avewright/proofwiki-math huggingface dataset:
Theorem: Rational Numbers under Addition form Infinite Abelian Group

Let $\Q$ be the set of rational numbers.
The structure $\struct {\Q, +}$ is a countably infinite abelian group.

Proof:
The rational numbers are, by definition, the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
Hence by definition, $\struct {\Q, +, \times}$ is a field.
The fact that $\struct {\Q, +}$ forms an abelian group follows directly from the definition of a field.
From Rational Numbers are Countably Infinite, we have that $\struct {\Q, +}$ is countably infinite.
{{Qed}}
Theorem: Non-Zero Rational Numbers under Multiplication form Infinite Abelian Group

Let $\Q_{\ne 0}$ be the set of non-zero rational numbers:
:$\Q_{\ne 0} = \Q \setminus \set 0$
The structure $\struct {\Q_{\ne 0}, \times}$ is a countably infinite abelian group.

Proof:
From the definition of rational numbers, the structure $\struct {\Q, + \times}$ is constructed as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
Hence from Multiplicative Group of Field is Abelian Group, $\struct {\Q_{\ne 0}, \times}$ is an abelian group.
From Rational Numbers are Countably Infinite, we have that $\struct {\Q_{\ne 0}, \times}$ is a countably infinite group.
{{qed}}
Theorem: Integers under Multiplication form Countably Infinite Commutative Monoid

The set of integers under multiplication $\struct {\Z, \times}$ is a countably infinite commutative monoid.

Proof:
First we note that Integers under Multiplication form Monoid.
{{qed|lemma}}
Then we have:
Theorem: Additive Group of Rationals is Normal Subgroup of Reals

Let $\struct {\Q, +}$ be the additive group of rational numbers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Then $\struct {\Q, +}$ is a normal subgroup of $\struct {\R, +}$.

Proof:
From the definition of real numbers, $\Q$ is a subset of $\R$.
As $\struct {\R, +}$ is a group, and $\struct {\Q, +}$ is a group, it follows from the definition of subgroup that $\struct {\Q, +}$ is a subgroup of $\struct {\R, +}$.
As $\struct {\R, +}$ is abelian, it follows from Subgroup of Abelian Group is Normal that $\struct {\Q, +}$ is normal in $\struct {\R, +}$.
{{Qed}}
Theorem: Additive Group of Integers is Subgroup of Rationals

Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {\Q, +}$ be the additive group of rational numbers.
Then $\struct {\Z, +}$ is a subgroup of $\struct {\Q, +}$.

Proof:
Recall that Integers form Integral Domain.
The set $\Q$ of rational numbers is defined as the field of quotients of the integers.
The fact that the integers are a subgroup of the rationals follows from the work done in proving the Existence of Field of Quotients from an integral domain.
{{qed}}
Theorem: Additive Group of Reals is Normal Subgroup of Complex

Let $\struct {\R, +}$ be the additive group of real numbers.
Let $\struct {\C, +}$ be the additive group of complex numbers.
Then $\struct {\R, +}$ is a normal subgroup of $\struct {\C, +}$.

Proof:
From Additive Group of Reals is Subgroup of Complex, $\struct {\R, +}$ is a subgroup of $\struct {\C, +}$.
Then from Complex Numbers under Addition form Infinite Abelian Group, $\struct {\C, +}$ is abelian.
The result follows from Subgroup of Abelian Group is Normal.
{{Qed}}
Category:Additive Group of Real Numbers
Category:Additive Group of Complex Numbers
Category:Examples of Normal Subgroups
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Theorem: Multiplicative Group of Reals is Normal Subgroup of Complex

Let $\struct {\R_{\ne 0}, \times}$ be the multiplicative group of real numbers.
Let $\struct {\C_{\ne 0}, \times}$ be the multiplicative group of complex numbers.
Then $\struct {\R_{\ne 0}, \times}$ is a normal subgroup of $\struct {\C_{\ne 0}, \times}$.

