´ ´ ´ P to prove that A A A : C1 for every´ . We proceed by induction on : ¨ ° 1 P1 0 If , then ( ) yields´ C. ˚ 1 If , then we may assume that´ ´C` by the induction´ hypothesis. Hence multiplying ( ) by yields“´ ´ ´ ´ ´ P 1 1 ´ 1 ´ 1 0 ´ P C 1 ñ “ r s 1 0 because C. Z ñ r sÑ “ ` ` P P ° § § (b) (c): Set S : A . “ “ 1 P 1“p q 0 ˆ (c) (a): By assumption A S A A , where B, Tr . Thus for each 1 “ ´p ∞q P p q 1 we have 1 for certain A. Set : and consider the -matrix M : I M S . Hence “ ñ p q “ p q M 0 adj M M 0 where adj M is the adjugate matrix of M (i.e. the transpose of its cofactor matrix). By the properties of the determinant (GDM), wep haveq “ p q p q “ § §adj M M detp Mq I“ P P p q“ Hencep ¨det´pM q qP0 forr eachs 1 , and so det M 0 for every S. As 1 S this gives us det M 0. It now follows from the definition of M that is a root of the monic polynomial det X I A X , thus integral over A. CorollaryÑ D.3 t P | u Let A B be a subring of a commutativeP ring. Then B integral` over¨ A is a subring of B. ´ ´ Z r s“ ` ` ` r s“ ` ` ` P ° Proof : We need to prove that if B are integral 1 over A, then so are and 1. By Theorem D.2(b) “ r s t | § § § § u 0 and its proof both `A A ¨ A A and A A A A for some . Thus S : A is finitely generated as an A-module by 0 0 . Theorem D.2(c) now yields that and are integral over A because they belong to S. Z Z r s Example 17 Z ` P r s All the elements of the2 ring of Gaussian intergers are integral over , hence algebraic integers, since is a root of X 1 X . Q Z Z LemmaP D.4 P If is integral“ over , then . • Z ´ P Proof : We may write , where and are coprime´ integers and 1. By the hypothesis there exist 0 1 such that ` ´ ´1 ` ` ` “ 1 1 1 0 ´ ´ 0 ` ´ ` ` ` “ hence 1 1 1 1 0 looooooooooooooooooooomooooooooooooooooooooon 0 Z | p q“ • divisible by “ P Thus . As gcd 1 and 1 this is only possible if 1, and we deduce that . Clearly, the aforementionnend lemma can be generalised to integral domains (=Integritätsring) and their field of fractions..

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