Tressl's Structure Theorem for Separable Algebras
Gabriel Ng


§ ABSTRACT 
    This note presents a generalisation of Tressl's structure theorem for differentially finitely generated algebras over differential rings of characteristic 0 to the case of separable algebras over differential rings of arbitrary characteristic.


Quantum reservoir computing on random regular graphs
    Tapio Ala-Nissila
    
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§ INTRODUCTION


Tressl's structure theorem for differential algebras <cit.> is a result which describes the structure of differentially finitely generated algebras over differential rings of characteristic 0. It has been applied by León Sanchez and Tressl in the study of differentially large fields of characteristic 0 in order to characterise these fields in terms of the existence of points of certain algebras <cit.>.

In this note, we adapt Tressl's proof to generalise this theorem to the case of separable algebras over differential rings of arbitrary characteristic. We assume throughout that all rings are commutative and unital, and differential rings are equipped with m commuting derivations.



§ CHARACTERISTIC SETS


Let us recall the notion of a characteristic set, and some facts regarding differential ideals. For this section, we fix the following notation and objects:

    
  * (R, ) is a differential ring in m commuting derivations (of arbitrary characteristic), where = (δ_0,...,δ_m-1);
    
  * Y = (Y_0,...,Y_n-1) is a tuple of n indeterminates;
    
  * 𝒟 = {^α : α∈^m} is the free abelian monoid of differential operators generated by .
    
  * 𝒟Y = {δ Y_i : δ∈𝒟, i < n} is the set of derivatives of variables from Y;
    
  * 𝒟Y^* = {y^p : p ∈_>0} is the set of powers of elements from 𝒟Y;
    
  * R{Y} is the differential polynomial ring in indeterminates Y, identified with R[𝒟Y] as an R-algebra and equipped with the natural derivations  extending those on R. 




    We define the rank of 𝒟Y^* to be the map
    
    : 𝒟Y^*    →× n ×^m ×_>0
    
            (^α Y_i)^p    ↦ (|α|, i, α_m-1,...,α_0, p).

    We equip Ø = × n ×^m ×_>0 with the lexicographic order, which is well ordered.



    For a differential polynomial f ∈ R{Y}∖ R, say that a variable y ∈ appears in f if y appears in f considered as an algebraic polynomial in R[]. The leader of f, denoted u_f, is the variable y ∈ of maximal rank which appears in f. Denote by u_f^* the highest power of u_f which appears in f. We extend the notion of rank from ^* to R{Y} by setting
    
    (f) = (u_f^*) ∈Ø.




    Let f, g ∈ R{Y} be differential polynomials, with g ∉R. We say that f is partially reduced with respect to g (also called weakly reduced), if no proper derivative of u_g appears in f. Say that f is reduced with respect to g if f is partially reduced with respect to g, and _u_g(f) < _u_g(g).

    We say that f is (partially) reduced with respect to G, where G ⊆ R{Y}∖ R is nonempty, if f is (partially) reduced with respect to every g ∈ G.

    We say that a nonempty subset G ⊆ R{Y} is autoreduced if every f ∈ G is reduced with respect to every g ≠ f ∈ G (including the case where G is a singleton).


It is easy to see that if G is autoreduced, then for any distinct f, g ∈ G, u_f ≠ u_g. 


    Every autoreduced set is finite. 



    Let ∞ be an element larger than every element of Ø, and equip (Ø∪{∞})^ with the lexicographic order.  The rank of an autoreduced set G is defined as follows: Let G = {g_0,...,g_l-1} with (g_0) < ... < (g_l-1). Define
    
    (G) = ((g_0),... , (g_l-1), ∞, ∞,...)




    There is no infinite strictly rank-decreasing sequence of autoreduced sets.



    Let 𝔨⊆ R{Y} be a differential ideal not contained in R. Then, by the previous proposition,
    
    {(G) : G ⊆ M  is autoreduced such that  S(g) ∉𝔨 for all  g ∈ G}

    has a minimum. We call an autoreduced subset G of M of minimal rank a characteristic set of 𝔨.


 
    Let G be a characteristic set of M ⊆ R{Y}, and f ∈ M∖ R. Then, f is not reduced with respect to G. 


 
    Let f ∈ R{Y}∖ R, and write f = f_d u_f^d + ... + f_1 u_f + f_0, where f_i ∈ R[y ∈, y ≠ u_f], and f_d ≠ 0. The initial of f, denoted I(f) is:
    
    I(f) = f_d.

    The separant of f, denoted S(f) is:
    
    S(f) = ∂/∂ u_f f = df_du_f^d-1 + ... + f_1.