Proof:
Let $x, y \in \C_{\ne 0}$ such that $x = x_1 + 0 i, y = y_1 + 0 i$.
As $x$ and $y$ are wholly real, we have that $x, y \in \R_{\ne 0}$.
Then $x + y = x_1 y_1 - 0 \cdot 0 + \left({0 \cdot y_1 + x_1 \cdot 0}\right) i = x_1 y_1 + 0i$ which is also wholly real.
Also, the inverse of $x$ is $\dfrac 1 x = \dfrac 1 {x_1} + 0 i$ which is also wholly real.
Thus by the Two-Step Subgroup Test, $\struct {\R_{\ne 0}, \times}$ is a subgroup of $\struct {\C_{\ne 0}, \times}$.
From Non-Zero Complex Numbers under Multiplication form Infinite Abelian Group, $\struct {\C_{\ne 0}, \times}$ is abelian.
The result follows from Subgroup of Abelian Group is Normal.
{{Qed}}
Theorem: Multiplicative Group of Rationals is Normal Subgroup of Reals

Let $\struct {\Q_{\ne 0}, \times}$ be the multiplicative group of rational numbers.
Let $\struct {\R_{\ne 0}, \times}$ be the multiplicative group of real numbers.
Then $\struct {\Q_{\ne 0}, \times}$ is a normal subgroup of $\left({\R_{\ne 0}, \times}\right)$.

Proof:
From the definition of real numbers, it follows that $\Q_{\ne 0}$ is a subset of $\R_{\ne 0}$.
As $\struct {\R_{\ne 0}, \times}$ is a group, and $\struct {\Q_{\ne 0}, \times}$ is a group, it follows from the definition of subgroup that $\struct {\Q_{\ne 0}, \times}$ is a subgroup of $\struct {\R_{\ne 0}, \times}$.
As $\struct {\R_{\ne 0}, \times}$ is abelian, it follows from Subgroup of Abelian Group is Normal that $\struct {\Q_{\ne 0}, \times}$ is normal in $\struct {\R_{\ne 0}, \times}$.
{{qed}}
Theorem: Circle Group is Infinite Abelian Group

The circle group $\struct {K, \times}$ is an uncountably infinite abelian group.

Proof:
From Circle Group is Group, we have that $\struct {K, \times}$ is a group.
As $K \subseteq \C$, it follows by definition that $\struct {K, \times}$ is a subgroup of $\struct {\C, \times}$.
From Complex Multiplication is Commutative, $\times$ is commutative on $\C$.
From Restriction of Commutative Operation is Commutative, it follows that $\times$ is commutative on $K$.
Hence, by definition, $\struct {K, \times}$ is an abelian group.
Finally we have that the Circle Group is Uncountably Infinite.
{{qed}}
Category:Circle Group
Category:Complex Numbers
Category:Circle Group is Infinite Abelian Group
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Theorem: Homomorphism from Reals to Circle Group

Let $\struct {\R, +}$ be the additive group of real numbers.
Let $\struct {K, \times}$ be the circle group.
Let $\phi: \struct {\R, +} \to \struct {K, \times}$ be the mapping defined as:
:$\forall x \in \R: \map \phi x = e^{i x}$
Then $\phi$ is a (group) homomorphism.

Proof:
Let $x, y \in \R$.
Then:
{{begin-eqn}}
{{eqn | l = \map \phi x \times \map \phi y
      | r = e^{i x} e^{i y}
      | c = 
}}
{{eqn | r = e^{i \paren {x + y} }
      | c = 
}}
{{eqn | r = \map \phi {x + y}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Integers under Addition form Infinite Cyclic Group

The additive group of integers $\struct {\Z, +}$ is an infinite cyclic group which is generated by the element $1 \in \Z$.

