Suppose I have a module $M$ being acted on by a group $G$.
- $0 + a = a$
- $a + 0 = a$
- $--a = a$
- $a - a = 0$
- $a + (b + c) = (a + b) + c$
- $a + b = b + a$
- $r(a+b) = ra + rb$
- $(r+s)(a) = ra + sa$
- $(-r)(a) = -ra$
- $(rs)(a) = rsa$
- $rsa = sra$
- $g(ra + sb) = gra + gsb$
- $(gh)(ra) = g(hra)$
Let $M$ be $\mathbb{H}$ as a real module.
Let $G$ be $\mathbb{H} \setminus \{0\}$ as a group.
$G$-module.
modal formula: $\lozenge^{(n+1)}p \to \lozenge^n p$ enforces a maximum path length of $n$.
modal formula: $\left(\bigwedge_{1 \le i \le (n+1)}\lozenge p_i\right) \to \lozenge \left( \bigvee_{1 \le i \neq j \le (n+1)} p_i \land p_j \right)$
The above enforces a maximum degree of $n$
Let $M$ be a simple $R$-module. Let $\Lambda$ be the lattice of submodules of $M$.
We can define a function $f : \Lambda \to \Lambda$ as:
$$ f(N) = \{x_1 : x \in N\} \times \{x_2 : x \in N\} \times \cdots \times \{x_n : x \in N \} $$
I.e. $f$ takes an ideal and sends it to the product of its components.
What are the ideals of non-associative rings?
I'm looking for an example of a non-UFD GCD besides the algebraic integers and the ring of entire functions on $\mathbb{C}$ or the monoid ring $(k, \mathbb{Q}_{\ge 0})$.
I'm curious whether this example works.
Consider the ring $R = \mathbb{Z}[a,b,c]/(b^2-a, c^2-a)$. The idea being that $a$ is a propertyless element with two square roots, $b$ and $c$.
$R$ is not a UFD. $a$ factors as $b \cdot b$ and as $c \cdot c$ and $b$ and $c$ are not equal.
Acting on a complex number:
$k$ is a real number.
$i$ is $\sqrt{-1}$.
$T_z$ is a shift operator.
$I_z$ is a shift by $iz$.
$F$ is conjugation.
- $k \triangleright (a+ib) = ka + ikb$
- $i \triangleright (a+ib) = -b + ia$
- $T_{k} \triangleright (a+ib) = (k+a)+ib$
- $I_{k} \triangleright (a+ib) = a + i(b+k) $
- $F \triangleright (a+ib) = a - ib$
- $g - h \triangleright (a+ib) = (g \triangleright a+ib) - (h \triangleright a+ib)$
- $gh \triangleright (a+ib) = g \triangleright h \triangleright (a+ib) $
Definition of higher-level operators, rotations and reflections.
- $R_{p, \theta}$ is $T_{p} R_\theta T_{-p}$.
- $X_{0, 0}$ is $F$.
- $X_{p, 0}$ is $T_p F T_{-p}$
- $X_{p, \theta}$ is $T_p R_{\theta} F R_{-\theta} T_{-p}$
Let a matrix $M$ be quasisymmetric if there exists permutation matries $U$ and $V$ such that $UMV = (UMV)^T$.
If I look at character tables of finite cyclic groups, it appears that the character table, considered as a matrix, if quasisymmetric when the group has prime order.
Some groups like $\mathbb{Z}_2 \times \mathbb{Z}_2$ are also quasisymmetric, but they are not cyclic and I'm only interested in cyclic groups at the moment.
Consider noncommutative, nonunital rings $R$ where the underlying group $(R, +)$ is cyclic.
If $n$ is finite, there is one isomorphism class of ring $R$ with a cycligc additive grouop for each divisor of $n$.
How do you characterize the rings with exactly one cool ideal.
Let's define a cool ideal $I \subset R$ as a prime ideal that is not principal and is generated by two elements.
So, in the ring $\mathbb{C}[x, y]$, $(x, y)$ is a cool ideal. However, it is not the only one, since $(x, y+1)$ is also a cool ideal.
