Towards a homological generalization of the direct summand theorem

Remark 1

Let R ↪ S be a finite extension of Noetherian rings, where ( R , m ) is local. Then, it is an elementary fact to see that the maximal spectrum of S, Spec m S , is equal to V ( m S ) ⊆ Spec S (see for example [18, Corollary 5.8]).

Lemma 1

Let ( R , m , k ) be a local complete ring and R ↪ S a finite extension. Assume that Spec S = V ( m S ) = { η 1 , … , η n } . Then S is naturally isomorphic, as a ring, to S η 1 × ⋯ × S η n .

Proof

This is a well-known consequence of the fact that a local complete ring is Henselian and one of the equivalence forms of being Henselian is essentially the statement of the lemma. See, for example [19, Lemma 10.152.3].^[1]□

Corollary 2

Let ( R , m , k ) be a complete local ring with algebraically closed residue field k and B = R [ x ] / ( F ( x ) ) , where F ( x ) is a monic polynomial of degree n. Then there exist monic polynomials G i ( x ) of degree n i such that ∑ i = 1 r n i = n , B = ⊕ i = 1 r R [ x ] / ( G i ( x ) ) as rings, and

where all a i j ∈ m .

Proof

Since k is algebraically closed we can factor f ( x ) ≔ F ¯ ( x ) = ∏ i = 1 r ( x − b i ) n i ∈ k [ x ] , for some b i ∈ k and n i ∈ ℕ , with ∑ i = 1 r n i = n . Besides, B / m B ≅ k [ x ] / ( f ( t ) ) is an Artinian ring with maximal ideals η i ¯ = ( x − b i ) for i = 1 , … , r ; therefore, B / m B ≅ ⊕ i = 1 r k [ x ] / ( ( x − b i ) n i ) (see [18, Theorem 8.7 and proof]). Now, by Hensel’s Lemma there exist monic polynomials F i ( x ) ∈ R [ x ] such that F [ x ] = ∏ n = 1 r F i ( x ) and F i ¯ ( x ) = ( x − b i ) n i . By Remark 1, Spec m B = { η 1 , … , η r } = V ( m B ) .

Now, let B i ∈ R such that B i ¯ = b i . Since η i ¯ = ( x − b i ) in B/mB, we see by the correspondence between the ideals of B/mB and the ideals of B containing mB that η i = ( x − B i ) + m B . Now, ( F ( x ) ) R [ x ] η j = ( F i ( x ) ) R [ x ] η i , because ( F j ( x ) ) + η i = R [ x ] for all i ≠ j , since mod m R [ x ] this ideal corresponds to ( ( x − b j ) n j ) + ( x − b i ) = rad ( x − b i , x − b j ) = k [ x ] for all i ≠ j . Therefore, F j ( x ) is a unit in R [ x ] η i . Besides, the ring R [ x ] / ( F i ( x ) ) is local with maximal ideal ( x − B i ) + m e , where m e denotes the expansion of m in R [ x ] (which we denote again by η i ). This is because any maximal ideal should contain the expansion of m (the extension R ↪ R [ x ] / ( F i ( x ) ) is finite) and this ring module m e is the local ring ( k [ x ] / ( ( x − b i ) n i ) , ( x − b i ) e ) . So,

Thus, by the previous lemma, B ≅ ⊕ i = 1 r B η i ≅ ⊕ i = 1 r R [ x ] / ( F i ( x ) ) .

Finally, the isomorphism of rings θ i : R [ x ] → R [ t ] , sending x ↦ t + B i , induces an isomorphism of rings θ i ¯ : R [ x ] / ( F i ( x ) ) → R [ t ] / ( F i ( t + B i ) ) . But the reduction of F i ( x ) mod m R [ x ] is exactly ( ( t + B i ) − B i ) n i = t n i , because translation and reduction mod m commutes. But that means exactly that G i ( t ) = t n i + a i , 1 t n i − 1 + ⋯ + a i , n i ∈ R [ t ] with a i , j ∈ m , for any indices i, j. In conclusion, we get an isomorphism of rings between B and ⊕ i = 1 r R [ x ] / ( G i ( t ) ) satisfying the conditions of our corollary.□

