# Krull Dimension and Going-Down in Fixed Rings

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1 David Dobbs Jay Shapiro April 19, 2006

2 Basics R will always be a commutative ring and G a group of (ring) automorphisms of R. We let R G denote the fixed ring, that is, Thus R G is a subring of R R G = {a R : g(a) = a for all g G}.

3 Basics R will always be a commutative ring and G a group of (ring) automorphisms of R. We let R G denote the fixed ring, that is, Thus R G is a subring of R R G = {a R : g(a) = a for all g G}. Example Let R = K[x, y]; let G = g where g : R R via g(x) = x, g(y) = y and g fixes the elements of K. Then R G = K[x 2, xy, y 2 ]

4 Types of Questions 1. What properties of R are inherited by the ring R G? 2. What is the relation between R and the subring R G?

5 Types of Questions 1. What properties of R are inherited by the ring R G? 2. What is the relation between R and the subring R G? An important example of a type 1 question.

6 Types of Questions 1. What properties of R are inherited by the ring R G? 2. What is the relation between R and the subring R G? An important example of a type 1 question. Hilbert s XIV th Problem Let K be a field, x 1, x 2,..., x n algebraically independent elements over K, and G a subgroup of GL(n, K). Is the fixed ring K[x 1, x 2,..., x n ] G (or ring of invariants) finitely generated over K?

7 Solution There were some partial positive answers by Zariski and Noether. However in 1958 Nagata showed that the answer in general was no.

8 Solution There were some partial positive answers by Zariski and Noether. However in 1958 Nagata showed that the answer in general was no. A simple example of a type two question and answer:

9 Solution There were some partial positive answers by Zariski and Noether. However in 1958 Nagata showed that the answer in general was no. A simple example of a type two question and answer: Definition The group G is said to be locally finite on R if for each a R, the orbit of a under the action of G is finite, i.e., for each a R, the set {g(a) : g G} is finite.

10 Proposition If G is locally finite, then R G R is an integral extension. That is, every element of R satisfies a monic polynomial over R G.

11 Proposition If G is locally finite, then R G R is an integral extension. That is, every element of R satisfies a monic polynomial over R G. Proof Let a R, then a satisfies the polynomial f (x) = Π b O(a) (x b), where O(a) denotes the orbit of a under the action of G. The coefficients are in R G.

12 Type 1 Question: Does R Noetherian imply R G Noetherian under the assumption that G is finite?

13 Type 1 Question: Does R Noetherian imply R G Noetherian under the assumption that G is finite? Answers 1 Nagarajan (1968) constructed an example of a Noetherian ring R (R = F [[x, y]]) of characteristic 2, and a group G of order 2 acting on R such that R G was not Noetherian. 2 Bergman (1971) showed that if the order of G was a unit of R, then R Noetherian implies that R G is Noetherian.

14 A Question We Considered When does R Artinian imply that R G Artinian?

15 A Question We Considered When does R Artinian imply that R G Artinian? Definitions 1 Recall that R is Artinian if R has the descending chain condition on ideals. 2 The Krull dimension of R (which we denote dim(r)) is the length of the longest chain of prime ideals.

16 A Question We Considered When does R Artinian imply that R G Artinian? Definitions 1 Recall that R is Artinian if R has the descending chain condition on ideals. 2 The Krull dimension of R (which we denote dim(r)) is the length of the longest chain of prime ideals. Basic Facts 1 Artinian Noetherian & dim(r) = 0. Also note that R Artinian implies that R has only finitely many maximal ideals. 2 If G is locally finite, then R is integral over R G. This in turn implies that dim(r G ) = dim(r).

17 Transfer of Krull Dimension 1 Without any assumptions on G we have examples such that dim(r) dim(r G ) is any integer we want (positive or negative). We can even have dim(r) = and dim(r G ) = 0 2 On the other hand we have no examples where the dimensions differ if dim(r) = 0.

18 Transfer of Krull Dimension 1 Without any assumptions on G we have examples such that dim(r) dim(r G ) is any integer we want (positive or negative). We can even have dim(r) = and dim(r G ) = 0 2 On the other hand we have no examples where the dimensions differ if dim(r) = 0. However with an assumption on R we have the following (Note: No assumptions on G.):

19 Transfer of Krull Dimension 1 Without any assumptions on G we have examples such that dim(r) dim(r G ) is any integer we want (positive or negative). We can even have dim(r) = and dim(r G ) = 0 2 On the other hand we have no examples where the dimensions differ if dim(r) = 0. However with an assumption on R we have the following (Note: No assumptions on G.): Theorem If dim(r) = 0 (so all prime ideals are maximal) and R has n maximal ideals, then R G has at most n maximal ideals and dim(r G ) = 0.

20 Corollary If R is Artinian, then dim(r G ) = 0 and R G has finitely many maximal ideals.

21 Corollary If R is Artinian, then dim(r G ) = 0 and R G has finitely many maximal ideals.

22 Corollary If R is Artinian, then dim(r G ) = 0 and R G has finitely many maximal ideals. We saw how to modify Nagarajan (R Noetherian, G finite, yet R G not Noetherian) to provide an example of R Artinian and G finite, yet R G is not Artinian.