The ring $\mathbb{C}[x, y]/S$, where $S$ is the set of all elements $j(x, y)$ such that $x \not \mid j(x, y)$ and $y \not \mid j(x, y)$, does have exactly one cool ideal. (I think this construction is $\mathbb{C}[x, y]$ localized at $(x, y)$?)
I'm wondering if there's a nice characterization of all rings with exactly one cool ideal.
What follows is what I've done so far to try to solve this question on my own. It consists of:
- A proof that $\mathbb{C}[x, y]$ contains the cool ideal $(x, y)$ .
- A proof that $\mathbb{C}[x, y]$ contains the additional cool ideal $(x, y+1)$ .
- A proof that $\mathbb{C}[x, y]/S$ contains exactly one cool ideal.
Definition 31: polynomial with no constant term
A polynomial $p(\vec{x})$ has no constant term if and only if it is $0$ or has at least one term, but none of the terms are constant.
This definition says very little, I'm just emphasizing that $0$ is considered to have no terms at all instead of having a constant term which is zero.
Lemma 41: $p(x, y)$ is in $(x, y)$ if and only if $p$ has no constant term.
RTL: Suppose $p(x, y)$ has no constant term. We can collect all the terms with at least one $x$ into $f(x)x$ and all the remaining terms into $g(y)y$.
For example, $x^2 + xy + 3y^2 = (x+y)x + (3y)y$.
Thus, $p(x, y)$ is in $(x, y)$. Note that when $p(x, y)$ is $0$, we choose $f(x) = 0$ and $g(y) = 0$.
LTR: Suppose $p(x, y)$ is in $(x, y)$.
Thus, $p(x, y)$ has the form $f(x, y)x + g(x, y)y$. There are no constant terms in $f(x, y)x$ and there are no constant terms in $g(x, y)y$. Therefore $p(x, y)$ has no constant terms, as desired.
End of proof of Lemma 41.
Lemma 51: $(x, y)$ is a cool ideal of $\mathbb{C}[x, y]$.
$(x, y)$ is an ideal.
$(x, y)$ is generated by two elements.
Let's prove that $(x, y)$ is a prime ideal by noting that its complement is multiplicatively closed. Suppose we have not necessarily distinct elements of the complement of $(x, y)$; call them $f(x, y)$ and $g(x, y)$. By Lemma 41, $f(x, y)$ and $g(x, y)$ each have no constant term.
If $a(x, y)$ is a term of $f(x, y)$ and $b(x, y)$ is a term of $g(x, y)$, then $(ab)(x, y)$ is nonconstant because it is a product of two nonconstant terms.
Therefore $(fg)(x, y)$ has no nonconstant terms and, by Lemma 41, is an element of the complement of $(x, y)$.
Thus the complement of $(x, y)$ is multiplicatively closed and $(x, y)$ is a prime ideal.
Suppose for contradiction that $(x, y)$ is principal and generated by $h(x, y)$. If $h(x, y)$ contains a term that contains $x$, then $h(x, y) \not\mid y$. If $h(x, y)$ contains a term that contains $y$, then $h(x, y) \not\mid x$. If $h(x,y)$ has a constant term, then $(h) = (1) = \mathbb{C}[x, y]$. If $h$ has no terms at all, then $h=0$ which is a contradiction since $(0) \neq (x, y)$.
Lemma 71: $\varphi$ which sends $x$ to $p(x, y)$ and $y$ to $q(x, y)$ is a ring homomorphism.
Every element of $\mathbb{C}[x, y]$ is equivalent to a sum of the form $\sum_{m, n}a_{m,n}x^m y^n$, and, as usual, all but finitely many values of $a_{m,n}$ are zero.