Proof:
By Epimorphism from Integers to Cyclic Group and integer multiplication:
:$\forall n \in \Z: n = \map {+^n} 1 \in \gen 1$
Thus:
:$\struct {\Z, +} = \gen 1$
and thus, by the definition of a cyclic group, is cyclic.
{{qed}}
Theorem: Generators of Additive Group of Integers

The only generators of the additive group of integers $\struct {\Z, +}$ are $1$ and $-1$.

Proof:
From Integers under Addition form Infinite Cyclic Group, $\struct {\Z, +}$ is an infinite cyclic group generated by $1$.
From Generators of Infinite Cyclic Group, there is only one other generator of such a group, and that is the inverse of that generator.
The result follows.
{{qed}}
Theorem: Inverse of Generator of Cyclic Group is Generator

Let $\gen g = G$ be a cyclic group.
Then:
:$G = \gen {g^{-1} }$
where $g^{-1}$ denotes the inverse of $g$.
Thus, in general, a generator of a cyclic group is not unique.

Proof:
Let $\gen g = G$.
Then from Set of Words Generates Group:
:$\map W {\set {g, g^{-1} } } = G$
But:
:$\gen {g^{-1} } = \map W {\set {g, g^{-1} } }$
and the result follows.
{{qed}}
Theorem: Inverse of Generator of Cyclic Group is Generator

Let $\gen g = G$ be a cyclic group.
Then:
:$G = \gen {g^{-1} }$
where $g^{-1}$ denotes the inverse of $g$.
Thus, in general, a generator of a cyclic group is not unique.

Proof:
Let $C_n = \gen g$ be the cyclic group of order $n$.
By definition, $g^n = e$.
We have that $n - 1$ is coprime to $n$.
So it follows from that Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order that:
: $C_n = \gen {g^{n - 1} }$
But from Inverse Element is Power of Order Less 1:
:$g^{n - 1} = g^{-1}$
{{Qed}}
Theorem: Generators of Infinite Cyclic Group

Let $\gen g = G$ be an infinite cyclic group.
Then the only other generator of $G$ is $g^{-1}$.
Thus an infinite cyclic group has exactly $2$ generators.

Proof:
By definition, the infinite cyclic group with generator $g$ is:
:$\gen g = \set {\ldots, g^{-2}, g^{-1}, e, g, g^2, \ldots}$
where $e$ denotes the identity $e = g^0$.
The fact that $g^{-1}$ generates $G$ is shown by Inverse of Generator of Cyclic Group is Generator.
Futhermore:
:$\gen e = \set e \ne G$
By definition of infinite cyclic group:
:$g^i \ne g^j$ for all $i \ne j$
Let $n \in Z \setminus \set {-1, 0, 1}$.
Then:
:$\gen {g^n} = \set {\ldots, g^{-2 n}, g^{-n}, e, g^n, g^{2 n}, \ldots}$
But since $\order n > 1$, none of these elements is equal to $g$, because $1 \notin n \Z$.
So:
:$g \notin \gen {g^n} \implies \gen {g^n} \ne \gen g$
{{qed}}
Theorem: Subgroup of Integers is Ideal

Let $\struct {\Z, +}$ be the additive group of integers.
Every subgroup of $\struct {\Z, +}$ is an ideal of the ring $\struct {\Z, +, \times}$.

Proof:
Let $H$ be a subgroup of $\struct {\Z, +}$.
Let $n \in \Z, h \in H$.
Then from the definition of cyclic group and Negative Index Law for Monoids:
:$n h = n \cdot h \in \gen h \subseteq H$
The result follows.
{{Qed}}
Theorem: Additive Group of Integers is Subgroup of Reals

Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Then $\struct {\Z, +}$ is a subgroup of $\struct {\R, +}$.

Proof:
From Additive Group of Integers is Subgroup of Rationals, $\struct {\Z, +}$ is a subgroup of $\struct {\Q, +}$.
From Additive Group of Rationals is Normal Subgroup of Reals, $\struct {\Q, +}$ is a subgroup of $\struct {\R, +}$.
Thus $\struct {\Z, +}$ is a subgroup of $\struct {\R, +}$.
{{qed}}
Theorem: Quotient Group of Reals by Integers is Circle Group

Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Let $K$ be the circle group.
Then the quotient group of $\struct {\R, +}$ by $\struct {\Z, +}$ is isomorphic to $K$.

