On differential algebra, 2:
We henceforth take our differential fields to be of characteristic zero (so that we avoid complications arising from trying to extend derivations to inseparable extensions, c.f. the previous proposition) and work in the language of rings enriched by the function symbol \(\partial\). As in the case of algebraically closed fields, the definable sets cut out by the atoms in this language are precisely the zerosets of polynomials lying in the ring \[K[T, T’, \dots, T^{(n)}, \dots]\] and if \(a\) lies over \(K\) then \[K[a]_d \overset{\text{df}}{=} K[a, a’, \dots, a^{(n)}, \dots]\] is the differential ring generated by \(a\) over \(K\). If a polynomial of the form displayed above can be found such that it vanishes on \(a\), then \(a\) is said to be differentially algebraic over \(K\); otherwise (i.e. the set of all of \(a\)’s derivatives is algebraically independent over \(K\)) \(a\) is said to be differentially transcendental over \(K\). As in the case of plain fields, we have minimal polynomials and the analogous associated results:
Lemma. Let \(K\) be a differential field, with \(a,b\) lying in some differential field extension of \(K\), both differentially algebraic with the same minimal polynomial \(f \in K[T_{i \leq n}]\); without loss \(T_n\) appears in \(f\). Then there is a differential ring isomorphism \[K[a]_d \overset{\sim}{\to} K[b]_d\] fixing \(K\) and taking \(a \mapsto b\).
Proof. Write \(\vec{a} = \left(a, \dots, a^{(n)}\right)\), resp. \(\vec{b}\). The evaluation morphism \[\operatorname{eval}(\vec{a}): K[T_{i \leq n}] \to K[\vec{a}]\] has kernel \(\left(f\right)\) in \(K[T_{i \leq n}]\) which induces an isomorphism of the underlying ring structures, so that for each \(0 \leq i \leq n\), \[a^{(i)} \mapsto b^{(i)}.\] To check compatibility of the differential structure, note that by a previous lemma, \[f(\vec{a}) ‘ = g^{\partial}(\vec{a}) + \displaystyle \sum_{i = 0}^n \left(\dfrac{\partial h}{\partial T_i} \right) (\vec{a}) \cdot a^{(i + 1)}\] for \(h \in K[T_{i \leq n}].\) As we’re in characteristic zero, minimal polynomials are separable, hence \(\left(\dfrac{\partial f}{\partial T_n}\right)(\vec{a}) \neq 0\), which allows us to manipulate the displayed equation to obtain \[a^{(n+1)} = \dfrac{f^{\partial}(\vec{a}) + \sum_{i = 0}^{n-1} \left(\frac{\partial f}{\partial T_i} \right) (\vec{a}) \cdot a^{(i + 1)}}{\frac{\partial f}{\partial T_n} (\vec{a})} \in K[\vec{a}],\] hence \(K[\vec{a}]’ \subseteq K[\vec{a}] \implies K[\vec{a}] \subseteq K[a]_d \subseteq K(\vec{a})\) as rings. This argument works just as well with \(b\); then \(\partial : K \to K\) extends to a derivation \(K[\vec{a}] \to K(\vec{b})\), which in particular factors through the induced isomorphism of rings \(K[\vec{a}]_d \to K[\vec{b}]_d\). As the extension of \(\partial\) is unique, it must agree with the \(\partial\)-structure on \(K[\vec{b}]_d\), which gives us the differential ring isomorphism we require. \(_\square\)
We can also realize the roots of a prescribed minimal polynomial, i.e.
Lemma. Let \(f \in K[T_0, \dots, T_n]\) be irreducible such that \(T_n\) appears in \(f\). Then there is an element \(a\) in a differential field extension of \(K\) such that \(a\) is differentially algebraic over \(K\) with minimal polynomial \(f\) over \(K\).