Here is the definition of $\varphi$, based on how it acts on the monomials $x$ and $y$.
$$ \varphi(\sum_{m,n}a_{m,n}x^m y^n) = \sum_{m,n}a_{m,n} \varphi(x)^m \varphi(y)^n $$
$\varphi$ respects $-$.
$$ \varphi((\sum_{m,n}a_{m,n}x^m y^n) - (\sum_{m,n}b_{m,n}x^my^n)) = \varphi(\sum_{m,n}(a-b)_{m,n}x^my^n) = \sum_{m,n} (a-b)_{m,n} \varphi(x)^m \varphi(y)^n = \varphi(\sum_{m,n} a_{m,n} x^m) $$
$\varphi$ respects multiplication.
$$ \varphi((\sum_{m,n}a_{m,n}x^m y^n)(\sum_{k,l} b_{k,l}x^k y^l)) $$
Lemma 81: $\varphi$ preserves ideals when $\varphi$ is a surjective ring homomorphism.
like $I$ be an ideal. I want to show that $\varphi I$ is an ideal.
Suppose $a$ and $b$ are in $I$, then $\varphi(a) + \varphi(b) = \varphi(a + b)$, therefore $\varphi I$ is additively closed.
By hypothesis, $I$ is an ideal. Thus is satisfies $ra \in I$ for each $r$ in $R$ and each $a$ in $I$. Therefore, $\varphi(ra) \in \varphi I$ for each $r$ in $R$ and $i$ and $I$. $\varphi(r)$ hits every element of $S$ by surjectivity of $\varphi$, thus $\varphi I$ is an ideal of $S$.
End of proof of Lemma 81.
Lemma 91: $\varphi$ preserves cool ideals when $\varphi$ is a surjective homomorphism.
TBD.
End of proof of Lemma 91.
Lemma 61: $(x, y+1)$ is a cool ideal of $\mathbb{C}[x, y]$.
The map $\varphi : \mathbb{C}[x, y] \to \mathbb{C}[x, y]$ that sends $x$ to $x$ and $y$ to $y+1$ is surjective.
$$ \varphi((x, y)) = (\varphi(x), \varphi(y)) = (x, y+1) $$
Therefore $(x, y+1)$ is a cool ideal.
End of proof of Lemma 61.
Lemma 101: the elements of $\mathbb{C}[x, y] / S$ has the form $ux^m y^n$ where $u$ is a unit.
Let $p(x, y)$ be a polynomial in $\mathbb{C}[x, y]$ such that $x \not\mid p(x, y)$ and $y \not\mid p(x, y)$ . By construction of $S$, $p(x, y)$ is in $S$. Therefore $\frac{1}{p(x, y)}$ is in $\mathbb{C}[x, y] / S$.
Again by construction of $S$, $x$ and $y$ do not have inverses.
$0$ doesn't have an inverse and isn't expected to. Nonzero constants in $\mathbb{C}$ do have inverses.
Every element of $\mathbb{C}[x, y]/S$ thus has the form $\frac{p(x, y)x^n y^m}{q(x, y)}$ where $p(x, y)$ and $q(x, y)$ are in $S$.
$\frac{p(x, y)}{q(x, y)}$ is also in $S$, therefore an arbitrary element of $\mathbb{C}[x, y]$ has the form $ux^n y^m$.
How do you characterize the rings with exactly one cool ideal.
Let's define a cool ideal $I \subset R$ as a prime ideal that is not principal and is generated by two elements.
So, in the ring $\mathbb{C}[x, y]$, $(x, y)$ is a cool ideal. However, it is not the only one, since $(x, y+1)$ is also a cool ideal.
The ring $\mathbb{C}[x, y]/S$, where $S$ is the set of all elements $j(x, y)$ such that $x \not \mid j(x, y)$ and $y \not \mid j(x, y)$, does have exactly one cool ideal. (I think this construction is $\mathbb{C}[x, y]$ localized at $(x, y)$?)
I'm wondering if there's a nice characterization of all rings with exactly one cool ideal.
What follows is what I've done so far to try to solve this question on my own. It consists of:
- A proof that $\mathbb{C}[x, y]$ contains the cool ideal $(x, y)$ .
- A proof that $\mathbb{C}[x, y]$ contains the additional cool ideal $(x, y+1)$ .
- A proof that $\mathbb{C}[x, y]/S$ contains exactly one cool ideal.
Definition 31: polynomial with no constant term
A polynomial $p(\vec{x})$ has no constant term if and only if it is $0$ or has at least one term, but none of the terms are constant.
This definition says very little, I'm just emphasizing that $0$ is considered to have no terms at all instead of having a constant term which is zero.
