aa r X i v : . [ m a t h . A C ] M a r SOME REMARKS ON WATANABE’S BOLD CONJECTURE
CHRIS MCDANIELA bstract . At the 2015 Workshop on Lefschetz Properties of ArtinianAlgebras, Junzo Watanabe conjectured that every graded Artinian com-plete intersection algebra with the standard grading can be embeddedinto another such algebra cut out by quadratic generators. We verify thisconjecture in the case where the defining polynomials split into linearfactors.
1. I ntroduction
Let F be any field, and let R = F [ x , . . . , x n ] be the polynomial ring in n variables, endowed with the standard grading, i.e. deg( x i ) = i . Aregular sequence is a sequence of homogeneous polynomials f , . . . , f n ∈ R such that f , ≤ i ≤ n , f i is a non-zero divisor in thequotient R / h f , . . . , f i − i . The quotient ring R / h f , . . . , f n i is known as agraded Artinian complete intersection algebra with the standard grading,but in this paper we shall just use the term complete intersection . A com-plete intersection in which the generators f , . . . , f n are all quadratic formsis called a quadratic complete intersection . Recall that a complete inter-section is always Gorenstein, hence its socle is a one dimensional gradedsubspace. A graded F algebra homomorphism between two complete inter-section algebras is called an embedding if it maps a socle generator onto asocle generator. Note that a complete intersection can embed into anothercomplete intersection only if the two have the same socle degree.In their recent paper [2], Harima, Wachi, and Watanabe have shown thatin a “generic” quadratic complete intersection equipped with an appropri-ate action of a symmetric group, that the invariant subring under a Youngsubgroup is again a complete intersection algebra with the standard grad-ing. At the 2015 Workshop on Lefschetz Properties of Artinian Algebrasat University of G¨ottingen, Junzo Watanabe asked to what extent completeintersections arise in this way. Specifically he posed the following “ratherbold conjecture” which we shall heretofore refer to as Watanabe’s bold con-jecture: Watanabe’s Bold Conjecture.
Every complete intersection algebra em-beds into some quadratic complete intersection algebra.
In this paper we prove that this conjecture holds for complete intersec-tions cut out by polynomials that split into linear factors. To wit
Theorem 1.1.
Suppose that f , . . . , f n form a regular sequence in R, andlet I = h f , . . . , f n i be the ideal they generate. Further suppose that foreach ≤ i ≤ n the polynomial f i splits into a product of linear factors,i.e. f i = ℓ i , , . . . , ℓ i , N i for some linear forms ℓ i , j ∈ V ∗ . Then the completeintersection R / I satisfies Watanabe’s bold conjecture.
To prove Theorem 1.1 we first show that a regular sequence as in The-orem 1.1 has a “normal form”. Then we find a quadratic complete inter-section into which our given complete intersection embeds, in its normalform. 2. P reliminaries
Regular Sequences.
Let R = F [ x , . . . , x n ] be the polynomial ring in n variables with the standard grading. Definition 2.1.
A sequence of homogeneous polynomials of positive degreef , . . . , f k ∈ R is called a partial regular sequence if f , and for each ≤ i ≤ k, f i is not a zero divisor in the quotient ring R / h f , . . . , f i − i . A regular sequence is a partial regular sequence of length n. Remark 2.1.
What we refer to as a partial regular sequence here is usuallyreferred to as just a regular sequence. We use the adjective “partial” heresince we want all of our regular sequences to have length n.
Lemma 2.1.
If f , . . . , f k is a partial regular sequence, then f σ (1) , . . . , f σ ( k ) is also a partial regular sequence for any permutation σ ∈ S k .Proof. See Matsumura [3][Corollary pg. 127]. (cid:3)
Lemma 2.2.
A sequence of homogeneous polynomials of positive degreef , . . . , f n is a regular sequence if and only if the quotient R / h f , . . . , f n i isa finite dimensional vector space over F .Proof. This follows from the graded analogues of Theorems 14.1 and 17.4in Matsumura [3]. (cid:3)
Lemma 2.3.