Proof:
Define $\phi: \R / \Z \to K$ by:
:$\map \phi {x + \Z} = \map \exp {2 \pi i x}$
Then $\phi$ is well-defined.
For, if $x + \Z = y + \Z$, then $y = x + n$ for some $n \in \Z$, and:
:$\map \exp {2 \pi i \paren {x + n} } = \map \exp {2 \pi i x}$
by Complex Exponential Function has Imaginary Period.
Moreover, by Exponential of Sum:
:$\map \exp {x + y + \Z} = \map \exp {2 \pi i \paren {x + y} } = \map \exp {2 \pi i x} \, \map \exp {2 \pi i y}$
meaning $\phi$ is a group homomorphism.
By Euler's Formula:
:$\map \exp {2 \pi i x} = \map \cos {2 \pi i x} + i \, \map \sin {2 \pi i x}$
so that, by Sine and Cosine are Periodic on Reals:
:$\map \phi x = 1$ {{iff}} $x + \Z = 0 + \Z$
Hence, by Kernel is Trivial iff Group Monomorphism, $\phi$ is a monomorphism.
From polar form for complex numbers, it follows that all $z \in \C$ with $\cmod z = 1$ are of the form:
:$z = \map \exp {2 \pi i x}$
for some $x \in \R$.
Hence $\phi$ is also an epimorphism.
Thus $\phi: \R / \Z \to K$ is a group isomorphism.
{{qed}}
Theorem: Integers Modulo m under Addition form Cyclic Group

Let $\Z_m$ be the set of integers modulo $m$.
Let $+_m$ be the operation of addition modulo $m$.
Let $\struct {\Z_m, +_m}$ denote the additive group of integers modulo $m$.
Then $\struct {\Z_m, +_m}$ is a cyclic group of order $m$, generated by the element $\eqclass 1 m \in \Z_m$.

Proof:
From the definition of integers modulo $m$, we have:
:$\Z_m = \dfrac \Z {\RR_m} = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$
It is established that Modulo Addition is Well-Defined:
:$\eqclass a m +_m \eqclass b m = \eqclass {a + b} m$
The group axioms are fulfilled:
:'''{{Group-axiom|0}}''': Addition modulo $m$ is closed.
:'''{{Group-axiom|1}}''': Addition modulo $m$ is associative.
:'''{{Group-axiom|2}}''': The identity element of $\struct {\Z_m, +_m}$ is $\eqclass 0 m$.
:'''{{Group-axiom|3}}''': The inverse of $\eqclass k m \in \Z_m$ is $-\eqclass k m = \eqclass {-k} m = \eqclass {n - k} m$.
:Commutativity: Addition modulo $m$ is commutative.
From Integers under Addition form Infinite Cyclic Group and Quotient Group of Cyclic Group, $\struct {\dfrac \Z {\RR_m}, +_m}$ is cyclic order $m$.
{{Qed}}
Theorem: Integers Modulo m under Multiplication form Commutative Monoid

Let $\struct {\Z_m, \times_m}$ denote the algebraic structure such that:
:$\Z_m$ is the set of integers modulo $m$
:$\times_m$ denotes the operation of multiplication modulo $m$.
Then $\struct {\Z_m, \times_m}$ is a commutative monoid.

Proof:
Multiplication modulo $m$ is closed.
Multiplication modulo $m$ is associative.
Multiplication modulo $m$ has an identity:
:$\forall k \in \Z: \eqclass k m \times_m \eqclass 1 m = \eqclass k m = \eqclass 1 m \times_m \eqclass k m$
This identity is unique.
Multiplication modulo $m$ is commutative.
Thus all the conditions are fulfilled for this to be a commutative monoid.
{{qed}}
Theorem: Multiplicative Inverse in Ring of Integers Modulo m

Let $\struct {\Z_m, +_m, \times_m}$ be the ring of integers modulo $m$.
Then $\eqclass k m \in \Z_m$ has an inverse in $\struct {\Z_m, \times_m}$ {{iff}} $k \perp m$.

