For a polynomial with $n$ distinct roots, the criterion isWrite $\delta ={\prod }_{i<j}({\alpha }_i-{\alpha }_j)\ne 0$. Part (a) gives $\sigma \delta =\mathrm{sgn}(\sigma )\delta$. If every Galois permutation is even, $\delta$ is fixed by the entire Galois group and therefore belongs to its fixed field $\mathbb{Q}$. Thus ${\Delta }_f={\delta }^2$ is a rational square. Conversely, if ${\Delta }_f=q^2$ with $q\in {\mathbb{Q}}^{\times }$, then ${\delta }^2=q^2$ in a field implies $\delta =\pm q$. Every automorphism fixes this nonzero element, so its permutation sign must be one.

The distinct-root hypothesis is necessary in the statement's intended degree-$n$ permutation action. The source says “arbitrary degree” without stating that hypothesis. If repeated roots were allowed, zero discriminant would carry no sign information: $(X-1{)}^2(X^2-2)$ has discriminant zero, while its nontrivial automorphism swaps $\sqrt{2},-\sqrt{2}$ and fixes the two rational-root positions, an odd permutation. The criterion above applies to the squarefree polynomial, or to the squarefree part with its degree and root action adjusted accordingly.

The discriminant is a symmetric polynomial in the four roots, so the Fundamental theorem of symmetric polynomials expresses it polynomially in the elementary symmetric polynomial. For this depressed quartic they are $e_1=e_2=0$, $e_3=-r$, $e_4=s$. Scaling each root by $t$ multiplies the discriminant by $t^{12}$, while $r,s$ scale by $t^3,t^4$. Thus only weighted-degree-twelve monomials can appear, giving ${\Delta }_f=A{r}^4+B{s}^3$.

For $r=0$, use $f^{\prime }=4X^3$ and the monic discriminant identity ${\Delta }_f=(-1{)}^{n(n-1)/2}{\prod }_i{f}^{\prime }({\alpha }_i)$. At $n=4$ this gives $4^4({\prod }_i{\alpha }_i{)}^3=256s^3$, so $B=256$. For $s=0$, the roots are zero together with those of $g=X^3+r$. The cubic's discriminant is $-\prod g^{\prime }({\beta }_i)=-27r^2$. The extra squared differences from zero multiply this by $(\prod {\beta }_i{)}^2=r^2$, giving $-27r^4$. HenceThese computations are polynomial identities, so they extend to parameter values with repeated roots as well.

The alternating polynomial given by the Vandermonde product is alternating: interchanging two variables changes its sign. One proof writes it as a Vandermonde determinant, up to a fixed sign, and interchanging two variables interchanges the corresponding determinant rows. Since transpositions generate $S_n$,Thus every even permutation fixes $\delta$, while a transposition negates the nonzero formal polynomial, so it is not invariant under all of $S_n$. This uses $n\ge 2$ and characteristic different from two, as in the subsequent rational setting. For $n=1$ both groups are trivial and the last distinction does not arise.

The sum of squared roots is $c_1^2-2c_2$. The sum of their pairwise products isTheir product is $c_3^2$. Thus the required monic polynomial is

Reciprocal roots require $c_3\ne 0$, equivalently every original root is nonzero. Their monic polynomial is the normalized reciprocal polynomialIf $c_3=0$, at least one reciprocal in the source is undefined, so that part has no polynomial interpretation without the nonzero-root qualification.

Scaling every root scales its elementary symmetric polynomials by the corresponding powers of $d$. Therefore the monic polynomial isThis remains valid for $d=0$, when all three roots are zero; deriving it by symmetric coefficients avoids dividing by $d$.

Let $P_1$ and $P_3$ be the sets of primes congruent to one and three modulo four. Both are infinite. For $P_1$, if a finite list contained them all, put $N=(2p_1\cdots p_r{)}^2+1$. Any prime divisor $q$ of this odd number is new and has minus one as a square modulo $q$, hence belongs to $P_1$ by the given criterion. For $P_3$, use $N=4p_1\cdots p_r-1$. It is three modulo four, so at least one of its odd prime divisors is three modulo four; none of the listed primes divides $N$. These are elementary Euclidean proofs of infinitely many primes in each odd residue class modulo four.

On the set of all primes, the cofinite sets together with $P_1$ have the finite intersection property; so do the cofinite sets together with $P_3$. Extend the generated filters to ultrafilters ${\mathcal{U}}_1$ and ${\mathcal{U}}_3$ using the ultrafilter lemma. They are nonprincipal because they contain every cofinite set, including the complement of any singleton. FormEvery sentence of $\Sigma$ has a cofinite truth set, hence belongs to both ultrafilters. The Łoś theorem shows $F_1,F_2\models \Sigma$. In particular they are fields, since each field axiom holds in every factor.