Suppose that f , . . . , f k is a partial regular sequence, and thatf k = g · h for some homogeneous positive degree polynomials g and h. Thenf , . . . , f k − , g is also a partial regular sequence.Proof. With f , . . . , f k , g , and h as above, it su ffi ces to show that g is nota zero divisor in the quotient ring R / h f , . . . , f k − i . But of course if g can-not be a zero divisor, since otherwise f k would be one, contradicting theregularity assumption on f , . . . , f k . (cid:3) OLD CONJECTURE 3
Corollary 2.1.
If f , . . . , f k is a partial regular sequence, and f i splits intoa product of linear factors, i.e. f i = ℓ i , · · · ℓ i , N i for each ≤ i ≤ k, then ℓ , j , . . . , ℓ k , j k is also a partial regular sequence for each ≤ j i ≤ N i ≤ i ≤ k.Proof. Start with f k = ℓ k , · · · ℓ k , N k . By Lemma 2.3, we know that f , . . . , f k − , ℓ k , j k is a partial regular sequence for each 1 ≤ j k ≤ N k . By Lemma 2.1, we knowthat ℓ k , j k , f , . . . , f k − is also a partial regular sequence for each 1 ≤ j k ≤ N k .Now repeat. (cid:3) The following proposition gives us a normal form for a split regular se-quence as described in the statement of Theorem 1.1.
Proposition 2.1.
Let f , . . . , f n be a regular sequence such that each f i splitsas a product of linear forms, and let I be the ideal generated by them. Thenafter a linear change of coordinates, we can writef i = x i N i − Y j = ( x i − X j , i λ i , j x j ) , for some λ i , j ∈ F . Proof.
For each 1 ≤ i ≤ n write f i = ℓ i , · · · ℓ i , N i for 1 ≤ i ≤ n . By Corollary2.1, the linear forms ℓ , N , . . . , ℓ n , N n are a regular seqence hence they mustbe linearly independent, hence ℓ i x i is a linear change of coordinates.Now we can rewrite f i = x i · Q N i − j = ˆ ℓ i , j , where ˆ ℓ i , j = P nk = a ki , j x k for somescalars a ki , j . We would like to know that the coe ffi cient a ii , j , ≤ j ≤ N i −
1. Again by Corollary 2.1 the sequence x , . . . , ℓ i , j , . . . , x n must be regular for each 0 ≤ j ≤ N i −
1. On the other hand, if a ii , j = ℓ i , j would be zero in the quotient R / h x , . . . , ˆ x i , . . . , x n i (wherethe “hat” indicates omission), contradicting regularity. Therefore a ii , j mustbe non-zero. Finally we can define another change of coordinates x i Q Ni − j = a ii , j x i to get our desired form. (cid:3) We also will make use of the following lemma, lifted from a recent paperof Abedelfatah [1]
Lemma 2.4.
Suppose that f = x L , . . . , f n = x n L n is a regular sequencein F [ x , . . . , x n ] , where L i is a linear form for ≤ i ≤ n. Then the quotient F [ x , . . . , x n ] / h f , . . . , f n i is spanned (as an F vector space) by the equivalence classes of square freemonomials from F [ x , . . . , x n ] .Proof. See Lemma 3.1 and Lemma 3.2 in the aforementioned paper [1]. (cid:3)
CHRIS MCDANIEL
Corollary 2.2.
With notation as in Lemma 2.4, a socle generator for thequotient F [ x , . . . , x n ] / h f , . . . , f n i is L · · · L n .Proof. By Corollary 2.1, the linear forms L , . . . , L n must form a regularsequence, hence must be linearly independent over F . Hence the poly-nomial rings F [ x , . . . , x n ] and F [ L , . . . , L n ] are identical. Now applyingLemma 2.4 to the regular sequence f = x · L , . . . , f n = x n · L n , regarding L , . . . , L n as the coordinate functions, we see that F [ L , . . . , L n ] / h f , . . . , f n i is spanned by square free monomials in the L , . . . , L n . In particular, the so-cle is therefore generated by the unique square free monomial in degree n ,namely L · · · L n . (cid:3) Monomial Orderings.
