diff --git a/iii/alggeom/01_introduction.tex b/iii/alggeom/01_introduction.tex index 550e018..57a9551 100644 --- a/iii/alggeom/01_introduction.tex +++ b/iii/alggeom/01_introduction.tex @@ -142,3 +142,95 @@ \subsection{Limitations of classical algebraic geometry} If \( D_\delta = \mathbb V(Y + \delta) \) for \( \delta \in k \), \( C \cap D_\delta \) is two points unless \( \delta = 0 \). This breaks a continuity property. Therefore, the intersection of two affine varieties is not naturally an affine variety. + +\subsection{Spectrum of a ring} +Let \( A \) be a commutative unital ring. +\begin{definition} + The \emph{Zariski spectrum} of \( A \) is \( \Spec A = \qty{\mathfrak p \trianglelefteq A \text{ prime}} \). +\end{definition} +\begin{remark} + Given a ring homomorphism \( \varphi : A \to B \), we have an induced map of sets \( \varphi^{-1} : \Spec B \to \Spec A \) given by \( \mathfrak q \mapsto \varphi^{-1}(\mathfrak q) \), as the preimage of a prime ideal is always prime. + Note, however, that this property would fail if we only considered maximal ideals, because the preimage of a maximal ideal need not be maximal. + % TODO: Example + + Given \( f \in A \) and a point \( \mathfrak p \in \Spec A \), we have an induced \( \overline f \in \faktor{A}{\mathfrak p} \) obtained by taking the quotient. + We can think of this operation as `evaluating' an \( f \in A \) at a point \( \mathfrak p \in \Spec A \), with the caveat that the codomain of this evaluation depends on \( \mathfrak p \). +\end{remark} +\begin{example} + \begin{enumerate} + \item Let \( A = \mathbb Z \). + Then \( \Spec A = \Spec \mathbb Z \) is the set \( \qty{(p) \mid p \text{ prime}} \cup \qty{(0)} \). + Consider an element of \( \mathbb Z \), say, \( 132 \). + Given a prime \( p \), we can `evaluate it at \( p \)', giving \( 132 \text{ mod } p \in \faktor{\mathbb Z}{p\mathbb Z} \). + Thus \( \Spec Z \) is a space, \( 132 \) is a function on \( \Spec Z \), and \( 132 \text{ mod } p \) is the value of this function at \( p \). + \item Let \( A = \mathbb R[X] \). + Then \( \Spec A \) is naturally \( \mathbb C \) modulo complex conjugation, together with the zero ideal. + \item If \( A = \mathbb C[X] \), then \( \Spec A \) is naturally \( \mathbb C \), together with the zero ideal. + \end{enumerate} +\end{example} +% TODO: Draw Spec Z[X]. Draw Spec k[X]. +\begin{definition} + Let \( f \in A \). + Then we define + \[ \mathbb V(f) = \qty{\mathfrak p \in \Spec A \mid f = 0 \text{ mod } \mathfrak p \text{, or equivalently, } f \in \mathfrak p} \subseteq \Spec A \] + Similarly, for \( J \trianglelefteq A \) an ideal, + \[ \mathbb V(J) = \qty{\mathfrak p \in \Spec A \mid \forall f \in J,\, f \in \mathfrak p} = \qty{\mathfrak p \in \Spec A \mid J \subseteq \mathfrak p} \] +\end{definition} +\begin{proposition} + The sets \( \mathbb V(J) \subseteq \Spec A \) ranging over all ideals \( J \trianglelefteq A \) form the closed sets of a topology. +\end{proposition} +This topology is called the \emph{Zariski topology} on \( A \). +\begin{proof} + We have \( \varnothing = \mathbb V(1) \) and \( \Spec A = \mathbb V(0) \), so they are closed. + Note that + \[ \mathbb V\qty(\sum_\alpha I_\alpha) = \bigcap_\alpha \mathbb V(I_\alpha) \] + It remains to show \( \mathbb V(I_1) \cup \mathbb V(I_2) = \mathbb V(I_1 \cap I_2) \). + The