Proof:
First, suppose $k \perp m$.
That is:
:$\gcd \set {k, m} = 1$
Then, by Bézout's Identity:
:$\exists u, v \in \Z: u k + v m = 1$
Thus:
:$\eqclass {u k + v m} m = \eqclass {u k} m = \eqclass u m \eqclass k m = \eqclass 1 m$
Thus $\eqclass u m$ is an inverse of $\eqclass k m$.
Suppose that:
:$\exists u \in \Z: \eqclass u m \eqclass k m = \eqclass {u k} m = 1$.
Then:
:$u k \equiv 1 \pmod m$
and:
:$\exists v \in \Z: u k + v m = 1$
Thus from Bézout's Identity:
:$k \perp m$
{{qed}}
Theorem: Multiplicative Inverse in Ring of Integers Modulo m

Let $\struct {\Z_m, +_m, \times_m}$ be the ring of integers modulo $m$.
Then $\eqclass k m \in \Z_m$ has an inverse in $\struct {\Z_m, \times_m}$ {{iff}} $k \perp m$.

Proof:
From Ring of Integers Modulo m is Ring, $\left({\Z_m, +_m, \times_m}\right)$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity $\left[\!\left[{1}\right]\!\right]_m$]].
Thus by definition $\left({\Z_m, \times_m}\right)$ is a commutative monoid.
The result follows from Multiplicative Inverse in Monoid of Integers Modulo m.
{{qed}}
Theorem: Reduced Residue System is Subset of Set of All Residue Classes

Let $\Z_m$ be the set of set of residue classes modulo $m$.
Let $\Z'_m$ be the reduced residue system modulo $m$.
Then:
:$\forall m \in \Z_{> 1}: \O \subset \Z'_m \subset \Z_m$

Proof:
By definition of reduced residue system modulo $m$:
:$\Z'_m = \set {x \in \Z_m: x \perp m}$
From Subset of Set with Propositional Function:
:$\Z'_m \subseteq \Z_m$
We have that:
:$\gcd \set {m, 0} = m$
Thus it follows that:
:$m > 1 \implies \gcd \set {m, 0} \ne 1$
So:
:$\eqclass 0 m \notin \Z'_m$
However:
:$\eqclass 0 m \in \Z_m$
so:
:$\Z'_m \ne \Z_m$
Thus:
:$\Z'_m \subset \Z_m$
Then:
:$\eqclass 1 m = 1 \implies 1 \perp m$
So:
:$\forall m \in \Z: \eqclass 1 m \in \Z'_m$
Thus:
:$\Z'_m \ne \O$
and therefore:
:$\O \subset \Z'_m$
{{qed}}
Category:Residue Classes
Category:Reduced Residue Systems
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Theorem: Reduced Residue System under Multiplication forms Abelian Group

Let $\Z_m$ be the set of set of residue classes modulo $m$.
Let $\struct {\Z'_m, \times}$ denote the multiplicative group of reduced residues modulo $m$.
Then $\struct {\Z'_m, \times}$ is an abelian group, precisely equal to the group of units of $\Z_m$.

Proof:
From Ring of Integers Modulo m is Ring, $\struct {\Z_m, +, \times}$ forms a (commutative) ring with unity.
Then we have that the units of a ring with unity form a group.
By Multiplicative Inverse in Ring of Integers Modulo m we have that the elements of $\struct {\Z'_m, \times}$ are precisely those that have inverses, and are therefore the units of $\struct {\Z_m, +, \times}$.
The fact that $\struct {\Z'_m, \times}$ is abelian follows from Restriction of Commutative Operation is Commutative.
{{qed}}
Theorem: Reduced Residue System under Multiplication forms Abelian Group

Let $\Z_m$ be the set of set of residue classes modulo $m$.
Let $\struct {\Z'_m, \times}$ denote the multiplicative group of reduced residues modulo $m$.
Then $\struct {\Z'_m, \times}$ is an abelian group, precisely equal to the group of units of $\Z_m$.