Remark 2

Let i : R ↪ S = S 1 × ⋯ × S n be an extension of rings, where R is an integral domain. Then there exists an S i such that π i ∘ i : R ↪ S i is also an extension, where π i is the natural projection. Suppose by contradiction that for any i there exist a i ≠ 0 ∈ R such that π i ( i ( a i ) ) = 0 . Then, if a = ∏ i = 1 n a i ≠ 0 , for any j,

Therefore, i ( a ) = ( π 1 ( i ( a ) ) , … , π n ( i ( a ) ) ) = 0 , contradicting the hypothesis that i is an extension.

Theorem 3

Let ( R , m , k ) be a regular complete local ring with algebraically closed residue field k. Then, to prove the DSC for R it is enough to consider finite extensions R ↪ S , where S = T / J ,

f i ( y i ) = y i n i + a i , 1 y 1 n i − 1 + ⋯ + a i , n i with a i , j ∈ m and J ⊆ T is an ideal of height zero.

Proof

Let us fix a finite extension R ↪ S . By the discussion above, we know that S = T / J , where T = R [ y 1 , … , y r ] / ( f 1 ( y 1 ) , … , f r ( y r ) ) (the coefficients of the monic polynomials f i ( y i ) are not necessarily in m), and ht ( J ) = 0 . It is elementary to see that

Write B i = R [ y i ] / ( f i ( y i ) ) , then by Corollary 2 B i ≅ ⊕ α = 1 m i R [ y i ] / ( f i α ( y i ) ) with f i α ( y i ) = y i n i α + a i α 1 y i n i α − 1 + ⋯ + a i α n i α and a i α j ∈ m . Furthermore, by the distributive law between tensor products and direct sums (see [20]) we get

where w = ( α 1 , … , α r ) , 1 ≤ α i ≤ m i , and

where each f i α i ( y i ) has lower coefficients in m, as desired. Besides, J = ⊕ w J w and then S ≅ ⊕ w T w / J w .

Finally, by Remark 2, there is an α such that R ↪ T w / J w is an extension. But if there is a retraction ρ w : T w / J w ↪ R , then ρ = π w ∘ ρ w : S → R is also a retraction. Besides ht ( J w ) = 0 , because dim T w = dim R = dim T w / J w . In conclusion, S w = T w / J w has the desired form of our proposition, and then it is enough to prove the DSC in this case.□

Proposition 4

Let ( R , m , k ) be a regular local ring of dimension d, and

f i ( y i ) = y i n i + a i , 1 y 1 n i − 1 + ⋯ + a i , n i , with a i , j ∈ m . Then T is a Gorenstein local ring with maximal ideal η = m + ( y 1 ¯ , … , y r ¯ ) .

Proof

First, we see that T is a local Cohen-Macaulay (C-M) ring. In fact, let m 1 be any maximal ideal of T. Then m 1 ∩ R = m , because dim ( R / ( m 1 ∩ R ) ) = dim ( T / m 1 ) = 0 and m 1 ∩ R ∈ Spec ( R ) . Therefore, R / ( m 1 ∩ R ) is a field. Thus, y i n i ∈ m 1 and therefore y i ∈ m 1 . In conclusion, m 1 = η . Now, R [ y 1 , … , y r ] is a C-M ring (see [21, Proposition 18.9]). Thus, B = R [ y 1 , … , y r ] η is also C-M. Let m = ( x 1 , … , x d ) , where d = dim R . Then { f 1 ( y 1 ) , … , f r ( y r ) , x 1 , … , x d } ⊆ B is a system of parameters because dim B = ht ( η ) = dim R [ y 1 , … , y r ] = dim R + r = d + r , and since R [ y 1 , … , y r ] is C-M, then by previous comments is equidimensional, and rad ( f 1 , … , f r , x 1 , … , x d ) = η B . Then { f 1 , … , f r , x 1 , … , x d } is a regular sequence in B. Thus, B / ( f 1 , … , f r ) ≅ T is C-M, due to the fact that dim B / ( f 1 , … , f r ) = d and { x 1 ¯ , … , x d ¯ } ⊆ B / ( f 1 , … , f r ) is a regular sequence.