    Finally, for any autoreduced subset G = {g_0,...,g_l-1} of R{Y}, define
    
    H(G) = ∏_i < l I(g_i)· S(g_i),

    i.e. H(G) is the product of all initials and separants of elements of G, and
    
    H_G = {∏_i < l I(g_i)^n_i S(g_i)^m_i : n_i, m_i ∈},

    i.e. H_G is the set of all products of powers of initials and separants of elements of G.



    Let G be an autoreduced set in R{Y} and f ∈ R{Y}. There exists f̃∈ R{Y} which is reduced with respect to G, and H ∈ H_G such that
    
    H · f ≡f̃[G].

    If in addition f is partially reduced with respect to G, then there is h ∈ H_G such that
    
    H · f ≡f̃(G).





§ (QUASI-)SEPARABLE RINGS AND IDEALS


We recall the notion of (quasi-)separable R-algebras and ideals from <cit.>. 


    Let R_0 ⊆ R be rings. We say that R is separable over R_0 if either R is zero, or R is nonzero and:
    
        
  * R contains no nonzero nilpotent elements;
        
  * For any a_0 ∈ R_0 and b ∈ R with a_0, b ≠ 0, then a_0b ≠ 0 (in particular, R_0 is a domain);
        
  * Either R_0 has characteristic 0, or R_0 has characteristic p > 0 and R^p and R_0 are linearly disjoint over R_0^p.
    
    Let 𝔨⊆ R be an ideal, and let ϕ: R → R/𝔨 denote the quotient map. We say that 𝔨 is separable over R_0 if ϕ(R) is separable over ϕ(R_0).



    In particular, we note the special case when L/K is a separable field extension, any K-subalgebra of L is a separable K-algebra.


From now, let S = R{Y} be a finitely generated differential polynomial algebra over R. For a set A ⊆ S, denote the differential ideal generated by A by [A]. For an ideal 𝔨⊆ S and s ∈ S, define the saturated ideal of 𝔨 over s as

    𝔨 : s^∞ = {h ∈ S : s^n h ∈𝔨 for some  n ∈}

It is verify that if 𝔨 is a differential ideal, then 𝔨:s^∞ is also differential.


    Let G be an autoreduced set in S, and 𝔨 be an ideal (not necessarily differential) of S. We say that G is  𝔨-coherent if the following hold:
    
        
  * 𝔨 has a set of generators partially reduced with respect to G;
        
  * [𝔨] ⊆ ([G] + 𝔨) : H(G)^∞
        
  * For any f, g ∈ G, and v a common derivative of u_f, u_g, say v = θ_f u_f = θ_g u_g, we have that
        
    S(g)θ_f f - S(f)θ_g g ∈ ((G_v) + 𝔨) : H(G)^∞,

        where G_v denotes he set of all differential polynomials of the form τ h, where h ∈ G, τ∈Θ, and τ u_h is of lower order than v.
    


 
    Let 𝔭 be a prime differential ideal of S which is quasi-separable over R, and let G be a characteristic set of 𝔭. Then, there exists a finite set Y' of derivatives of the indeterminates Y, each partially reduced with respect to G, such that setting 𝔭_1 = 𝔭∩ R[Y'], G is S𝔭_1-coherent and 𝔭 = ([G] + S𝔭_1) : H(G)^∞. The set Y' may be replaced by any larger finite set of derivatives of Y partially reduced with respect to A.



    Recall that by the definition of a characteristic set, for each a ∈ G, S(a) ∉𝔭. Further, for any g ∈ G, I(g) ∉𝔭 by <cit.>. Since 𝔭 is a prime ideal, then we also have that H_G does not contain any element of 𝔭, as for any g ∈ G, neither I(g) nor S(g) lie in . In particular, H_G does not contain 0.


We will also require an additional lemma which strengthens the above result for partially reduced polynomials:


    Let G be a 𝔨-coherent autoreduced set in S, and suppose that for each g ∈ G, S(g) is not a zero-divisor in S. Then, every element of ([G] + 𝔨) : H(G)^∞ which is partially reduced with respect to G lies in ((G) + 𝔨) : H(G)^∞.


In particular, in the case where G is a characteristic set of a prime differential ideal ⊆ S, we have the following:

 
    Let , G and _1 be as in Lemma <ref>. Then, any f ∈ partially reduced with respect to G lies in ((G) + S_1) : H(G)^∞.