containment \( \mathbb V(I_1) \cup \mathbb V(I_2) \subseteq \mathbb V(I_1 \cap I_2) \) is clear. + Conversely, note \( I_1 I_2 \subseteq I_1 \cap I_2 \). + If \( I_1 \cap I_2 \subseteq \mathfrak p \), then by primality of \( \mathfrak p \), either \( I_1 \subseteq \mathfrak p \) or \( I_2 \subseteq \mathfrak p \). +\end{proof} +\begin{example} + Consider \( \Spec \mathbb C[X, Y] \). + The point \( (0) \in \Spec \mathbb C[X, Y] \) is dense in the Zariski topology, so \( \overline{\qty{(0)}} = \Spec \mathbb C[X, Y] \). + This is because all prime ideals in integral domains contain the zero ideal. + \( (0) \) is sometimes called the \emph{generic point}. + + Consider the prime ideal \( (Y^2 - X^3) \), and consider a maximal ideal \( \mathfrak m_{a,b} = (X - a, Y - b) \) corresponding to the point \( (a, b) \). + Then one can show that + \[ \mathfrak m_{a,b} \in \overline{\qty{(Y^2 - X^3)}} \iff b^2 = a^3 \] + In general, points are not closed. +\end{example} +% TODO: lowercase variables? + +\subsection{Functions on open sets} +\begin{definition} + Let \( f \in A \). + Define the \emph{distinguished open} corresponding to \( f \) to be + \[ U_f = \Spec A \setminus \mathbb V(f) \] +\end{definition} +\begin{example} + \begin{enumerate} + \item Let \( A = \mathbb C[x] \), and recall that \( \Spec A \) is \( \mathbb C \cup \qty{(0)} \), where the complex number \( a \) represents the maximal ideal \( (x - a) \). + Let \( f = x \) and consider + \[ \mathbb V(x) = \qty{\mathfrak p \mid x \in \mathfrak p} = \qty{(x)} \] + Hence \( U_x = \Spec A \setminus \qty{(x)} \), which is \( \Spec A \) without the complex number 0. + \item More generally, suppose we fix \( a_1, \dots, a_r \in \mathbb C \). + Then + \[ U = \Spec A \setminus \qty{(x - a_i)}_{i=1}^r = U_f;\quad f = \prod_{i=1}^r (x-a_i) \] + \end{enumerate} +\end{example} +\begin{lemma} + The distinguished opens \( U_f \), taken over all \( f \in A \), form a basis for the Zariski topology on \( \Spec A \); that is, every open set in \( \Spec A \) is a union of some collection of the \( U_f \). +\end{lemma} +% Proof: ex sheet 1 + +\subsection{Localisations} +\begin{definition} + Let \( f \in A \). + The \emph{localisation} of \( A \) at \( f \) is + \[ A_f = \faktor{A[x]}{(xf - 1)} \] +\end{definition} +Informally, we adjoin \( \frac{1}{f} \) to \( A \). +\begin{lemma} + The distinguished open \( U_f \subseteq \Spec A \) is naturally homeomorphic to \( \Spec A_f \). +\end{lemma} diff --git a/iii/cat/01_definitions_and_examples.tex b/iii/cat/01_definitions_and_examples.tex index caa93fa..f23a512 100644 --- a/iii/cat/01_definitions_and_examples.tex +++ b/iii/cat/01_definitions_and_examples.tex @@ -207,3 +207,114 @@ \subsection{Natural transformations} This is a natural transformation \( \pi_n \to H_n U \) where \( U \) is the forgetful functor \( \mathbf{Top}_\star \to \mathbf{Top} \). \end{enumerate} \end{example} + +\subsection{Equivalence of categories} +There is a notion of isomorphism of categories, namely, isomorphism in the category \( \mathbf{Cat} \). +For example, \( \mathbf{Rel} \cong \mathbf{Rel}^\cop \) via the functor +\[ A \mapsto A;\quad R \mapsto R^\circ = \qty{(b, a) \mid (a, b) \in R} \] +However, there is a weaker notion that is often more useful in practice, called equivalence. +To define this, we need a