Finally, let Q = ( y 1 , … , y r , x 1 , … , x d ) be the ideal generated by the system of parameters { y 1 , … , y r , x 1 , … , x d } . Then T / Q ≅ k [ y 1 , … , y r ] / ( y 1 n 1 , … , y r n r ) = k [ w 1 , … , w r ] , where w i = y i ¯ . Let us see that

In fact, if h ¯ ∈ Ann T / Q ( ( w 1 , … , w r ) ) and c ∏ i = 1 r w i m i ≠ 0 is a monomial of h ¯ such that there exists m j ∈ ℕ with m j < n j − 1 , then w j h ¯ ≠ 0 ∈ T / Q , because the monomial term w j c ∏ i = 1 r w i m i has the power m j + 1 < n j on w j , which is a contradiction, by the reason that h is on the socle. Therefore, m i ≥ n i − 1 for all i and so h ¯ ∈ ∏ i = 1 r w i m i . The other contention is clear, and then the socle has dimension one. In conclusion, ( T , η ) is a Gorenstein local ring.

Another approach for proving this proposition can be summarized as follows: since f i ( y i ) involve different variables (for each i = 1 , … , r ), one can use this fact explicitly to see that { f 1 ( y 1 ) , … , f r ( y r ) } forms a regular sequence. Now, due to the fact that R is Gorenstein, we see that R [ y 1 , … , y r ] so is. Moreover, one can also prove that when one mods out a Gorenstein ring by a regular sequence, the corresponding quotient ring remains Gorenstein. So, this proves that T = R [ y 1 , … , y n ] / ( f 1 ( y 1 ) , … , f r ( y r ) is Gorenstein. Finally, T is local, due to the fact that T/mT so is (because it is isomorphic to k [ y 1 , … , y n ] / ( y 1 n 1 , … , y r n r ) , with maximal ideal ( y 1 , … , y r ) ) and due to the fact that all the maximal ideals of T contract to m by module-finiteness.□

Proposition 5

Let ( A , η , k ) be a local Gorenstein ring of dimension zero (i.e. dim k ( Ann A η ) = 1 ). Let u ∈ η such that ( u ) = Ann A η . Then u ∈ I for any ideal I ≠ ( 0 ) ⊆ A .

Proof

Clearly, we can assume that I ⊆ η . We know nil ( A ) = η ; therefore, there exists n ∈ ℕ such that η n = 0 . Let x ≠ 0 ∈ I . Then η n − 1 x ⊆ η n = ( 0 ) . Let r ∈ ℕ be such that η r x = ( 0 ) but η r − 1 x ≠ ( 0 ) . Hence, η ( η r − 1 x ) = ( 0 ) and so η r − 1 x ⊆ Ann A η = ( u ) . Then ( u ) contains a nonzero element of the form bx, where b ∈ η r − 1 . This means that there exists c ∈ A \ η with c u = x b , so u = ( c − 1 b ) x ∈ I .□

Remark 3

If h : R ↪ T is any homomorphism of rings, we can consider T ∗ = Hom R ( T , R ) as a T -module with the following action: fix t ∈ T and define ( t ⋅ ϕ ) ( x ) ≔ ϕ ( t x ) , for ϕ ∈ Hom R ( T , R ) and x ∈ T .

Remark 4

Let ( R , m ) be a local ring, T a finitely generated R-free module, and θ : T → T an R-homomorphism. Then θ is an isomorphism of R-modules if and only if θ ¯ : T / m T → T / m T is an isomorphism of K-vector spaces. In fact, if A ∈ M n × n ( R ) is the matrix defining θ , then θ is an isomorphism if and only if det A is an unit, which means that det A ∉ m . But that is equivalent to saying that det A ¯ ≠ 0 ∈ k , where A ¯ is the reduction of A mod m. Finally, since A ¯ is the matrix defining θ ¯ , the last condition is equivalent to saying that θ ¯ is an isomorphism of k - vector spaces.