For each fixed positive integer $n$, the sentence $n\cdot 1\ne 0$ holds in ${\mathbb{F}}_p$ unless $p$ divides $n$. Its truth set is cofinite, so both fields satisfy it. Therefore $\boxed{\mathrm{char}{F}_1=\mathrm{char}{F}_2=0}$.

Finally the sentence $\exists x(x^2+1=0)$ holds at exactly the primes in $P_1\cup \{2\}$. This set belongs to ${\mathcal{U}}_1$, whereas its complement contains $P_3$ and belongs to ${\mathcal{U}}_3$. Łoś therefore givesThus minus one is a square in the first characteristic-zero first-order model of $\Sigma$ and is not a square in the second. The construction also shows why eventual truth over all primes need not decide a sentence whose truth splits along two infinite prime subsequences.

The Łoś theorem states that, for every first-order formula $\phi (\mathbf{x})$ and product representatives ${\mathbf{a}}^1,\dots ,{\mathbf{a}}^r$,It extends the defining relation-symbol rule to arbitrary first-order formulas, including quantified ones.

Identify the countable index set with $\mathbb{N}$. A nonprincipal ultrafilter on it contains every cofinite set. Let $p(x)$ be a type over a finite parameter set in the ultraproduct $\mathcal{M}$, finitely satisfiable in $\mathcal{M}$. Because the language and parameter set are countable, enumerate it as ${\phi }_0(x),{\phi }_1(x),\dots$, with chosen product representatives for the parameters. By finite satisfiability and Łoś,Make these large sets decrease while forcing their intersection to be empty:For each index $i$, let $r(i)$ be the largest $n\le i$ with $i\in Y_n$, if one exists. Choose $b_i\in A_i$ satisfying the first $r(i)+1$ first-order formulas; this is possible because $i\in X_{r(i)}$. If no such $n$ exists, choose any $b_i$.

For each fixed $n$, every $i\in Y_n$ has $r(i)\ge n$, so $b_i$ satisfies ${\phi }_n$. Its coordinate truth set contains $Y_n$ and belongs to $\mathcal{U}$. Another use of Łoś gives $\mathcal{M}\models {\phi }_n([\mathbf{b}])$ for every $n$. Thus $[\mathbf{b}]$ realizes the type. This proves the countable saturation of an ultraproduct over omega, and in particular the ultraproduct is omega-saturated. The construction chooses a finite number of requirements at each coordinate; it does not presume the whole infinite type has a witness in any factor.

Write the ultrafilter as $\mathcal{U}$ to distinguish it from first-order structure domains. On the direct product ${\prod }_{i\in I}{A}_i$, putThe domain of the ultraproduct is the quotient $({\prod }_i{A}_i)/\sim$. Its elements are classes $[\mathbf{a}]$. A constant $c$ is interpreted by $[(c^{{\mathcal{A}}_i}{)}_i]$, and a $k$-ary function symbol byFor a $k$-ary relation symbol,These definitions are independent of representatives: if finitely many input representatives are changed, the coordinates where all old and new inputs agree form a set in the filter. At those coordinates the interpreted outputs and relation truth values agree. Filter closure and upward closure therefore preserve the resulting quotient elements and truth conditions.

A filter on a set $I$ means a proper family $\mathcal{F}\subseteq \mathcal{P}(I)$ with $I\in \mathcal{F}$, $\varnothing \notin \mathcal{F}$, closure under finite intersections, and upward closure under inclusion. An ultrafilter is a maximal proper filter; equivalently, for every $X\subseteq I$, exactly one of $X$ and $I\setminus X$ belongs to it. A principal ultrafilter is $\{X\subseteq I:i\in X\}$ for some $i\in I$. A nonprincipal ultrafilter is one not of that form, equivalently one containing no finite set.

To prove the ultrafilter lemma using Zorn's lemma, order all proper filters containing $\mathcal{F}$ by inclusion. The union of any chain is a proper filter: finitely many members lie together in one filter of the chain, and no chain member contains the empty set. Thus every chain has an upper bound, and Zorn's lemma gives a maximal member $\mathcal{U}$.

For completeness, its decision property follows from maximality. If $X\notin \mathcal{U}$, adjoining $X$ cannot generate a proper filter. Hence some $F\in \mathcal{U}$ has $F\cap X=\varnothing$. Upward closure gives $I\setminus X\in \mathcal{U}$. Both complements cannot be members, since their intersection is empty. Therefore $\boxed{\mathcal{F}\subseteq \mathcal{U}\text{ for an ultrafilter }\mathcal{U}}$. Properness is essential; a family containing the empty set cannot extend to an ultrafilter.