We can endow the set of monomials in thepolynomial ring F [ x , . . . , x n ] with a total ordering by declaring that x a · · · x a n n < x b · · · x b n n if a + · · · + a n < b + · · · + b n or if a + · · · + a n = b + · · · + b n and a n = b n , . . . , a n − j + = b n − j + and a n − j < b n − j for some 1 ≤ j ≤ n . Note thatwith this order, we have x < · · · < x n . This is called the graded lexicographic or grlex monomial ordering. Proposition 2.2.
The grlex ordering has the following properties.(i) For any monomials m , m , m ∈ F [ x , . . . , x n ] we havem ≤ m ⇔ m · m ≤ m · m (ii) For any fixed monomial m ∈ F [ x , . . . , x n ] there are only finitelymany monomials less than m.Proof. The proof of (1) is obvious. To see (2), fix a monomial m ∈ F [ x , . . . , x n ]and suppose that m ′ ≤ m is any lesser monomial. Then by definition,deg( m ′ ) ≤ deg( m ), hence m ′ belongs to the finite set n monomials in F [ x , . . . , x n ]of degree ≤ deg( m ) o . (cid:3) We shall use monomial orderings to argue that certain quotients of poly-nomial rings are Artinian.
OLD CONJECTURE 5
3. T he M ain R esult Fix a field F , fix integers N , . . . , N n , and suppose f , . . . , f n is a regularsequence in F [ x , . . . , x n ] of degrees N , . . . , N n , respectively, and let I = h f , . . . , f n i be the ideal they generate. We will further assume that each f i splits as a product of linear forms, so that we may write the quotient A ≔ F [ x , . . . , x n ] / I in its normal form A = F [ x , . . . , x n ] ,* x i N i − Y k = x i − X j , i λ ki , j x j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ i ≤ n + for some λ ki , j ∈ F for 1 ≤ i ≤ n , 1 ≤ j ≤ N i and 0 ≤ k ≤ N i −
1, byProposition 2.1 above.Now consider the polynomial ring F (cid:2) Z i , k | ≤ i ≤ n , ≤ k ≤ N i (cid:3) . Foreach 1 ≤ i ≤ n define the linear form L i , N i ≔ Z i , + · · · + Z i , N i − − Z i , N i + X j , i λ i , j Z j , N j and for 1 ≤ k ≤ N i − L i , k ≔ Z i , + · · · + Z i , k − X j , i ( λ N i − ki , j − λ i , j ) Z j , N j . Define the quotient ringˆ A ≔ F h Z i , j | ≤ i ≤ n , ≤ j ≤ N i i . h Z i , j · L i , j | ≤ i ≤ n , ≤ j ≤ N i i Proposition 3.1.
There is a well defined F algebra map φ : A → ˆ A definedby φ ( x i ) = Z i , N i , ≤ i ≤ n.Proof. We need to show that for each 1 ≤ i ≤ n we have the followingrelation in ˆ A : Z i , N i N i − Y k = Z i , N i − X j , i λ ki , j Z j , N j ≡ . Note that in ˆ A we have the relations(3.1) Z i , N i · Z i , N i − X j , i λ i , j Z j , N j ≡ Z i , N i · ( Z i , + · · · + Z i , N i − )and for 1 ≤ k ≤ N i − Z i , N i · Z i , N i − X j , i λ ki , j Z j , N j ≡ Z i , N i · (cid:0) L i , N i − k + Z i , N i − k + + · · · + Z i , N i − (cid:1) . CHRIS MCDANIEL
Combining Equivalences (3.1) and (3.2) we see that(3.3) Z i , N i · N i − Y k = Z i , N i − X j , i λ ki , j Z j , N j ≡ Z i , N i (cid:0) Z i , + · · · + Z i , N i − (cid:1) N i − Y k = (cid:0) L i , N i − k + Z i , N i − k + + · · · + Z i , N i − (cid:1) . Note that the product on the RHS of Equivalence (3.3) collapses, i.e.(3.4) N i − Y k = (cid:0) L i , N i − k + Z i , N i − k + + · · · + Z i , N i − (cid:1) ≡ N i − Y k = L i , N i − k since L i , N i − is killed by Z N i − and inductively L i , N i − · · · L i , N i − k is killed by Z i , N i − k + · · · + Z i , N i − . Plugging Equivalence (3.4) into Equivalence (3.3) wefinally get the relation Z i , N i · N i − Y k = Z i , N i − X j , i λ ki , j Z j , N j ≡ Z i , N i (cid:0) Z i , + · · · + Z i , N i − (cid:1) N i − Y k = L i , N i − k ≡ Z i , + · · · + Z i , N i − kills the product Q N i − k = L i , N i − k in ˆ A . (cid:3) Corollary 3.1.