Theorem 6

Let ( R , m ) and ( T , η ) be local rings. Assume that T/mT is Gorenstein. Let h : R → T be a finite homomorphism of local rings, such that T is R - free. Then there exists a T - isomorphism β : T → T ⁎ such that for any ideal J ⊆ T , β − 1 ( ( T / J ) ⁎ ) = Ann T J .

Proof

We identify ( T / J ) ⁎ with { f ∈ T ⁎ : f ( J ) = 0 } . We know that dim T / m T = dim R / m = 0 , since T/mT is a finitely generated R / m - module. So, fix u 1 ∈ η such that ( u 1 ¯ ) = Ann T / m η ¯ , since T/mT is Gorenstein. Let { u 2 , … , u d } ⊆ T such that T / m T = ( u 1 ¯ , … , u d ¯ ) (where T ≅ R d ). Then, by the Lemma of Nakayama T = ( u 1 , … , u d ) , T is generated by u 1 , … , u d as an R -free module. In fact, let { w 1 , … , w d } ⊆ T be an R - basis for T. Define θ : T → T by w i → u i , then the induced θ ¯ : T / m T → T / m T is clearly an isomorphism of k - vector spaces. Since T is R - free, by Remark 4, θ is an isomorphism which means just that { u 1 , … , u d } ⊆ T is an R - basis for T. Let u 1 ⁎ ∈ T ⁎ be the dual element and define β : T → T ⁎ by t → t ⋅ u 1 ⁎ , where ( t ⋅ u 1 ⁎ ) ( t 1 ) ≔ u 1 ⁎ ( t t 1 ) , for all t 1 ∈ T . By definition, it is clear that β is a T - homomorphism. Now, we can make the natural identifications T ⁎ = Hom R ( R d , R ) ≅ ( Hom R ( R , R ) ) d = R d . Therefore, T ⁎ is an R - free module of dimension d and β is an R - isomorphism if and only if β ¯ : T / m T → T ⁎ / m T ⁎ is so, due to Remark 4 ( T ≅ T ⁎ as R - free modules). But β ¯ is an isomorphism of k - vector spaces if it is injective. Suppose by contradiction that ker β ¯ ≠ ( 0 ) . Then, by Proposition 5, u 1 ¯ ∈ ker β ¯ . This means that β ( u 1 ) ∈ m T ⁎ , so there exist m 1 , … , m d ∈ m such that β ( u 1 ) = u 1 u 1 ⁎ = ∑ i = 1 d m i u i ⁎ . But, evaluating at 1 we get: 1 = u 1 u 1 ⁎ ( 1 ) = ∑ i = 1 d m i u i ⁎ ( 1 ) ∈ m , a contradiction. In conclusion, β is a T - isomorphism.

For the last part, let a ∈ Ann T J . Then, for any j ∈ J , β ( a ) ( j ) ≔ ( a u 1 ⁎ ) ( j ) = u 1 ⁎ ( a j ) = u 1 ⁎ ( 0 ) = 0 . Therefore, β ( a ) ∈ ( T / J ) ⁎ and thus a = β − 1 ( β ( a ) ) ∈ β − 1 ( ( T / J ) ⁎ ) .

Conversely, take ϕ 0 ∈ β − 1 ( ( T / J ) ⁎ ) . Then there exists a ϕ ∈ ( T / J ) ⁎ such that ϕ 0 = β − 1 ( ϕ ) , i.e. ϕ = ϕ 0 u 1 ⁎ and ϕ ( J ) = 0 , which means that u 1 ⁎ ( ϕ 0 j ) = 0 for all j ∈ J . Now, let us fix j 0 ∈ J . Then, for all t ∈ T , β ( ϕ 0 j 0 ) ( t ) = u 1 ⁎ ( ϕ 0 j 0 t ) = u 1 ⁎ ( ϕ 0 ( j 0 t ) ) = 0 , because j 0 t ∈ J . Therefore, β ( ϕ 0 j 0 ) ≡ 0 and then by the first part ϕ 0 j 0 = 0 , which means that ϕ 0 ∈ Ann T J .□

Theorem 7

Let h : R → T / J be a finite homomorphism of local rings such that ( T , η ) is a local R-free ring, with T/mT Gorenstein, and J ⊆ T an ideal. Assume that the structure of R-module of T/J inherited by the R-structure of T is the same as the one induced by h. Then R ↪ T / J splits if and only if Ann T J ⊈ m T , where m is the maximal ideal of m.