Lemma 41: $p(x, y)$ is in $(x, y)$ if and only if $p$ has no constant term.
RTL: Suppose $p(x, y)$ has no constant term. We can collect all the terms with at least one $x$ into $f(x)x$ and all the remaining terms into $g(y)y$.
For example, $x^2 + xy + 3y^2 = (x+y)x + (3y)y$.
Thus, $p(x, y)$ is in $(x, y)$. Note that when $p(x, y)$ is $0$, we choose $f(x) = 0$ and $g(y) = 0$.
LTR: Suppose $p(x, y)$ is in $(x, y)$.
Thus, $p(x, y)$ has the form $f(x, y)x + g(x, y)y$. There are no constant terms in $f(x, y)x$ and there are no constant terms in $g(x, y)y$. Therefore $p(x, y)$ has no constant terms, as desired.
End of proof of Lemma 41.
Lemma 51: $(x, y)$ is a cool ideal of $\mathbb{C}[x, y]$.
$(x, y)$ is an ideal.
$(x, y)$ is generated by two elements.
Let's prove that $(x, y)$ is a prime ideal by noting that its complement is multiplicatively closed. Suppose we have not necessarily distinct elements of the complement of $(x, y)$; call them $f(x, y)$ and $g(x, y)$. By Lemma 41, $f(x, y)$ and $g(x, y)$ each have no constant term.
If $a(x, y)$ is a term of $f(x, y)$ and $b(x, y)$ is a term of $g(x, y)$, then $(ab)(x, y)$ is nonconstant because it is a product of two nonconstant terms.
Therefore $(fg)(x, y)$ has no nonconstant terms and, by Lemma 41, is an element of the complement of $(x, y)$.
Thus the complement of $(x, y)$ is multiplicatively closed and $(x, y)$ is a prime ideal.
Suppose for contradiction that $(x, y)$ is principal and generated by $h(x, y)$. If $h(x, y)$ contains a term that contains $x$, then $h(x, y) \not\mid y$. If $h(x, y)$ contains a term that contains $y$, then $h(x, y) \not\mid x$. If $h(x,y)$ has a constant term, then $(h) = (1) = \mathbb{C}[x, y]$. If $h$ has no terms at all, then $h=0$ which is a contradiction since $(0) \neq (x, y)$.
Lemma 71: $\varphi$ which sends $x$ to $p(x, y)$ and $y$ to $q(x, y)$ is a ring homomorphism.
Every element of $\mathbb{C}[x, y]$ is equivalent to a sum of the form $\sum_{m, n}a_{m,n}x^m y^n$, and, as usual, all but finitely many values of $a_{m,n}$ are zero.
Here is the definition of $\varphi$, based on how it acts on the monomials $x$ and $y$.
$$ \varphi(\sum_{m,n}a_{m,n}x^m y^n) = \sum_{m,n}a_{m,n} \varphi(x)^m \varphi(y)^n $$
$\varphi$ respects $-$.
$$ \varphi((\sum_{m,n}a_{m,n}x^m y^n) - (\sum_{m,n}b_{m,n}x^my^n)) = \varphi(\sum_{m,n}(a-b)_{m,n}x^my^n) = \sum_{m,n} (a-b)_{m,n} \varphi(x)^m \varphi(y)^n = \varphi(\sum_{m,n} a_{m,n} x^m) $$
$\varphi$ respects multiplication.
$$ \varphi((\sum_{m,n}a_{m,n}x^m y^n)(\sum_{k,l} b_{k,l}x^k y^l)) $$
Lemma 81: $\varphi$ preserves ideals when $\varphi$ is a surjective ring homomorphism.
like $I$ be an ideal. I want to show that $\varphi I$ is an ideal.
Suppose $a$ and $b$ are in $I$, then $\varphi(a) + \varphi(b) = \varphi(a + b)$, therefore $\varphi I$ is additively closed.
By hypothesis, $I$ is an ideal. Thus is satisfies $ra \in I$ for each $r$ in $R$ and each $a$ in $I$. Therefore, $\varphi(ra) \in \varphi I$ for each $r$ in $R$ and $i$ and $I$. $\varphi(r)$ hits every element of $S$ by surjectivity of $\varphi$, thus $\varphi I$ is an ideal of $S$.
