MA 526 Commutative Algebra / JanApril 2018 (MSc and Ph. D. - - PDF document

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 z1,...,zm ∈ R such that z1,...,zm are algebraically independent over A and Rf is finite over Af [z1,...,zm]. 1.2 Let R := A[X1,...,Xn], 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) = Af for some f = 0 in A. (Hint : If Q(A) = Af

then (fX1 −1,X2,...,Xn) 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 z1,...,zm in the field L := R/M such that z1,...,zm are algebraically independent over A and L is finite over Af [z1,...,zm]. Therefore Af [z1,...,zm] is also a field which means m = 0 and Af = Q(A). — For example, in a principal ideal domain A, there exists an element f ∈ A, f = 0, with Q(A) = Af 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 ∈ SpmA[X1,...,Xn] the ideal m := M ∩ A is maximal in A and A[X1,...,Xn]/M is finite over A/m. In particular, for every maximal ideal M in Z[X1,...,Xn], the residue class field Z[X1,...,Xn]/M is finite. In

• ther 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 CharK = 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 Cn, 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 Rm ×Cn, 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 fm ∈ K[X1,...,Xm] whose zero-set in Km is singleton {0 = (0,...,0)}, i.e. VK( f) = {(0,...,0)}. (Hint : Induction on m. For m ≥ 2, put fm+1 =

f2( fm,Xm+1).)

(b) Every K-algebraic set V ⊆ Kn ,n ≥ 1, is a hypersurface in Kn, i.e. it is the zero-set of a single polynomial, in sympols : V = VK( f) with f ∈ K[X1,...,Xn]. (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[X1,...,Xn] that have no zeros in Kn, i.e. S := { f ∈ K[X1,...,Xn] | VK(f ) = / 0} and let a be an ideal in K[X1,...,Xn]. If S∩a = / 0, then VK(a) = /

• 0. (Hint : Use the Exercise 1.6.)

1.8 ( H N S 4 ) Let K be an algebraically closed field. Then the map Kn → Spm K[X1,...,Xn], a → ma = X1 − a1,...,Xn − an is bijective. Moreover, for any ideal a ∈ I(K[X1,...,Xn], a ∈ VK(a) if and only if a ⊆ ma. 1.9 Let E |K be an arbitrary field extension and a K[X1,...,Xn] be a non-unit ideal. Then the extended ideal aE[X1,...,Xn] E[X1,...,Xn] 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.

• D. P. Patil/IISc

2018MA526-IITBombay-CA-ex01.tex January 23, 2018 ; 2:38 p.m.

1/3

Page 2

MA 526 Commutative Algebra / Jan–April 2018

Exercise Set 1

Supplements / Test Exercises — I n t e g r a l E x t e n s i o n s

T1.1 Let K be a field. (a) Let B = K[x] be a cyclic k-algebra. Then every K-subalgebra A of B is a finite type K-algebra. (Hint : If

f ∈ A, f = ∑m

i=0 aixi, am = 0, m ≥ 1, then B = ∑m−1 i=0 K[f]xi is a finite over K[f] ⊆ A.)

(b) Let B = K[N2] be the monoid algebra over k of the additive monoid N2 and let X := e(1,0), Y := e(0,1). Then B = K[X,Y], and the monomials XiY j = e(i, j), (i, j) ∈ N2, form a K-basis of B. Let A be the K- subalgebra of B generated by the monomials X,X2Y,...,Xn+1Y n,.... Then A is not a noetherian ring, much less than a finite type K-algebra. (Hint : Note that B is the polynomial algebra in two indeterinates X,Y over K and

Xn+1Y n does not belong to the ideal (in A) generated by X,...,XnY n−1, for every n ∈ N.)

T1.2 (a) Every surjective ring endomorphism of a noetherian ring A is an automorphism. (b) Let A be a finite type (commutative) algebra over the ring R. Then every surjective R-algebra endomor- phism ϕ of A is an automorphism. (Hint : Suppose that ϕ(x) = 0 and x1,...,xm ia a R-algebra generating system for

• A. The construct a finitely generated Z-subalgebra R′ of R such that R′[x1,...,xm] contain x as well as ϕ is a surjective

endomorphism R′[x1,...,xm]. — Note that the assertion does not hold for arbitrary commutative algebra. Examples!)

(c) Let V be a module over a ring A and U be a submodule = 0 of V. If V noetherian or if V finite, then V and V/U are not isomorphic as A-modules. (Hint : If they are isomorphic then there is a surjective A-endomorphism of

V with kernel U.)

(d) Let a be a non-zero ideal is a noetherian ring A. Then A and A/a are not isomorphic rings. T1.3 Let A be a normal domain with quotient field Q. If an element x of a Q-algebra L ( not necessarily a field ) is integral over A then the minimal polynomial µx of x over Q has coefficients in A. (Hint : By extending

L, we may assume that µx splits into linear factors over L (see also Exercise 1.E.10 (1) below). Let f(x) = 0, f ∈ A[X], be an integral equation of x over A. Then µx divides f in Q[X] and so, every zero y of µx is integral over A (with integral equation f(y) = 0). Therefore the coefficients of µx are integral over A by 1.E.5 and hence elements of A, since A is normal.)