Proof

Assume that ρ : T / J ↪ R is a splitting R-homomorphism. Then ρ ∈ ( T / J ) ⁎ and ρ ( 1 ) = 1 . By the last theorem, there exists ρ 0 ∈ Ann T J such that ρ 0 = β − 1 ( ρ ) , which means, in particular, that 1 = ρ ( 1 ) = ρ 0 ⋅ u 1 ⁎ ( 1 ) = u i ⁎ ( ρ 0 ) ∉ m . Then ρ 0 ∉ m T , so Ann T J ⊈ m T .

Conversely, take a ∈ Ann T J \ m T . Then, by Proposition 5, u 1 ¯ ∈ ( a ¯ ) ⊆ T / m T , which means that there exists m i ∈ m , with u 1 − a = ∑ i = 1 d m i u i , where { u 1 , … , u d } ⊆ T is an R - basis for T as in the last Theorem. Then,

So ρ = β ( ( 1 − m i ) − 1 a ) satisfies that ρ ∈ ( T / J ) ⁎ because β − 1 ( ρ ) = ( 1 − m 1 ) − 1 a ∈ Ann T J , and by Theorem 6 β − 1 ( ( T / J ) ⁎ ) = Ann T J . Besides, ρ 1 = ( 1 − m i ) − 1 a u 1 ⁎ ( 1 ) = ( 1 − m 1 ) − 1 ( 1 − m 1 ) = 1 , which implies that ρ : T / J → R is the desired splitting R - homomorphism.□

Proposition 8

Let ( R 0 , m 0 , k 0 ) be a regular local ring and R 0 ↪ T 0 be a finite extension, where

and each f i ( y i ) = y i n i + a i 1 y i n i − 1 + ⋯ + a i n i , with a i j ∈ m , for all indices i, j. Let S = T 0 / J , with J = ( g 1 , . … , g s ) ⊆ T 0 such that J ∩ R 0 = ( 0 ) . Let x 1 , … , x s be new variables, and let R be R 0 [ x 1 , … , x s ] , and let m be m 0 + ( x 1 , … , x s ) , a maximal ideal of R. Write

Let g be the element x 1 g 1 + ⋯ + x s g s . Then:

1. ( R m , m R m , k 0 ) is a regular local ring.

2. ( g ) T m ∩ R m = ( 0 ) .

3. R m ↪ ( T / ( g ) ) m ≅ T m / ( g ) e is a finite extension, and g ∈ ( m R m ) T m .

4. R m ↪ T m / ( g ) splits if and only if R 0 ↪ T 0 / J splits.

Proof

1. In general, if R is regular, then so is the polynomial ring R [ T ] (see [20]). In particular, R m is a regular local ring and R m / m R m ≅ R / m ≅ R 0 / m 0 = k 0 .

2. R 0 ↪ R is an R - free extension, then, in particular, it is flat. Therefore, by tensoring R 0 ↪ T 0 / J with R, we see that R ↪ R ⊗ R 0 T 0 / J ≅ T / J e is also an extension, and since localization is flat too, we get an extension R m ↪ T m / J e . Because of g ∈ J e , we get g T m ∩ R m = 0 .

3. Clearly, by definition g ∈ ( m R m ) T m . Now, by the previous paragraph R m ↪ T m / ( g ) e is an extension, and it is finite because R 0 ↪ T 0 is finite, in fact free. Then, after tensoring with R m we get a module finite extension R m ↪ T m , so T m / ( g ) e is also a finitely generated R m - module.

4. Assume that ρ 0 : T 0 / J → R 0 is an R 0 - homomorphism such that ρ 0 ( 1 ) = 1 . Then, by tensoring with the flat R 0 - module R m , we get an R m - homomorphism ρ : T m / J ↪ R m , with ρ ( 1 ) = 1 . Now, composing ρ with the natural map T m / ( g ) → T m / J , we obtain a retraction from T m / ( g ) to R m .□