Proof. We have \[K[T_{i \leq n}] / (f) = K[t_0, \dots, t_n] = K[t]\] with \(t_i = T_i + (f)\), so that \(K[t]\) is a domain with fraction field \(K(t)\), such that \(f(t) = 0\) but (by separability) \(\dfrac{\partial f}{\partial T_n} (t) \neq 0\). The plan is to extend \(\partial\) to a derivation on \(K(t)\) such that \(t_i’ = t_{i + 1}\) for \(0 \leq i \leq n\). First, as in the previous lemma, set \[t_{n + 1} = \dfrac{f^{\partial}(t) + \sum^{n-1}_{i = 0} \frac{\partial f}{\partial T_i} (t) \cdot t_{i + 1}}{\frac{\partial f}{\partial T_n} (t)} \in K[t].\] By a previous lemma, this is the value \(t_{n}’\) must take if it has the property prescribed above. One checks that the map \[d: K[T_{i \leq n}] \to K(t) \hspace{5mm} \text{by} \hspace{5mm} h \mapsto h^{\partial} + \sum_{i = 0}^n \dfrac{\partial h}{\partial T_i} (t) \cdot t_{i + 1}\] is a derivation into \(K(t)\) which extends (hence, extends uniquely) to a derivation on \(K(t)\). This must extend \(\partial\), and setting \(a = t_0\) we have \(a^{(i)} = t_i\) for \(0 \leq i \leq n\), as required. \(_\square\)
Now, a differentially closed field is an existentially closed differential field (which is the proper model-theoretic generalization for “algebraically closed” in the sense of finite definable sets.) With quantifier elimination this reduces to throwing existentials in front of definable sets of the form \(A_1 \backslash A_2\) with \(A_1, A_2\) atoms, which in this case means that for every \(f \in K[T_{i \leq n}]\) and \(0 \neq g \in K[T_{i \leq n}]\) there exists an \(a\) such that \(f\) vanishes on \(a, a’, \dots, a^{(n)}\) and \(g\) is nonzero on \(a, a’, \dots, a^{(n - 1)}.\) Every differential field embeds into a differentially closed field, and the theories of (characteristic zero) differentially closed fields are first-order axiomatizable by a theory \(\mathsf{DCF}\).
Theorem. \(\mathsf{DCF}\) admits quantifier-elimination and is complete.
Proof. Completeness follows from substructure-completeness by taking \(\mathbb{Z}\) equipped with the trivial derivation as a prime substructure. Set up the quantifier-elimination criterion developed earlier. Let \(\mathbf{M}\) and \(\mathbf{N}\) be models of \(\mathsf{DCF}\) with the latter \(|M|^+\)-saturated; let \(\mathbf{K} \subseteq \mathbf{M}\) be a proper substructure of \(\mathbf{M}\), with \(\iota : \mathbf{K} \to \mathbf{N}\) an embedding. Due to the uniqueness of extensions of derivations on integral domains, we may assume \(\mathbf{K}\) is a proper differential subfield. Let \(a \in M \backslash K\). If \(a\) is differentially transcendental over \(K\), we can find via saturation a corresponding \(b \in N\) saturated over \(\iota(K)\) in \(\mathbf{N}\); then \(a \mapsto b\) induces an \(\iota’\) extending \(\iota\) where \(\iota’ : \mathbf{K}[a]_d \to \mathbf{N}\). If \(a\) is differentially algebraic with minimal polynomial \(f \in K[T_{i \leq n}]\) with \(T_n\) appearing in \(f\), then for every \(g \in K[T_{i \leq n}] \backslash K,\) \[f(a,a’, \dots, a^{(n)}) = 0 \land g(a, a’, \dots, a^{(n-1)}) \neq 0.\] Via saturation we can find a corresponding \(b\) satisfying the same formulas with all parameter (sets) from \(K\) replaced by their images under \(\iota\). By the first lemma in this post, \(\iota\) extends to an embedding \(\mathbf{K}[a]_d \to \mathbf{N}\) taking \(a \mapsto b\), which gives us QE. \(_\square\)
Now, a natural question to ask, given the notion of differential rings, is whether there exist differential analogues of modules over plain rings.