End of proof of Lemma 81.
Lemma 91: $\varphi$ preserves cool ideals when $\varphi$ is a surjective homomorphism.
TBD.
End of proof of Lemma 91.
Lemma 61: $(x, y+1)$ is a cool ideal of $\mathbb{C}[x, y]$.
The map $\varphi : \mathbb{C}[x, y] \to \mathbb{C}[x, y]$ that sends $x$ to $x$ and $y$ to $y+1$ is surjective.
$$ \varphi((x, y)) = (\varphi(x), \varphi(y)) = (x, y+1) $$
Therefore $(x, y+1)$ is a cool ideal.
End of proof of Lemma 61.
Lemma 101: the elements of $\mathbb{C}[x, y] / S$ has the form $\sum ux^m y^n$ where the elements $u$ are units.
Let $p(x, y)$ be a polynomial in $\mathbb{C}[x, y]$ such that $x \not\mid p(x, y)$ and $y \not\mid p(x, y)$ . By construction of $S$, $p(x, y)$ is in $S$. Therefore $\frac{1}{p(x, y)}$ is in $\mathbb{C}[x, y] / S$.
Again by construction of $S$, $x$ and $y$ do not have inverses.
$0$ doesn't have an inverse and isn't expected to. Nonzero constants in $\mathbb{C}$ do have inverses.
Every element of $\mathbb{C}[x, y]/S$ thus has the form $\frac{p(x, y)x^n y^m}{q(x, y)}$ where $p(x, y)$ and $q(x, y)$ are in $S$.
$\frac{p(x, y)}{q(x, y)}$ is also in $S$, therefore an arbitrary element of $\mathbb{C}[x, y]$ has the form $ux^n y^m$.
Lemma 111: $\mathbb{C}[x, y]/S$ has exactly one cool ideal $(x, y)$.
$(x, y)$ is a prime ideal.
$(x, y)$ is not principal.
Suppose $I$ is a cool ideal. It is generated by $(ux^my^n, vx^ky^l)$.
$u$ and $v$ are both units and don't affect the ideal.
It can't be the case that $m = n = 0$ or likewise for $k = l = 0$ because then the ideal generated would be $(1)$.
Let $F$ be an algebraic signature and $T$ be an equational $F$-theory. Let $A$ be an $F$-structure such that $A \models T$.
Let's call the relation $=$ on $A$ equality and define a proper congruence as one that is not equality.
I want to show that $A$ has a maximal proper congruence.
Let $(\Gamma)$ where $\Gamma$ is a set of pairs be the congruence generated by the pairs $\Gamma$ defined as:
$$ (\Gamma) = \bigcap \{R : R \in \Lambda \;\;\text{and}\;\; \Gamma \subset \Lambda \} $$
The full relation exists and is a congruence so $\{R: \cdots\}$ is never empty.
Let $\alpha$ be an ordinal such that $A_\alpha$ is the elements of $A$ indexed by $\alpha$.
How do you prove that $\sin(1)$ is not algebraic?
Name for extending a propositional logic to a first-order logic with constant and relation symbols.
We need the rules making $=$ an equivalence relation:
$$ \vdash a = a $$
$$ a = b \vdash b = a $$
$$ a = b, b = c \vdash a = c$$
And the following rule making $=$ a congruence.
$$ \vec{x} = \vec{y}, R(\vec{x}) \vdash R(\vec{y}) $$
Determinant algorithm.
$$ \begin{vmatrix} 1 & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \mapsto \begin{vmatrix} \begin{vmatrix} 1 & b \\ d & e \end{vmatrix} & \begin{vmatrix} 1 & c \\ d & f \end{vmatrix} \\ \begin{vmatrix} 1 & b \\ g & h \end{vmatrix} & \begin{vmatrix} 1 & c \\ g & i \end{vmatrix} \end{vmatrix} $$
Every representation besides the identity representation is reducible.
As proof, all permutations of dimension $n$ fix a subspace generated by $\begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix}$.