Proof:
Taking the group axioms in turn:
=== {{Group-axiom|0|nolink}} ===
From Modulo Multiplication on Reduced Residue System is Closed:
:$\struct {\Z'_m, \times}$ is closed.
{{qed|lemma}}
=== {{Group-axiom|1|nolink}} ===
We have that Modulo Multiplication is Associative.
{{qed|lemma}}
=== {{Group-axiom|2|nolink}} ===
From Modulo Multiplication has Identity, $\eqclass 1 m$ is the identity element of $\struct {\Z'_m, \times_m})$.
{{qed|lemma}}
=== {{Group-axiom|3|nolink}} ===
From Multiplicative Inverse in Monoid of Integers Modulo m, $\eqclass k m \in \Z_m$ has an inverse in $\struct {\Z_m, \times_m}$ {{iff}} $k$ is coprime to $m$.
Thus every element of $\struct {\Z'_m, \times_m}$ has an inverse.
{{qed|lemma}}
All the group axioms are thus seen to be fulfilled, and so $\struct {\Z'_m, \times_m}$ is a group.
{{qed}}
Theorem: Reduced Residue System under Multiplication forms Abelian Group

Let $\Z_m$ be the set of set of residue classes modulo $m$.
Let $\struct {\Z'_m, \times}$ denote the multiplicative group of reduced residues modulo $m$.
Then $\struct {\Z'_m, \times}$ is an abelian group, precisely equal to the group of units of $\Z_m$.

Proof:
Taking the finite group axioms in turn:
=== $\text {FG} 0$: Closure ===
From Modulo Multiplication on Reduced Residue System is Closed:
:$\struct {\Z'_m, \times}$ is closed.
{{qed|lemma}}
=== $\text {FG} 1$: Associativity ===
We have that Modulo Multiplication is Associative.
{{qed|lemma}}
=== $\text {FG} 2$: Finiteness ===
The order of $\struct {\Z'_m, \times}$ is $\map \phi n$ by definition, where $\map \phi n$ denotes the Euler $\phi$ function.
As $\map \phi n < n$ it follows that $\struct {\Z'_m, \times}$ is of finite order.
{{qed|lemma}}
=== $\text {FG} 3$: Cancellability ===
We have that Modulo Multiplication on Reduced Residue System is Cancellable.
{{qed|lemma}}
Thus all the finite group axioms are fulfilled, and $\struct {\Z'_m, \times}$ is a group.
It remains to note that Modulo Multiplication is Commutative to confirm that $\struct {\Z'_m, \times}$ is abelian.
{{qed}}
Theorem: Ring of Integers Modulo Prime is Field

Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$.
Then:
:$m$ is prime
{{iff}}:
:$\struct {\Z_m, +, \times}$ is a field.

Proof:
=== Prime Modulus ===
$\struct {\Z_m, +, \times}$ is a commutative ring with unity by definition.
From Reduced Residue System under Multiplication forms Abelian Group, $\struct {\Z'_m, \times}$ is an abelian group.
$\Z'_m$ consists of all the elements of $\Z_m$ coprime to $m$.
Now when $m$ is prime, we have, from Reduced Residue System Modulo Prime:
:$\Z'_m = \set {\eqclass 1 m, \eqclass 2 m, \ldots, \eqclass {m - 1} m}$
That is:
:$\Z'_m = \Z_m \setminus \set {\eqclass 0 m}$
where $\setminus$ denotes set difference.
Hence the result.
{{qed|lemma}}
=== Composite Modulus ===
Now suppose $m \in \Z: m \ge 2$ is composite.
From Ring of Integers Modulo Composite is not Integral Domain, $\struct {\Z_m, +, \times}$ is not an integral domain.
From Field is Integral Domain $\struct {\Z_m, +, \times}$ is not a field.
{{qed}}
Theorem: Ring of Integers Modulo Prime is Field

Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$.
Then:
:$m$ is prime
{{iff}}:
:$\struct {\Z_m, +, \times}$ is a field.