Definition. Let \((R, \partial)\) be a differential ring. A \(\partial\)-module over \(R\) is the data \((M, D_M)\) equipped with an additive map \(M \overset{D_M}{\longrightarrow} M\) (called a connection) which satisfies the product rule with respect to the \(R\)-action: for all \(r \in R\) and \(m \in M\), \[D_M(r \cdot m) = \partial(r) \cdot m + r \cdot D_M(m).\] Let \(M^{\partial}\) denote the subset of \(M\) where \(D_M\) vanishes.
Remark. The formula \[f(a)’ = f^{\partial}(a) + \dfrac{\partial f}{\partial T}(a) \cdot a’\] is then seen to arise naturally by equipping the polynomial ring \(K[T]\) with the structure of a \(\partial\)-module over \(K\), with the connection being precisely the formal derivative \(f \mapsto f’\).
Lemma. Let \((K, \partial)\) be a differential field, and \((V, D_V)\) a \(\partial\)-module over \(K\). Let \(C_K\) be \(K\)’s field of constants. Then \(V^{\partial}\) is a \(C_K\)-vector space, and if \(K\) is differentially closed then the dimension of \(V^{\partial}\) over \(C_K\) is the same as the dimension of \(V\) over \(K\), and there exist \(\{v_i \in V^{\partial}\}_{i \leq n}\) forming a \(K\)-basis of \(V\), hence also a \(C_K\)-basis of \(V^{\partial}\).
Proof. For the first item, let \(c \in K\). Then \(D_V(cv) = \partial(c) v + d D_V(v) = 0\) whenever \(c \in C_K\). Hence, whenever \(\partial(c) = 0\), the action by \(c\) on \(V\) restricts to an action on \(V^{\partial}\).
For the second item, fix a \(K\)-basis \(\{\beta_i\}_{i \leq n}\) of \(V\). The equation \(D_V(x) = 0\) becomes \[\sum_{i = 1}^n \partial(c_i) \beta_i \sum_{i = 1}^n c_i D_V(\beta_i) = 0 \iff \partial(\vec{c}) = \left(d^i_j\right)_{i, j \leq n} (\vec{c}).\] Claim: a set of solutions \(\{\vec{c}_i\}\) to the above matrix equation is linearly independent over \(C_K\) if and only if we also have linear independence over \(K\). The “if” direction is clear; in the other, suppose \(\{\vec{c}_1, \dots, \vec{c}_n\}\) is linearly independent over \(C_K\) but not over \(K\). Then we can write (say) \[\vec{c}_n = \sum_{i = 1}^{n-1} a_i \vec{c}_i\] where \(\{a_i\} \subseteq K\) proper. Without loss we can assume the \(\vec{c}_i\) are minimally dependent, that is, \(\{\vec{c}_i\}_{i \leq n - 1}\) are \(K\)-linearly independent. Applying \(D_V\) to the equation then yields \[0 = \sum_{i = 1}^{n-1} \partial(a_i)(\vec{c}_i) + \sum_{i = 1}^{n-1} a_i D_V(\vec{c}_i)\]\[\implies 0 = \sum_{i = 1}^{n-1} \partial(a_i)\vec{c}_i \implies \partial(a_i) = 0 \implies a_i \in C_k\] for \(0 \leq i \leq n -1\). As \(1 \in C_K\), this means that \(\{\vec{c}_i\}_{i \leq n}\) are linearly dependent over \(C_K\), a contradiction. Thus the claim is proved; as a fundamental matrix of solutions can be found in some extension of \((K, \partial)\) (and can thus be already found in \((K, \partial)\) due to differential i.e. existential closedness), this concludes the proof. \(_\square\)