I'm wondering whether it is the case that any irreducible representation of a finite group can be extended to a representation with dimension one greater. Conversely, I'm also wondering whether I can hit all the irreducible subrepresentations of a finite group by taking subrepresentations of permutation representations, but the $\textbf{NonPermutation} \longrightarrow \textbf{Permutation}$ direction is the focus of this particular question to keep it narrow.
My goal here, eventually, is to come up with a heuristic that might help me compute a row or two of the character table of a group by hand by looking at permutation representations that fix a one-dimensional subspace. It is logically a follow-up to this question that I asked in the recent past.
$S_3$ is generated by the derangement $(123)$ and the transposition $(12)$.
We have the sign representation that sends $(12)$ to $\begin{bmatrix} -1 \end{bmatrix}$ and $(123)$ to $\begin{bmatrix} 1 \end{bmatrix}$. This is a 1-dimensional representation.
We can "extend it" to the reducible permutation representation $(12) \mapsto \begin{bmatrix} & 1 \\ 1 & \end{bmatrix}\;\;,\;\; (123) \mapsto \begin{bmatrix} 1 & \\ & 1 \end{bmatrix} $, which has dimension 1 greater.
The faithful representation $(12) \mapsto \begin{bmatrix} & 1 \\ 1 & \end{bmatrix} \;\;,\;\; (123) \mapsto \begin{bmatrix} \omega & \\ & \omega^2 \end{bmatrix}$ with $\omega = \exp\left(\frac{2\pi}{3}\right)$ also has dimension 2.
And it can be extended to $(12) \mapsto \begin{bmatrix} & 1 & \\ 1 & & \\ & & 1 \end{bmatrix} \;\;,\;\; (123) \mapsto \begin{bmatrix} & 1 & \\ & & 1 \\ 1 & & \end{bmatrix}$, which has dimension 1 higher.
Let $R$ be a commutative ring with unity. Let $\Lambda$ be its lattice of ideals.
Let me define a function $f : \Lambda \to \Lambda$ using a transfinite definition.
Let $E_\lambda$ be the elements of $R$ placed in a well-order and indexed by the ordinal $\lambda$.
I will now compute $f(X)$, where $X$ is an ideal. Let $(S)$ be the ideal generated by a set $S$ and likewise for mixtures of sets and elements such as $(S, e)$.
Let $g(0) = X \;\,\text{if}\;\, (E_0, X) = R \;\,\text{else}\;\, (E_0, X)$.
For successor ordinals, let $g(\alpha+1) = g(\alpha) \;\,\text{if}\;\, (E_\alpha, g(\alpha+1)) = R \;\,\text{else}\;\, (E_\alpha, g(\alpha+1))$.
For limit ordinals, let $g(\alpha) = (g(\beta) : \beta < \alpha) \;\,\text{if}\;\, (E_\alpha, (g(\beta) : \beta < \alpha)) = R \;\,\text{else}\;\, (E_\alpha, (g(\beta) : \beta < \alpha))$.
And finally, define $f(X)$ as $g(\lambda)$.
If $X$ is a unit ideal, then $f(X) = X$ since we will never hit an else branch.
If $X$ is not a unit ideal, then $f(X)$ will be a maximal ideal containing $X$. Suppose for contradiction that $f(X)$
I was wondering the other day why interpretations from one structure $\mathfrak{A}$ to another $\mathfrak{B}$ preserve the truth of sentences in $\mathfrak{A}$. I think I have a simple proof of this fact, but it has led me to an expected conclusion.
First, let $E$ be the class of all structures whose domain is $\{c\}$, the denotation of every function and constant symbol is completely determined, but each relation symbol has two possible denotations.
It seems to me that we can always interpret any structure inside a structure within $E$ by sending every element of the domain to $c$.
This interpretation will preserve the truth of sentences in the original structure, but loses information about the relationship between, for example, $\forall x \mathop. \varphi(x) \land \psi(x)$ and $\varphi(x)$ and $\psi(x)$ since it identifies all satisfiable formulas with the same number of free variables.