§ THE STRUCTURE THEOREM FOR SEPARABLE ALGEBRAS


We state and prove the main theorem, which is adapted from <cit.>:


    Let (S, ) be a differential domain, and let (R, ) be a differential subring of S, with R Noetherian as a ring, such that:
    
        
  * S is separable over R; and,
        
  * S is differentially finitely generated as a differential R-algebra.
    
    Then, there exist (not necessarily differential) R-subalgebras P and B of S and an element h ∈ B with h ≠ 0 such that:
    
        
  * B is a finitely generated R-algebra, and B_h is a finitely presented R-algebra;
        
  * P is a polynomial algebra over R;
        
  * S_h = (B· P)_h, and S_h is a differentially finitely presented R-algebra;
        
  * The homomorphism B ⊗_R P → B · P induced by multiplication is an isomorphism of R-algebras.
    




 §.§ Proof of Theorem <ref>

    As S is a differentially finitely generated R-algebra, there is a surjective differential R-algebra homomorphism ϕ: R{Y_1,...,Y_n}→ S. Write Y = (Y_1,...,Y_n), and let 𝔭 = (ϕ). As S is prime, and ϕ is injective on R, we have that 𝔭 is a differential prime ideal of R{Y} with 𝔭∩ R = 0. Further, since S is separable over R, the ideal 𝔭 is separable over R also. Let G be a characteristic set of 𝔭.

    By Lemma <ref>, there is a finite set Y' ⊆, whose members are each partially reduced with respect to G, such that G is S𝔭_1-coherent and 𝔭 = ([G] + S𝔭_1) : H(G)^∞, where 𝔭_1 = 𝔭∩ R[Y'].

    Define the following:
    
    h    ϕ(H(G)), 
    
            V    {y ∈ : y  is not a proper derivative of any  u_g ∈ G}, 
    
            V_B     Y' ∪{y ∈ V : y  appears in some  g ∈ G}, 
    
            B    ϕ(R[V_B]), 
    
            P    ϕ(R[V ∖ V_B]).


    
        Observe that Y' is a subset of V: if a differential indeterminate is a proper derivative of some u_g, then it is not partially reduced with respect to G and thus does not lie in Y'.
    
    
        Since G is autoreduced, f ∈ R{Y} is partially reduced with respect to G if and only if f ∈ R[V].
    

    
        The restriction of ϕ to P' = R[V ∖ V_B] is injective.
    
    
        Suppose we have f ∈ P' ∩. We claim that f is reduced with respect to G. By the previous observation, we have that f is partially reduced with respect to G. Since the leader u_g of each g ∈ G is in V_B, no leader of any g ∈ G appears in f, and thus f is reduced with respect to G. By Lemma <ref>, f ∈ R, and since ∩ R = 0, f = 0, as required.
    

    
        h ≠ 0.
    
    
        By the remark following Lemma <ref> and writing G = {g_0,...,g_l-1}, we observe that
        
    H(G) = ∏_i < l I(g_i)· S(g_i)

        does not lie in the prime ideal , as no I(g_i) nor S(g_i) lies in . Thus h = ϕ(H(G)) is not 0 in S.
    

    
         S_h = (B · P)_h.
    
    
        Let f ∈ R{Y}. By Proposition <ref>, there is f̃∈ R{Y} reduced with respect to G, and H ∈ H_G such that H · f ≡f̃[G]. Since G ⊆, we have that [G] ⊆, thus ϕ(f) ·ϕ(H) = ϕ(f̃) in S.
        
        Since f̃ is reduced with respect to G, we have that f̃∈ R[V], and in particular ϕ(f̃) ∈ B · P.
        Further, for each g ∈ G, ϕ(I(g)) and ϕ(S(g)) are units in (B · P)_h, so ϕ(H) is a unit also. Thus, ϕ(f) = ϕ(H)^-1ϕ(f̃) ∈ (B · P)_h, as required.
    

    
        S_h is a differentially finitely presented R-algebra.
    
    
        We extend the projection ϕ: R{Y}→ S to ψ: R{Y}[H(G)^-1] → S_h by defining ψ(H(G)^-1) = h^-1. Thus S_h is differentially finitely generated.

        Let 𝔮 = (ψ). We show that 𝔮 is finitely generated as a differential ideal. As R is Noetherian, R[Y'] is also Noetherian, and thus _1 is finitely generated as an ideal. Let A be such an generating set.
        Clearly, since ψ extends ϕ, G ∪ A ⊆𝔮. We claim that G ∪ A generates 𝔮 as a differential ideal.
        