If A is Artinian, then ˆ A is also Artinian.Proof.
Suppose that A is Artinian. Rewriting the defining relations for ˆ A weget that for each 1 ≤ i ≤ nZ i , N i ≡ Z i , N i Z i , + · · · + Z i , N i − + X j , i λ i , j Z j , N j Z i , N i − ≡ Z i , N i − X j , i (cid:16) λ i , j − λ i , j (cid:17) Z j , N j − (cid:0) Z i , + · · · + Z i , N i − (cid:1) ... Z i , N i − k ≡ Z i , N i − k X j , i (cid:16) λ ki , j − λ i , j (cid:17) Z j , N j − (cid:0) Z i , + · · · + Z i , N i − k − (cid:1) ... Z i , ≡ Z i , X j , i (cid:16) λ N i − i , j − λ i , j (cid:17) Z j , N j Now endow the set of monomials in F [ Z i , j | ≤ i ≤ n , ≤ j ≤ N i ] with thegrlex ordering stemming from the following ordering of the variables: Z , N ≺ · · · ≺ Z n , N n ≺ Z , ≺ · · · ≺ Z , N − ≺ · · · ≺ Z n , ≺ · · · ≺ Z n , N n − . OLD CONJECTURE 7
Then we see that by using the above relations, we can express the squaresof each of the variables, with the exception of Z , N , . . . , Z n , N n , as monomi-als of strictly lesser order. Therefore, by virtue of Proposition 2.2, everymonomial in ˆ A is equivalent to an F linear combination of monomials of theform Z A , N · · · Z A n n , N n n Y i = N i − Y j = Z ǫ i , j i , j , A i ∈ N , ǫ i , j ∈ { , } . Hence to show that ˆ A is Artinian, i.e. a finite dimensional F vector space, itsu ffi ces to show that the elements Z j , N j are nilpotent in ˆ A . But Z j , N j = φ ( x j )and if A is Artinian, x j is nilpotent for all j , and the result now follows. (cid:3) Now we see that if A is Artinian, then so is ˆ A , and hence by Lemma 2.2,the sequence Z i , j · L i , j , 1 ≤ i ≤ n , 1 ≤ j ≤ N i must be a regular sequence in F [ Z i , j | ≤ i ≤ n , ≤ j ≤ N i ]. Hence in this case, Corollary 2.2 tells us thata socle generator for ˆ A is given by Q ni = Q N i j = L i , j . Proposition 3.2.
If A is Artinian, then so is ˆ A and the map φ : A → ˆ Adefined above is an embedding.Proof.