Here fixed means fixed setwise: $\sigma [D]=D$ for every structure automorphism $\sigma$, not that each point must be fixed individually. A parameter-free definable set has this invariance by automorphism invariance of parameter-free definability.

For the converse, first establish the needed saturation implies homogeneity step. In a countable omega-saturated model, a finite partial elementary map can be extended to include any prescribed additional element in its domain. Transfer the element's complete type over the old finite domain along the map. Each finite part of this transported type has a witness because the old map is elementary; the omega-saturated model realizes the full transported type. The inequalities in the type maintain injectivity. Apply the same construction to the inverse map to enlarge the range. Alternating these steps through an enumeration of the countable domain is the back-and-forth method, giving a structure automorphism. In particular, any two elements with the same complete type over the empty set are related by a structure automorphism.

Assume now that $D$ is finite and invariant, and fix $a\in D$. All realizations in $A$ of $p_a=\mathrm{tp}(a/\varnothing )$ belong to $D$, by the preceding homogeneity and the invariance assumption. Suppose that no parameter-free first-order formula ${\theta }_a\in p_a$ has all its realizations in $D$. Then every finite subset ofis satisfiable in $A$: conjoin the finitely many first-order formulas from $p_a$, and use the assumed absence of a defining neighbourhood inside $D$ to find a witness outside $D$. This is a partial type over the finite parameter set $D$. The omega-saturated model would realize it by an element outside $D$ with the same empty-set type as $a$, a contradiction. Hence such a ${\theta }_a$ does exist.

There are only finitely many $a\in D$, soFor $D=\varnothing$, use $x\ne x$. Thus finite invariant sets in a countable saturated structure are definable. Finiteness is used twice: the extra parameters form a finite set, and the final disjunction is a first-order formula rather than an infinite disjunction.

An omega-saturated model realizes every complete one-type over every finite subset of its domain that is consistent with the first-order structure's first-order theory with those named parameters. Equivalently, every partial type in a finite tuple over finitely many parameters that is finitely satisfiable in the first-order structure has a realization there. The equivalence follows by extending a partial type to a complete type; finite tuples can be handled one coordinate at a time.

For an example, take a countably infinite set in the language of equality alone. Over finitely many named elements, a complete one-type either requires equality with one of them or inequality with all of them. Both possibilities are realized. quantifier elimination for an infinite pure set follows by choosing a quantified variable either equal to one of the finitely many named values or distinct from all of them. Indeed every permutation fixing the parameters is a structure automorphism, and all other elements have the same type, so there are no further one-types. This first-order structure is omega-saturated.

For a counterexample, take $(\mathbb{N};0,S)$, where $S(n)=n+1$. The parameter-free partial typeis finitely satisfiable: choose an integer beyond the finitely many excluded numerals. It is not realized in the first-order structure, because every natural number is one of those numerals. Any complete extension of this partial type is therefore omitted as well. Thus the standard successor first-order structure is not omega-saturated.

The inclusion $\mathcal{C}\subseteq \mathcal{D}$ is a structure embedding between first-order models of the same complete theory, but it is not an elementary embedding, as the existential partner first-order formula at $u$ shows. Therefore the first-order theory is not model-complete. The facts $\mathcal{B}⪯\mathcal{C}$ and $\mathcal{B}⪯\mathcal{D}$ do not make the intervening inclusion elementary.

Take a countable elementary substructure ${\mathcal{B}}^{\prime }$ of ${\mathcal{A}}^{\prime }$ containing the named constant, using the Downward Lowenheim-Skolem theorem. Its reduct $\mathcal{B}$ has infinitely many singleton and doubleton classes. In particular it contains the partner of each of its non-singleton elements, and its interpretation of $P$ is exactly the singleton-class predicate.

Choose new distinct elements $u,v\notin B$. DefineIn $\mathcal{C}$, make $u$ a new singleton class. In $\mathcal{D}$, make $\{u,v\}$ a new doubleton class; keep every old class of $\mathcal{B}$ unchanged. All three reducts are first-order models of the complete theory, and $\mathcal{B}\subseteq \mathcal{C}\subseteq \mathcal{D}$.