notion of `natural isomorphism'. +There are two obvious definitions, which we show are equivalent. +\begin{lemma} + Let \( \alpha : F \to G \) be a natural transformation between functors \( \mathcal C \rightrightarrows \mathcal D \). + Then \( \alpha \) is an isomorphism in the functor category \( [\mathcal C, \mathcal D] \) if and only if each component \( \alpha_A \) is an isomorphism in \( \mathcal D \). +\end{lemma} +\begin{proof} + The forward direction is clear as composition in \( [\mathcal C, \mathcal D] \) is pointwise; if \( \beta \) is an inverse for \( \alpha \), then \( \beta_A \) is an inverse for \( \alpha_A \). + Suppose \( \beta_A \) is an inverse for \( \alpha_A \) for each \( A \). + We show the \( \beta \) collectively form a natural transformation by verifying the naturality squares. + Given \( f : A \to B \) in \( \mathcal C \), consider + \[\begin{tikzcd} + FA & FB \\ + GA & GB + \arrow["Ff", from=1-1, to=1-2] + \arrow["{\alpha_A}", shift left=2, from=1-1, to=2-1] + \arrow["Gf"', from=2-1, to=2-2] + \arrow["{\alpha_B}"', shift right=2, from=1-2, to=2-2] + \arrow["{\beta_B}"', shift right=2, from=2-2, to=1-2] + \arrow["{\beta_A}", shift left=2, from=2-1, to=1-1] + \end{tikzcd}\] + Then + \[ (Ff)\beta_A = \beta_B \alpha_B (Ff) \beta_A = \beta_B(Gf) \alpha_A \beta_A = \beta_B (Gf) \] + using naturality of \( \alpha \). + Thus \( \beta \) is natural, and an inverse for \( \alpha \). +\end{proof} +\begin{definition} + Let \( \mathcal C, \mathcal D \) be categories. + An \emph{equivalence} between \( \mathcal C \) and \( \mathcal D \) is a pair of functors + \[ F : \mathcal C \to \mathcal D;\quad G : \mathcal D \to \mathcal C \] + and a pair of natural isomorphisms + \[ \alpha : 1_{\mathcal C} \to GF;\quad \beta : FG \to 1_{\mathcal D} \] + If \( \mathcal C \) and \( \mathcal D \) are equivalent, we write \( \mathcal C \simeq \mathcal D \). +\end{definition} +The reason the natural isomorphisms point in opposite directions will be clarified later. +A property \( P \) of categories that is called \emph{categorical} if whenever \( \mathcal C \) satisfies \( P \) and \( \mathcal C \simeq \mathcal D \), then \( \mathcal D \) satisfies \( P \). +For example, the properties of being a preorder or being a groupoid are categorical. +Being a partial order or being a group are not categorical. +Generally, properties that rely on equality of objects, not isomorphism, will not be categorical. +\begin{example} + \begin{enumerate} + \item Let \( \mathbf{Set}_\star \) be the category of pointed sets and functions preserving the base point. + Then \( \mathbf{Set}_\star \simeq \mathbf{Part} \) by + \[ F : \mathbf{Set}_\star \to \mathbf{Part};\quad F(A, a) = A \setminus \qty{a};\quad F((A, a) \xrightarrow f (B, b))(x) = f(x) \] + and + \[ G : \mathbf{Part} \to \mathbf{Set}_\star;\quad G(A) = A \cup \qty{A};\quad G(A \xrightarrow f B \text{ partial})(x) = \begin{cases} + f(x) & \text{if } f \text{ is defined at } x \\ + V & \text{otherwise} + \end{cases} \] + Note that \( FG = 1_{\mathbf{Part}} \), but \( GF \) is not equal to \( 1_{\mathbf{Set}_\star} \). + It is not possible for these two categories to be isomorphic, because there is an isomorphism class of \( \mathbf{Part} \) that has only one member, namely \( \qty{\varnothing} \), but this cannot occur in \( \mathbf{Set}_\star \). + \item Let \( \mathbf{fdVect}_k \) be the category of finite-dimensional vector spaces over \( k \). + This category is equivalent to its opposite category \( \mathbf{fdVect}_k^\cop \) via the dual space functors in both directions. + The natural isomorphisms \( \alpha \) and \( \beta \) are both as in the double dual example given above. + \item We show \( \mathbf{fdVect}_k \simeq \mathbf{Mat}_k \). + Define + \[ F : \mathbf{Mat}_k \to \mathbf{fdVect}_k;\quad F(n) = k^n \] + and sending a matrix \( A \) to the linear map it represents in the standard basis. + For each finite-dimensional vector space \( V \), choose a particular basis. + Define + \[ G : \mathbf{fdVect}_k \to \mathbf{Mat}_k;\quad G(V) = \dim V \] + and let \( G(\theta) \) be the matrix representing \( \theta \) with respect to the particular bases chosen above. + Then \( GF = 1_{\mathbf{Mat}_k} \), as long as we chose the bases above in such a way that the \( k^n \) have the standard basis. + Further, \( FG \) is naturally isomorphic to \( 1_{\mathbf{fdVect}_k} \), since the chosen bases define isomorphisms \( k^{\dim V} \to V \), which are natural in \( V \). + \end{enumerate} +\end{example} +In line with the idea that we do not want to consider equality of objects but only equality of morphisms, we make the following definitions. +\begin{definition} + Let \( F : \mathcal C \to \mathcal D \) be a functor. + We say that \( F \) is + \begin{enumerate} + \item \emph{faithful}, if for each \( f, g \in \mor \mathcal C \) with equal domain and codomain, \( Ff = Fg \) implies \( f = g \); + \item \emph{full}, if for each \( FA \xrightarrow g FB \), there exists a morphism \( A \xrightarrow f B \) such that \( Ff = g \); + \item \emph{essentially surjective}, if every \( B \in \ob \mathcal D \) is isomorphic to some \( FA \) for \( A \in \ob \mathcal C \). + \end{enumerate} +\end{definition} +Note that if \( F \) is full and faithful, it is \emph{essentially injective}: if \( FA \xrightarrow g FB \) is an isomorphism, the unique \( A \xrightarrow f B \) with \( Ff = g \) is an isomorphism, because its inverse is the unique \( B \to A \) mapped to \( g^{-1} \). +\begin{lemma} + Let \( F : \mathcal C \to \mathcal D \) be a functor. + Then \( F \) is part of an equivalence \( \mathcal C \simeq \mathcal D \) if and only if \( F \) is full, faithful, and essentially surjective. +\end{lemma} +\begin{proof} + Suppose \( G, \alpha, \beta \) make \( F \) into an equivalence. + The existence of \( \beta \) ensures that \( B \simeq FGB \) for any \( B \in \ob \mathcal D \), giving essential surjectivity. + For faithfulness, for any \( A \xrightarrow f B \) in \( \mathcal C \), we have \( f = \alpha_B^{-1} (GFf) \alpha_A \), allowing us to reproduce \( f \) from its domain, codomain, and image under \( F \). + For fullness, consider \( FA \xrightarrow g FB \), and define \( f = \alpha_B^{-1} (Gg) \alpha_A : A \to B \). + Then, \( GFf = Gg \). + As \( G \) is faithful by symmetry, \( Ff = g \). + + For the converse, for each object \( B \in \mathcal D \), we choose an isomorphism \( \beta_B : FGB \to B \), thus defining the action of \( G \) on objects. + Then we define \( G \) on morphisms by letting \( G(B \xrightarrow g C) \) be the unique \( GB \to GC \) whose image under \( F \) is + \[ FGB \xrightarrow{\beta_B} B \xrightarrow g C \xrightarrow{\beta_C^{-1}} FGC \] + This is functorial: given \( h : C \to D \), we can form \( G(hg) \) and \( (Gh)(Gg) \) which have the same image under \( F \), so must be equal. + By construction, \( \beta \) is a natural isomorphism \( FG \to 1_{\mathcal D} \). + It suffices to construct the natural isomorphism \( \alpha : 1_{\mathcal C} \to GF \). + Its component at \( A \), is the unique isomorphism whose image under \( F \) is + \[ FA \xrightarrow{\beta_{FA}} FGFA \] + The naturality squares for \( \alpha \) are mapped by \( F \) to naturality squares for \( \beta^{-1} \), so they commute. +\end{proof} +We call a subcategory full if its inclusion functor is full. +\begin{definition} + A category of \emph{skeletal} if every isomorphism class has a single member. + A \emph{skeleton} of \( \mathcal C \) is a full subcategory \( \mathcal C' \) containing exactly one object for each isomorphism class. +\end{definition} +Note that an equivalence of skeletal categories is bijective on objects, and hence is an isomorphism of categories. diff --git a/iii/mtncl/01_substructures.tex b/iii/mtncl/01_substructures.tex index 8e577d1..6c39059 100644 --- a/iii/mtncl/01_substructures.tex +++ b/iii/mtncl/01_substructures.tex @@ -19,6 +19,7 @@ \subsection{Homomorphisms and substructures} \end{definition} \begin{definition} If \( \mathcal M \subseteq \mathcal N \) and the inclusion map is an \( \mathcal L \)-homomorphism, we say that \( \mathcal M \) is a \emph{substructure} of \( \mathcal N \), and that \( \mathcal N \) is an \emph{extension} of \( \mathcal M \). + We will typically use the notation \( \mathcal M \subseteq \mathcal N \) to indicate that \( \mathcal M \) is a substructure of \( \mathcal N \) when both are \( \mathcal L \)-structures, not just that it is a subset. \end{definition} \begin{example} \begin{enumerate} @@ -176,3 +177,84 @@ \subsection{Tarski--Vaught test} It now remains to show that the substructure \( \mathcal M \) of \( \mathcal N \) is elementary. This follows from induction over the number of quantifiers in formulae, noting that the truth values of quantifier-free formulae are always preserved under any extension. \end{proof} + +\subsection{Universal theories and the method of diagrams} +\begin{definition} + A formula \( \varphi \) is \emph{universal} if it is of the form \( \forall \vb x.\, \psi(\vb x, \vb y) \) where \( \psi \) is quantifier-free. + A theory is \emph{universal} if all its axioms are universal sentences. +\end{definition} +\begin{definition} + Let \( \mathcal N \) be an \( \mathcal L \)-structure. + We define the \emph{diagram} of \( \mathcal N \) to be the set + \[ \operatorname{Diag} \mathcal N = \qty{\varphi(n_1, \dots n_k) \mid \varphi \text{ is a quantifier-free } \mathcal L_{\mathcal N} \text{ formula}, \mathcal N \vDash \varphi(n_1, \dots, n_k)} \] + The \emph{elementary diagram} of \( \mathcal N \) is + \[ \operatorname{Diag}_{\text{el}} \mathcal N = \qty{\varphi(n_1, \dots n_k) \mid \varphi \text{ is an } \mathcal L_{\mathcal N} \text{ formula}, \mathcal N \vDash \varphi(n_1, \dots, n_k)} \] +\end{definition} +The diagram of a group is a slight generalisation of its multiplication table. +Note that a model of a diagram is the same as an extension, and a model of an elementary diagram is the same as