Proof:
Let $p$ be prime.
From Irreducible Elements of Ring of Integers, we have that $p$ is irreducible in the ring of integers $\struct {\Z, +, \times}$.
From Ring of Integers is Principal Ideal Domain, $\struct {\Z, +, \times}$ is a principal ideal domain.
Let $\ideal p$ denote the principal ideal of $\struct {\Z, +, \times}$ generated by $p$.
By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, $\ideal p$ is a maximal ideal of $\struct {\Z, +, \times}$.
Let $\Z / \ideal p$ denote the quotient ring of $\Z$ by $\paren p$.
By Maximal Ideal iff Quotient Ring is Field, $\Z / \ideal p$, is a field.
But $\Z / \ideal p$ is exactly $\struct {\Z_p, +, \times}$.
{{qed|lemma}}
Let $p$ be composite.
Then $p$ is not irreducible in $\struct {\Z, +, \times}$.
Thus by Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, $\ideal p$ is not a maximal ideal of $\struct {\Z, +, \times}$.
Hence by Maximal Ideal iff Quotient Ring is Field, $\Z / \ideal p$ is not a field.
{{qed}}
Theorem: Ring of Integers Modulo Prime is Field

Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$.
Then:
:$m$ is prime
{{iff}}:
:$\struct {\Z_m, +, \times}$ is a field.

Proof:
Let $m$ be prime.
From Ring of Integers Modulo Prime is Integral Domain, $\struct {\Z_m, +, \times}$ is an integral domain.
Let $\eqclass a m \ne \eqclass 0 m$ be a residue class modulo $m$.
We need to find a residue class modulo $m$ $\eqclass x m$ such that $\eqclass a m \eqclass x m = \eqclass 1 m$.
Because $m$ is prime and $m \nmid a$, we have:
:$\gcd \set {a, m} = 1$
Hence from Bézout's Identity:
:$1 = x a + y m$
for some $x, y \in \Z$.
Thus we have:
{{begin-eqn}}
{{eqn | l = \eqclass a m \eqclass x m
      | r = \eqclass {a x} m
      | c = 
}}
{{eqn | r = \eqclass {1 - y m} m
      | c = 
}}
{{eqn | r = \eqclass 1 m
      | c = 
}}
{{end-eqn}}
So every non-zero residue class modulo $m$ has an inverse.
So by definition $\struct {\Z_m, +, \times}$ is a field.
{{qed|lemma}}
Now suppose $m \in \Z: m \ge 2$ is composite.
From Ring of Integers Modulo Composite is not Integral Domain, $\struct {\Z_m, +, \times}$ is not an integral domain.
From Field is Integral Domain $\struct {\Z_m, +, \times}$ is not a field.
{{qed}}
Theorem: Ring of Integers Modulo Prime is Field

Let $m \in \Z: m \ge 2$.
Let $\struct {\Z_m, +, \times}$ be the ring of integers modulo $m$.
Then:
:$m$ is prime
{{iff}}:
:$\struct {\Z_m, +, \times}$ is a field.

Proof:
Let $m$ be prime.
From Ring of Integers Modulo Prime is Integral Domain, $\struct {\Z_m, +, \times}$ is an integral domain.
From Finite Integral Domain is Galois Field, $\struct {\Z_m, +, \times}$ is a field.
{{qed|lemma}}
Now suppose $m \in \Z: m \ge 2$ is composite.
From Ring of Integers Modulo Composite is not Integral Domain, $\struct {\Z_m, +, \times}$ is not an integral domain.
From Field is Integral Domain $\struct {\Z_m, +, \times}$ is not a field.
{{qed}}