Department of Mathematics, IISc, Bangalore, Prof. Dr. D. P . Patil MA 526 Commutative Algebra / Jan–April 2018 MA 526 Commutative Algebra / Jan–April 2018 (MSc and Ph. D. Programmes) Mobile : 07975952079 E-mails : patil@iisc.ac.in Lectures : Tuesday and Friday ; 15:30–17:00 Venue: Maths Building Midterm : Final Examination : Evaluation Weightage : Quizes : 20% MidTerm : 30% Final Examination : 50% 1 . N o e t h e r ’s N o r m a l i z a t i o n L e m m a / H i l b e r t ’s N u l l s t e l l e n s a t z 1.1 Prove the following partial generalization of Noether’s normalization lemma : Let A ⊆ R be an extension of integral domains such that R is an A -algebra of finite type. Then there exist an element f ∈ A , f � = 0 , and elements z 1 ,..., z m ∈ R such that z 1 ,..., z m are algebraically independent over A and R f is finite over A f [ z 1 ,..., z m ] . 1.2 Let R : = A [ X 1 ,..., X n ] , n ≥ 1 , be the polynomial algebra over an integral domain A . Then there exists a maximal ideal M ∈ Spm R with M ∩ A = 0 if and only if Q ( A ) = A f for some f � = 0 in A . ( Hint : If Q ( A ) = A f then ( fX 1 − 1 , X 2 ,..., X n ) is such a maximal ideal. Conversely, if M is such a maximal ideal then by the Exercise 1.1, there exists an element f ∈ A , f � = 0 , and elements z 1 ,..., z m in the field L : = R / M such that z 1 ,..., z m are algebraically independent over A and L is finite over A f [ z 1 ,..., z m ] . Therefore A f [ z 1 ,..., z m ] is also a field which means m = 0 and A f = Q ( A ) . — For example, in a principal ideal domain A , there exists an element f ∈ A , f � = 0 , with Q ( A ) = A f if and only if A has only a finite number of prime ideals.) 1.3 Let A be a principal ideal domain with infinitely many prime ideals. Then for every maximal ideal M ∈ Spm A [ X 1 ,..., X n ] the ideal m : = M ∩ A is maximal in A and A [ X 1 ,..., X n ] / M is finite over A / m . In particular, for every maximal ideal M in Z [ X 1 ,..., X n ] , the residue class field Z [ X 1 ,..., X n ] / M is finite. In other words, a field of characteristic zero is never a Z -algebra of finite type. ( Remark / Hint : — If a field K is finite type over Z , then K is finite. If Char K = 0, then show that Q is finite type over Z -algebra.) 1.4 (a) A finite commutative reduced C -algebra � = 0 is isomorphic to a product algebra C n , n ∈ N , where n is determined uniquely by the isomorphism type of the algebra. Every such a C -algebra is cyclic. (b) A finite commutative R -algebra � = 0 is isomorphic to a product algebra R m × C n , m , n ∈ N ,where the natural numbers m , n are determined uniquely by the isomorphism type of the algebra. Every such R -algebra is cyclic. 1.5 Let K be a field. If the unit group K × of K is finitely generated, then K is finite. (One can generalise this result to commutative rings which has only finitely many maximal ideals. — Such rings are called s e m i l o c a l. See “Bemerkungen über die Einheitengruppen semilokaler Ringe”, Math. Phys. Semesterberichte 17 , 168-181(1970).) 1.6 Let K be a field which is not algebraically closed. (a) For every m ∈ N + , there exists a non-constant polynomial f m ∈ K [ X 1 ,..., X m ] whose zero-set in K m is singleton { 0 = ( 0 ,..., 0 ) } , i.e. V K ( f ) = { ( 0 ,..., 0 ) } . ( Hint : Induction on m . For m ≥ 2 , put f m + 1 = f 2 ( f m , X m + 1 ) .) (b) Every K -algebraic set V ⊆ K n , n ≥ 1 , is a hypersurface in K n , i.e. it is the zero-set of a single polynomial, in sympols : V = V K ( f ) with f ∈ K [ X 1 ,..., X n ] . ( Hint : Use (a).) 1.7 ( G e n e r a l i s a t i o n o f H N S 1 ) Let K be an arbitrary field, S be the set of all polynomials in K [ X 1 ,..., X n ] that have no zeros in K n , i.e. S : = { f ∈ K [ X 1 ,..., X n ] | V K ( f ) = / 0 } and let a be an ideal in K [ X 1 ,..., X n ] . If S ∩ a = / 0, then V K ( a ) � = / 0 . ( Hint : Use the Exercise 1.6.) 1.8 ( H N S 4 ) Let K be an algebraically closed field. Then the map K n → Spm K [ X 1 ,..., X n ] , a �→ m a = � X 1 − a 1 ,..., X n − a n � is bijective. Moreover, for any ideal a ∈ I ( K [ X 1 ,..., X n ] , a ∈ V K ( a ) if and only if a ⊆ m a . 1.9 Let E | K be an arbitrary field extension and a � K [ X 1 ,..., X n ] be a non-unit ideal. Then the extended ideal a E [ X 1 ,..., X n ] � E [ X 1 ,..., X n ] is also a non-unit ideal. ( Hint : Apply HNS1 to the field extension E | K , where E denote an algebraic closure of E .) 1.10 Prove the equivalence of HNS4 and HNS1. ( Hint : Use the above Exercise.) Below one can see some Supplements / Test-Exercises to the results proved in the class. 1/ 3 D. P. Patil/IISc 2018MA526-IITBombay-CA-ex01.tex January 23, 2018 ; 2:38 p.m.
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