Basically, my question is about the definition of interpretations and whether we need to or want to rule out interpretations going to a structure in $E$ when defining "just an interpretation" and not a "definable interpretation" or otherwise.
Interpretability is defined in Marker's Model Theory: An Introduction on page 24, but interpretations themselves are not until we get to the notion of definable interpretability. Hodges' a shorter model theory does define an interpretation on page 107 and that definition is similar to the one below; it includes domain formulas and maps formulas to formulas. I'm, however, using the definition from Wikipedia, I'm just changing it slightly to be a map between definable sets subject to certain conditions rather than a map between elements that preserves definability.
Here's the definition from Wikipedia:
An interpretation of a structure $M$ in a structure $M$ with parameters
(or without parameters, respectively) is a pair $(n, f)$ where n is a natural number and $f$ is a surjective map from a subset of $N^n$ onto $M$ such that the $f$-preimage (more precisely the $f^k$ preimage of every $X \subset M^k$ definable in $M$ by a first-order formula without parameters is definable (in $N$) by a first-order formula with parameters (or without parameters, respectively).
The following definition below is intentionally agnostic with respect to whether parameters are allowed or not.
An interpretation in model theory $i : \mathfrak{A} \to \mathfrak{B}$ is a map from the definable sets of $\mathfrak{A}$ to the definable sets of $\mathfrak{B}$ subject to the following constraints below:
- There must exist a natural number $n$ and a function $f : A \to 2^{B^{\,n}}$ such that, for any definable set $X$ of dimension $k$, $i(X)$ is equal to $\cup \{ f(\vec{v}) : \vec{v} \in X\}$, where $f(\vec{v})$ is a tuple of length $kn$ defined by elementwise application of $f$ and then combining the results together using the Cartesian product.
- Destinations must be unique. For any distinct $a, b$ in $A$, $f(a)$ and $f(b)$ must be disjoint.
- Destinations must exist. $f(a)$ is not empty for any $a$.
Theorem: The Wikipedia definition and my definition are equivalent.
LTR: Let $(n, g)$ be an interpretation. Let $f(x)$ be defined as $g^{\leftarrow}(x)$, the preimage of $x$. Note that $x$ is in $A$ and $f(x)$ is in $B^n$.
$g$ is surjective, so $f$ never sends anything to $\varnothing$. $g$ is a function, so $f$ always sends distinct elements of $A$ to disjoint subsets.
RTL: Let $f : A \to 2^{B^{\,n}}$ be an interpretation. Let $C$ be the image of $f$ and let $D$ be $\bigcup C$. Let $g$ be a function $g : C \to A$ defined as: $g(x) = y$ if and only if $x \in f(y)$. Since $f$ sends distinct items to disjoint sets and doesn't send anything to the empty set, $g$ is a function. $f$ is a function, therefore $g$ is surjective.
Lemma: a sentence $\varphi$ is true if and only if its definable set is a singleton.
Suppose $\varphi$ is true in a structure $\mathfrak{M}$. Then $\{ \vec{v} : \mathfrak{M} \models \varphi(\vec{v}) \}$ will consist of exactly the empty variable map.
Suppose $\varphi$ is not true in a structure $\mathfrak{M}$. Then $\{ \vec{v} : \mathfrak{M} \models \varphi(\vec{v}) \}$ will be empty.
Theorem: $i$ preserves the truth of sentences.
Let $\varphi$ be a sentence. Suppose $\varphi$ is true. $X = \{ \vec{v} : \mathfrak{M} \models \varphi(\vec{v}) \}$ is a singleton. Then $f$, applied elementwise to the empty tuple, the sole element of $X$, will also produce an empty tuple and thus $i(X)$ will be a singleton.
Suppose $\varphi$ is not true. Then $X = \{ \vec{v} : \mathfrak{M} \models \varphi(\vec{v}) \}$ will be empty and thus $i(X)$ will also be empty.
Therefore $i$ preserves the truth of sentences.
Theorem: the map $i : \mathfrak{A} \to E_0$ sending every element of the domain to $c$ is an interpretation, where $E_0$ is a structure in $E$.
Let $f$ be the function $x \mapsto \{c\}$, it is a constant map