T1.4 Let K be a field and let A be a normal K-subalgebra of K[T], A = K. (a) Show that A is a polynomial algebra K[ f ] for some f ∈ A. (f is necessarily a non-constant polynomial in A of least degree.) (Hint : Let µT = Xn + fn−1Xn−1 +···+ f0 ∈ Q(A)[X] be the minimal polynomial of T over

Q(A). By the Test-Exercise T1.3, the coefficients f0,..., fn−1 ∈ A. But every non-constant coefficient f of µT generates the field Q(A) over K (see the proof of L ü r o t h ’s t h e o r e m1) Then K[f ] ⊆ A ⊆ Q(A) = K(f ) and K[f ] = A, since K[T] and hence A is integral over K[ f ] and K[f ] is normal.)

(b) The normalization A of A is a polynomial algebra K[f] for some non-constant polynomial f ∈ K[T].

(Hint : Use (a). — Note that every K-subalgebra of K[T] is a K-algebra of finite type. In general a K-algebra A of finite type is called r a t i o n a l if it is an integral domain and if the quotient field Q(A) of A is K-isomorphic to a rational function field K(T1,...,Tm) in m variables T1,...,Tm. The integer m is nothing but the transcendence degree of the field extension K ⊆ Q(A). By Lüroth’s theorem, any K-subalgebra A of K(T), A = K, of finite type is rational with m = 1.)

(c) Consider the K-algebra homomorphism ϕ : K[X,Y] → K[T] defined by x := ϕ(X) = T 2 − 1 and y := ϕ(Y) = T(T 2 −1) and the K-subalgebra A := Imgϕ of K[T]. Then Kerϕ is the principal ideal generated by Y 2 −X2 −X3 and A ∼ = K[X,Y]/

• Y 2 −X2(X +1)
• . Further, the polynomial algebra K[T] is the normalization
• f A. (Hint : If f ∈ K[X,Y] with f = 0 and degY f ≤ 1 then f ∈ kerϕ. Further, T = y/x belongs to the quotient field
• f A.)

T1.5 (a) Let A ⊆ B be a ring extension and let H,G ∈ B[X] be monic polynomials such that HG ∈ A[X]. Then the coefficients of H and G are integral over A. If A is integrally closed in B then H,G ∈ A[X]. (Hint :

There is a finite ring extension C of B (even a free one) such that H and G factor into monic linear factors in C[X].)

(b) Let R be an A-algebra. Then the integral closure of A[T1,...,Tn] in R[T1,...,Tn] is A[T1,...,Tn], where A is the integral closure of A in R. In particular, if A is a normal integral domain then A[T1,...,Tn] too.

(Hint : Assume n = 1. Let g ∈ R[T] be integral over A[T] with integral equation 0 = f(g) = gn + fn−1gn−1 + ··· + f1g+ f0. Let r be an integer larger than n and the degrees of fn−1,..., f0 and let g1 := g−T r. From (g1 +T r)n + fn−1(g1 +T r)n−1 +···+ f1(g1 +T r)+ f0 = 0

• ne gets (−g1)(gn−1

1

+hn−1gn−2

1

+···+h1) = (T r)n +(T r)n−1 fn−1 +···+(T r)f1 + f0 which is a monic polynomial in A[T]. Now use (a) to conclude that g1 and hence g has coefficients in A.)

1 L ü r o t h ’s t h e o r e m Let K be a field and K ⊆ E ⊆ K(X) be an intermediate field. Then E is simple over K, i.e. E = K( f)

for some rational function f ∈ K(X). Moreover, if E = K, then every non-constant coefficient f of the minimal monic polynomial µX of X over E generates the field E over K. For a proof see [van der Waerden, B. L. : Algebra, Part I, §73, p. 222]

• D. P. Patil/IISc

2018MA526-IITBombay-CA-ex01.tex January 23, 2018 ; 2:38 p.m.

2/3

Exercise Set 1

MA 526 Commutative Algebra / Jan–April 2018

Page 3

T1.6 ( G r a d e d i n t e g r a l e x t e n s i o n s ) (a) Let A ⊆ B be an extension of Z-graded rings. Then the integral closure A of A in B is a graded A-subalgebra of B. (Hint : Note that, by the Test-Exercise T1.5 (b),

A[T,T −1] is the integral closure of A[T,T −1] in B[T,T −1]. Now, use the fact that, for any graded ring C =

m∈ZCm,

the map C → C[T,T −1], ∑m cm → ∑m cmT m, is an injective graded ring homomorphism.)

(b) Let A be a Z-graded integral domain. Then the normalization A of A is a graded subalgebra of S−1A where S is the multiplicative system of non-zero homogeneous elements in A. If A is positively graded (i.e. Am = 0 for m < 0) then A is also positively graded.

• D. P. Patil/IISc

2018MA526-IITBombay-CA-ex01.tex January 23, 2018 ; 2:38 p.m.

3/3