Expand $\mathcal{C},\mathcal{D}$ by interpreting $P$ as singleton-class membership and retaining the original named constant from $B$. The resulting countable first-order structures ${\mathcal{C}}^{\prime },{\mathcal{D}}^{\prime }$ are each isomorphic to ${\mathcal{B}}^{\prime }$: all have infinitely many classes of each size, and the bijections of classes can be chosen to preserve the constant and its class. Hence they are first-order models of $\mathrm{Th}({\mathcal{A}}^{\prime })$. Both inclusions ${\mathcal{B}}^{\prime }\subseteq {\mathcal{C}}^{\prime }$ and ${\mathcal{B}}^{\prime }\subseteq {\mathcal{D}}^{\prime }$ are genuine structure embeddings, since no class meeting $B$ was changed. The given quantifier elimination, together with the preceding implication to first-order model completeness, yieldsTaking reducts gives $\boxed{\mathcal{B}⪯\mathcal{C},\mathcal{B}⪯\mathcal{D}}$. Butso $\boxed{\mathcal{C}⋠\mathcal{D}}$. There is no conflict with quantifier elimination in the expanded language: $P(u)$ changes truth value, so ${\mathcal{C}}^{\prime }$ is not a substructure of ${\mathcal{D}}^{\prime }$.

No. Let $S$ be the displayed infinite axiom set. Any finite subset $S_0\subseteq S$ mentions only finitely many of the two infinitude requirements. Choose $N\ge 1$ at least as large as every index mentioned. A finite equivalence relation with exactly $N$ singleton classes and $N$ doubleton classes satisfies $S_0$, but is not a first-order model of the complete theory.

If a finite axiomatization existed, let $\sigma$ be the conjunction of its axioms. Then $S\models \sigma$, so the compactness theorem gives a finite $S_0\subseteq S$ with $S_0\models \sigma$. Every first-order model of $S_0$ would then satisfy the full first-order theory, contradicting the finite first-order model above. This is the finite subaxiomatization from an equivalent finite theory argument: finite axiomatizability would force one finite fragment of the explicit axioms to suffice.

WriteThese are first-order formulas in the original language, not new predicate symbols. An axiomatization consists of the equivalence relation axioms for $E$, the boundand, for every $n\ge 1$, the two sentencesThe first scheme supplies infinitely many singleton classes. The second supplies infinitely many distinct non-singleton classes; the size bound makes these doubletons.

Every countable first-order model has countably infinitely many classes of each size. A bijection between singleton classes, and another between doubleton classes followed by a bijection inside each matched pair, gives an isomorphism between any two countable first-order models. To infer completeness, suppose a sentence had different truth values in two arbitrary first-order models. The Downward Lowenheim-Skolem theorem supplies countable elementary substructures of those first-order models, which would still disagree on the sentence, contradicting the isomorphism just constructed. Thus all first-order models satisfy the same sentences. Since the original first-order structure satisfies the axioms, these axioms form a complete axiomatization of its first-order theory. This is the theory of infinitely many singleton and doubleton equivalence classes.

A structure embedding preserves and reflects all atomic formulas, hence all quantifier-free formulas by induction on Boolean combinations. If $T$ has quantifier elimination, replace any $\phi (\mathbf{x})$ by a quantifier-free $\theta (\mathbf{x})$ equivalent over $T$. For $\mathcal{A}\subseteq \mathcal{B}\models T$ and a tuple $\mathbf{a}$ from $A$,Thus $\mathcal{A}⪯\mathcal{B}$, and the preceding criterion proves $\boxed{\text{quantifier elimination implies model completeness}.}$

If $T$ is model-complete, the inclusion of a substructure of a first-order structure $\mathcal{A}\subseteq \mathcal{B}$, both first-order models of $T$, is a structure embedding and hence elementary. Therefore $\mathcal{A}⪯\mathcal{B}$.

Conversely, assume every such inclusion is elementary, and take any structure embedding $f:\mathcal{A}\to \mathcal{B}$ between first-order models of $T$. Its image $f[A]$, with the induced first-order structure, is a substructure of $\mathcal{B}$ isomorphic to $\mathcal{A}$, so it is itself a first-order model of $T$. By hypothesis $f[A]⪯\mathcal{B}$. The isomorphism $\mathcal{A}\to f[A]$ preserves all first-order formulas, and composing it with this elementary inclusion shows that $f$ is an elementary embedding. Thus $\boxed{T\text{ is model-complete}⟺(\mathcal{A}\subseteq \mathcal{B}\models T\Rightarrow \mathcal{A}⪯\mathcal{B})}$.

A model-complete theory is a first-order theory for which every structure embedding between two of its first-order models is an elementary embedding. Thus, if $f:\mathcal{A}\hookrightarrow \mathcal{B}$ is such an embedding, then for every first-order formula $\phi (\mathbf{x})$ and tuple $\mathbf{a}$ from $A$,first-order model completeness does not assert that every two first-order models are elementarily equivalent: it constrains embeddings that exist between first-order models, whereas completeness constrains all first-order models.

A first-order theory has quantifier elimination if every first-order formula is equivalent over the first-order theory to a quantifier-free formula with the same free variables:Logical truth and falsity may be treated as the empty Boolean combinations when considering sentences in a language without closed terms. The assertion concerns first-order formulas with free variables, not merely sentences.