        
        For the reverse inclusion, suppose that f/H(G)^d ∈𝔮 for some f ∈ R{Y} and d ∈.
        Since S_h is a domain, and h^-d is nonzero, we have that ψ(f) = ϕ(f) = 0, i.e. f ∈. Since = ([G] + S_1) : H(G)^∞, there is some n ∈ with fH(G)^n ∈ [G] + S_1 ⊆ [G ∪ A]. Thus, we have that f lies in the differential ideal generated by G ∪ A in R{Y}[H(G)^-1].
    

    
        B is finitely generated, and B_h is finitely presented as R-algebras.
    
    
        This is clear.
    

    The above claims prove parts (a), (b) and (c) of Theorem <ref>. It remains to prove (d).

    
        Suppose b_1,...,b_m ∈ B are linearly dependent over P. Then, they are linearly dependent over R.
    
    
        Suppose we have f_i ∈ R[V_B] with ϕ(f_i) = b_i, and p_i ∈ R[V ∖ V_B], not all in , such that q ∑_i f_i p_i ∈. We may assume in particular that p_1 ∉. Since q ∈ R[V], q is partially reduced with respect to G.

        Let A again be a finite generating set of _1 as an ideal of R[Y']. Then, by Corollary <ref>, there are n ∈, and h_g for g ∈ G ∪ A such that
        
    H(G)^n q = ∑_g ∈ G ∪ A h_g g.

        Since H(G), q and every f lies in R[V], we may assume the coefficients h_f also lie in R[V]. 

        Since p_1 ≠ 0, there exists an R-algebra homomorphism ψ: R[V ∖ V_B] → R such that ψ(p_1) ≠ 0. This can be extended to an R[V_B]-algebra homomorphism ψ: R[V] → R[V_B] with ψ(p_1) ≠ 0.

        Since all p_i and h_g lie in R[V], we may now apply ψ to the following equation:
        
    H(G)^n( f_1 p_1 + ... + f_k p_k) = ∑_g ∈ G∪ A h_g g.

        Since H(G), every f_i and g ∈ G ∪ A lies in R[V_B], these are preserved by ψ, and thus we obtain that
        
    H(G)^n(ψ(p_1) f_1 + ... + ψ(p_k) f_k) = ∑_g∈ G∪ Aψ(h_g) g.

        Observe that the right hand side lies in the ideal ([G] ∪ S_1), and thus
        
    ψ(p_1) f_1 + ... + ψ(p_k) f_k ∈ ([G] ∪ S_1) : H(G)^∞ = .

        Recall that ψ(p_i) ∈ R, and ϕ is an R-algebra homomorphism. Applying ϕ to the above yields:
        
    ψ(p_1) b_1 + ... + ψ(p_k) b_k = 0.

        By previous assumption, we have that ψ(p_1) is not zero, and thus the b_i are linearly dependent over R.
    
    This implies (d) as follows:
    
        The homomorphism B ⊗_R P → B· P induced by multiplication is an isomorphism of R-algebras.
    
    
        The homomorphism is clearly surjective, and it remains to show injectivity. Let m ∈ be minimal such that there are b_1,...,b_m ∈ B and p_1,...,p_m ∈ P with ∑_i b_i p_i = 0 and x ∑_i b_i ⊗ p_i ≠ 0. Then, the b_i are linearly dependent over P. By Claim <ref>, the b_i are in fact linearly dependent over R. Thus, there are r_1,...,r_m ∈ R, not all zero, such that r_1b_1 + ... + r_mb_m = 0.

        Without loss, assume that r_1 ≠ 0. Then, m > 1, and observe that
        
    r_1b_1 ⊗ p_1 = -(r_2b_2 + ... + r_mb_m) ⊗ p_1,

        which gives
        
    r_1x = -(r_2b_2 + ... + r_mb_m) ⊗ p_1 + r_1b_2 ⊗ p_2 + ... + r_1b_m ⊗ p_m.

        Collecting terms and rearranging, we have
        
    r_1x = b_2 ⊗ (r_1p_2 - r_2p_1) + ... + b_m ⊗ (r_1p_m - r_mp_1).

        We necessarily have that r_1x = 0, otherwise this contradicts the minimality of the choice of m. Let F denote the quotient field of R. Then, in F ⊗_R (B ⊗_R P), we have that
        
    1 ⊗ x = 1/r_1⊗ r_1x = 0.

        By Claim <ref>, P is a polynomial R-algebra, hence flat. As the inclusion B → F ⊗_R B is injective, it follows that B ⊗_R P → F ⊗_R B ⊗_R P is injective. Thus, 1 ⊗ x = 0 in F ⊗_R B ⊗_R P implies that x = 0, which is a contradiction.
    
    

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