By the preceding discussion, we need only show that Q ni = Q N i j = L i , j is in the image of φ . The claim is that we have the following equivalence: n Y i = N i − Y k = Z i , N i − X j , i λ ki , j Z j , N j ≡ C · n Y i = N i Y j = L i , j for some C ∈ F × . Then since the LHS is clearly in the image of φ , this will prove the desiredresult. For a shorthand notation, we write M i ≔ N i − Y k = Z i , N i − X j , i λ ki , j Z j , N j and we shall write L i ≔ N i − Y k = L i , N i − k . In this notation we want to show that in ˆ A we have the relation n Y i = M i ≡ C · n Y i = L i for some C ∈ F × . In fact we will show that for each 1 ≤ i ≤ n we always have the equivalencein ˆ A M · · · M n ≡ ( − i · L · · · L i · M i + · · · M n . CHRIS MCDANIEL
We prove this by induction on i . The base case i = A M · · · M n ≡ − L · M · · · M n . To see this, we first recall some computations performed previously. Tostart recall that we had for each 1 ≤ i ≤ n and for each 1 ≤ k ≤ N i − Z i , N i · ( Z i , N i − X j , i λ ki , j Z j , N j ) ≡ Z i , N i · ( L i , N i − k + Z i , N i − k + + · + Z i , N i − ) . Also since the sum Z i , N i − j + · · · + Z i , N i − annihilates the product Q jk = L i , N i − k ,the entire product collapses, i.e.(3.6) N i − Y k = ( L i , N i − k + Z i , N i − k + + · · · + Z i , N i − ) ≡ N i − Y k = L i , N i − k , for each 1 ≤ i ≤ n . Finally we recall the equation(3.7) Z i , N i = Z i , + · · · + Z i , N i − − L i , N i + X j , i λ i , j Z j , N j . OLD CONJECTURE 9
Now combining Equivalences (3.5), (3.6) and Equation (3.7), we obtain thefollowing equivalence: M i = Z i , N i − X j , i λ i , j Z j , N j N i − Y k = Z i , N i − X j , i λ ki , j Z j , N j = Z i , N i N i − Y k = Z i , N i − X j , i λ ki , j Z ki , j − X j , i λ ki , j Z j , N j N i − Y k = Z i − X j , i λ ki , j Z j , N j ≡ Z i , N i N i − Y k = (cid:0) L i , N i − k + Z i , N i − k + + · · · + Z i , N i (cid:1) − X j , i λ ki , j Z j , N j N i − Y k = Z i − X j , i λ ki , j Z j , N j ≡ Z i , N i N i − Y k = L i , N i − k − X j , i λ ki , j Z j , N j N i − Y k = Z i − X j , i λ ki , j Z j , N j ≡ Z i , + · · · + Z i , N i − − L i , N i + X j , i λ i , j Z j , N j N i − Y k = L i , N i − k − X j , i λ ki , j Z j , N j N i − Y k = Z i , N i − X j , i λ ki , j Z j , N j ≡ − N i − Y k = L i , N i − k − X j , i λ ki , j Z j , N j N i − Y k = Z i , N i − X j , i λ ki , j Z j , N j − N i − Y k = L i , N i − k M i ≡ − L i − X j , i λ ki , j Z j , N j N i − Y k = Z i , N i − X j , i λ ki , j Z j , N j − N i − Y k = L i , N i − k (3.8) Plugging Equivalence (3.8) into the product M · · · M n we get − L − X j , λ k , j Z j , N j N − Y k = Z , N − X j , λ ki , j Z j , N j − N i − Y k = L i , N i − k · M · · · M n . But since Z j , N j annihilates M · · · M n for all j ,
1, we get the equivalence M · · · M n ≡ − L · M · · · M n which establishes the base case. Inductively assume that we have the equiv-alence M · · · M n ≡ ( − i − · L · · · L i − · M i · · · M n for some i >
1. Now we substitute Equivalence (3.8) into L · · · L i − · M i · · · M n to get( − i − L · · · L i − − L i − X j , i λ ki , j Z j , N j N i − Y k = Z i , N i − X j , i λ ki , j Z j , N j − N i − Y k = L i , N i − k · M i + · · · M n . Note again that for each j , i , Z j , N j annihilates the product L · · · L i − · M i + · · · M n , and so we again obtain M · · · M n ≡ − ( − i − L · · · L i · M i + · · · M n which completes the induction and proves the claim. Therefore we haveshown that in ˆ A we have the equivalence M · · · M n ≡ ( − n L · · · L n and therefore φ : A → ˆ A must be an embedding, as desired. (cid:3) R eferences [1] A. Abedelfatah. On the Eisenbud-Green-Harris Conjecture. ArXiv e-prints , December2012.[2] T. Harima, A. Wachi, and J. Watanabe. The quadratic complete intersections with theaction of the symmetric group.
ArXiv e-prints , January 2015.[3] Hideyuki Matsumura.
Commutative ring theory. Transl. from the Japanese by M. Reid. ept . of M ath . and C omp . S ci ., E ndicott C ollege , B everly , MA 01915 E-mail address ::