23 The Construction of Nagarajan Let F := Z 2 (a i, b i, i 1), where the a i, b i are commuting algebraically independent indeterminates over the field Z 2 with two elements. (Note F is the field of quotients of the integral domain Z 2 [a i, b i ].) Consider the formal power series ring S := F [[X, Y ]]. Then S has an automorphism g given g(x ) := X, g(y ) := Y and g(a i ) := a i + p i+1 Y, g(b i ) := b i + p i+1 X, where p i := a i X + b i Y. Let G := g. Since g 2 is the identity map, G = 2. It is well known that S is a Noetherian ring. However, Nagarajan has shown that R G is not a Noetherian ring.

24 Our Variation Consider the ideal J := (X 2, Y 2 ) of S. Since X 2 and Y 2 are each fixed by g, it is easy to see that J is G-invariant, and so G acts (via ring automorphisms) on R := S/J. Of course, R inherits the property of being a Noetherian ring from S. Moreover, it is easy to check that R is zero-dimensional and local. Thus R is an Artinian ring, yet we can show that S G is not Noetherian, hence not Artinian. The proof does not involve describing all the elements of R G explicitly (that seems too difficult). Rather, one shows that a certain family of elements are in R G, from which we are able to create a strictly ascending chain of ideals.

25 Basic Ideas of Proof Let x and y denote the canonical images of X and Y, respectively, in R. A degree argument shows that the set {1, x, y, xy} is a basis of the vector space R over the field F. Also, note that when a i is viewed as an element of R, then g(a i ) = a i + a i+1 xy (since y 2 = 0 R). Thus g(a i y) = g(a i )g(y) = (a i + a i+1 xy)y = a i y, and so a i y R G. We show that the sequence of ideals of R G given in is strictly ascending. (a 1 y) (a 1 y, a 2 y) (a 1 y, a 2 y, a 3 y)...

26 A Question of Type 2

27 A Question of Type 2 Recall If G is locally finite on R, then R is integral over R G. Note since R G R, there is a map (called the contraction map) Spec(R) Spec(R G ), given by P P R G.

28 Integrality Integrality has a number of consequences for this map. For example: 1. The map is onto. 2. If P Q are elements of Spec(R), then P R G Q R G. 3. Going-up (GU). If p q are elements of Spec(R G ) and P Spec(R) such that P R G = p, then there exists Q Spec(R) such that P Q and Q R G = q. In other words the diagram can be filled in: P? p q

29 Definition of a Related Property A ring injection S R satisfies going-down (GD) if the following diagram can be filled in:? Q p q Integral extensions do not have to satisfy GD. Nonetheless we were able to show a stronger property for fixed rings. First a definition. Definition of a Stronger Version of GD An inclusion map of rings S R is said to be universally going-down, if for any commutative R-algebra T, the canonical map T T S R satisfies going-down. S R satisfies universally going-down if and only if the canonical map S[x 1, x 2,..., x n ] R[x 1, x 2,..., x n ] satisfies going-down for each n.

30 Note that if G acts on R, then this action naturally extends to R[x 1, x 2,..., x n ]. Theorem If G is locally finite on R, then R G R is universally going-down. We did not do this at one time. Steps in Proof We first proved this when G is locally finite and R Noetherian (what we really needed was that there are only finitely many primes minimal over an arbitrary ideal).

31 Continuation Then we realized, using a criteria of Kaplansky, that the obstruction to going-down was a finite data set. Basically, if the inclusion R G into R does not satisfy universally going-down, then there exists finitely many elements a 1, a 2,..., a n R that screw things up. Take these elements and their orbits (which still leave you with a finite set) and take the ring generated by Z and these finite number of elements. This ring, call it T, is Noetherian, and G still acts on this ring. By construction T G T does not satisfies universally going-down. But by earlier result it does - a contradiction. For a while we did not have an example to show that if we dropped the locally finite assumption, then R G R does not satisfy GD. But finally we did come up with an example using a semigroup ring.

32 Outline of the Construction We construct an abelian semigroup S (under +) via generators and relations. We define an automorphism group G acting on S. We let R = K[S] = K[x s : s S], where K is a field. The action of G extends in a natural fashion to R. We show that R G = K[S G ] We construct the appropriate diagram and show that it cannot be completed.

33 The Semigroup Let M denote the free abelian monoid on the symbols {A, C i i Z}, written additively. We define a congruence relation on this monoid, as follows. Let na + C i1 + C i2 + + C ik and ma + C j1 + + C jt be arbitrary elements of M, where t is a nonnegative integer, the C k are not necessarily distinct, and n and m are nonnegative integers. Then we declare that na + C i1 + C i2 + + C ik ma + C j1 + + C jt if either na + C i1 + C i2 + + C ik = ma + C j1 + + C jt or [n = m 0 and k = t]. This is a congruence relation on M (that is, an equivalence relation on M that is compatible with the operation of addition on M). Let S denote the factor semigroup M/.

34 The Automorphism Group of S First we define G on M. Let g : M M be given by g(a) = A and g(c i ) = C i+1 and then extending linearly. It is clear that if x, y M satisfy x y, then g(x) g(y). Hence, g induces an automorphism of S, also denoted g. Then g has infinite order and G = g acts on S. The Example With R = K[S], we have that K[S G ] = R G R does not satisfy going-down (much less universally going-down). The End THE END

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