an elementary extension. +\begin{lemma} + Let \( \mathcal T \) be a consistent theory, and let \( \mathcal T_\forall \) for the theory of universal sentences proven by \( \mathcal T \). + If \( \mathcal N \) is a model of \( \mathcal T_\forall \), then \( \mathcal T \cup \operatorname{Diag} {\mathcal N} \) is consistent. +\end{lemma} +\begin{proof} + Suppose \( \mathcal T \cup \operatorname{Diag} {\mathcal N} \) is inconsistent. + As \( \mathcal T \) is consistent, by compactness there must be a finite number of sentences in the diagram \( \operatorname{Diag} {\mathcal N} \) that are inconsistent with \( \mathcal T \). + Taking the conjunction, we can reduce to the case where there is a single sentence \( \varphi(\vb n) \) that is inconsistent with \( \mathcal T \). + Then as \( \mathcal T \cup \qty{\varphi(\vb n)} \) is inconsistent, \( \mathcal T \vdash \neg\varphi(\vb n) \). + Since \( \mathcal T \) has nothing to say about the new constants \( \vb n \), we must in fact have \( \mathcal T \vdash \forall \vb x.\, \neg \varphi(\vb x) \). + This is a universal consequence of \( \mathcal T \), so by assumption \( \mathcal N \) models it, giving a contradiction. +\end{proof} +\begin{corollary}[Tarski, \L{}o\'s] + An \( \mathcal L \)-theory \( \mathcal T \) has a universal axiomatisation if and only if it is preserved under substructures. + That is, if \( \mathcal M \subseteq \mathcal N \) are substructures and \( \mathcal M \vDash \mathcal T \) then \( \mathcal N \vDash \mathcal T \). + Dually, a theory has an existential axiomatisation if and only if it is preserved under extensions. +\end{corollary} +\begin{proof} + One direction is clear. + Suppose \( \mathcal T \) is preserved under taking substructures. + If \( \mathcal N \vDash \mathcal T \), then \( \mathcal N \vDash \mathcal T_\forall \); we show that the converse also holds. + By the previous proposition, \( \mathcal T \cup \operatorname{Diag} \mathcal N \) is consistent. + Let \( \mathcal N^\star \) be a model of this theory. + So \( \mathcal N^\star \) is an extension of \( \mathcal N \), and also models \( \mathcal T \). + But as \( \mathcal T \) is preserved under substructures, \( \mathcal N \) must model \( \mathcal T \). +\end{proof} +We can show much more with the same method. +\begin{theorem}[elementary amalgamation theorem] + Let \( \mathcal M, \mathcal N \) be \( \mathcal L \)-structures, and \( \vb m \in \mathcal M, \vb n \in \mathcal N \) be tuples of the same size such that \( (\mathcal M, \vb m) \equiv (\mathcal N, \vb n) \). + Then there is an elementary extension \( \mathcal K \) of \( M \) and an elementary embedding \( g : \mathcal N \hookrightarrow \mathcal K \) mapping each \( n_i \) to \( m_i \). +\end{theorem} +\begin{proof} + Replacing \( \mathcal N \) with an isomorphic copy if required, we can assume \( \vb m = \vb n \). + We show that the theory + \[ \mathcal T = \operatorname{Diag}_{\text{el}} \mathcal M \cup \operatorname{Diag}_{\text{el}} \mathcal N \] + is consistent, using compactness. + Suppose that \( \Phi \) is a finite subset of sentences in \( \mathcal T \), which of course includes only finitely many sentences in \( \operatorname{Diag}_{\text{el}} \mathcal N \). + Let the conjunction of those sentences be written as \( \varphi(\vb m, \vb k) \), where \( \varphi(\vb x, \vb y) \) is an \( \mathcal L_{\mathcal N} \)-formula, and \( \vb k \) are pairwise distinct elements of \( \mathcal N \setminus \vb m \). + If \( \Phi \) is inconsistent, then \( \operatorname{Diag}_{\text{el}} \mathcal M \vdash \neg \varphi(\vb m, \vb k) \). + Since the elements of \( \vb k \) are distinct and not in \( \mathcal M \), we in fact have \( \operatorname{Diag}_{\text{el}} \mathcal M \vdash \forall \vb y.\, \neg \varphi(\vb m, \vb y) \). + % TODO: why not in \mathcal M? why not just not in \vb m? + In particular, \( (\mathcal M, \vb m) \vDash \forall y.\, \neg \varphi(\vb m, \vb y) \). + By hypothesis, \( (\mathcal N, \vb n) \vDash \forall y.\, \neg \varphi(\vb m, \vb y) \). + This is a contradiction, as \( \varphi(\vb m, \vb k) \in \operatorname{Diag}_{\text{el}} \mathcal N \). + Hence \( \mathcal T \) is consistent. + Take \( \mathcal K \) to be the \( \mathcal L \)-reduct of a model of \( \mathcal T \). +\end{proof} +We can also use this technique to constrain the size of a model. +\begin{theorem}[L\"owenheim--Skolem theorem] + Let \( \mathcal M \) be an infinite \( \mathcal L \)-structure. + Let \( \kappa \geq \abs{\mathcal L} \) be an infinite cardinal. + Then, + \begin{enumerate} + \item if \( \kappa < \abs{\mathcal M} \), there is an elementary substructure of \( \mathcal M \) of size \( \kappa \); + \item if \( \kappa > \abs{\mathcal M} \), there is an elementary extension of \( \mathcal M \) of size \( \kappa \). + \end{enumerate} +\end{theorem} +We postpone the proof of part (i). +\begin{proof} + Expand the language \( \mathcal L \) by adding constant symbols for each \( m \in \mathcal M \) and \( c \in \kappa \). + Let + \[ \mathcal T = \operatorname{Diag}_{\text{el}} \mathcal M \cup \bigcup_{c \neq c' \in \kappa} \qty{\neg(c = c')} \] + \( \mathcal T \) has a model by compactness, and this model must be an elementary extension of \( \mathcal M \) with size at least \( \kappa \). + We then apply the downward L\"owenheim--Skolem theorem if necessary to obtain a model of size exactly \( \kappa \). +\end{proof} +For example, if \( \mathcal L \) is countable, every infinite \( \mathcal L \)-structure has a countable elementary substructure. diff --git a/iii/mtncl/02_universal_theories.tex b/iii/mtncl/02_universal_theories.tex deleted file mode 100644 index 72b7754..0000000 --- a/iii/mtncl/02_universal_theories.tex +++ /dev/null @@ -1,5 +0,0 @@ -\subsection{Definition} -\begin{definition} - A formula \( \varphi \) is \emph{universal} if it is of the form \( \forall \vb x.\, \psi(\vb x, \vb y) \) where \( \psi \) is quantifier-free. - A theory is \emph{universal} if all its axioms are universal sentences. -\end{definition} diff --git a/iii/mtncl/main.tex b/iii/mtncl/main.tex index 2b04c9c..a9e8b15 100644 --- a/iii/mtncl/main.tex +++ b/iii/mtncl/main.tex @@ -12,8 +12,6 @@ \section{Substructures} \input{01_substructures.tex} -\section{Universal theories} -\input{02_universal_theories.tex} \end{document} diff --git a/util.sty b/util.sty index 3d24fb2..3aaaacc 100644 --- a/util.sty +++ b/util.sty @@ -213,6 +213,7 @@ \DeclareMathOperator{\ob}{ob} \DeclareMathOperator{\mor}{mor} \DeclareMathOperator{\cod}{cod} +\DeclareMathOperator{\Spec}{Spec} % https://github.com/wspr/unicode-math/issues/457 \AtBeginDocument{% \newcommand{\dashrightarrow}{\mathrel{\rightdasharrow}}