nyuuzyou/wikis huggingface dataset:
Schröder Rule

== Theorem ==
Let $A$, $B$ and $C$ be relations on a set $S$.
:$(1): \quad A \circ B \subseteq C$
:$(2): \quad A^{-1} \circ \overline C \subseteq \overline B$
:$(3): \quad \overline C \circ B^{-1} \subseteq \overline A$
where:
:$\circ$ denotes relation composition
:$A^{-1}$ denotes the inverse of $A$
:$\overline A$ denotes the complement of $A$.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
* This result is usually seen with the before it: the Schröder Rule.
== Sources ==
* : $\S 4$ Proposition $4.4$
Category:Schröder Rule
Category:Relation Theory
Schröder Rule/Proof 1

== Theorem ==
== Proof ==
=== $(1)$ iff $(2)$ ===
By the definition of relation composition and subset we have that statement $(1)$ may be written as:
:$(1'): \quad \forall x, y, z \in S: \paren {\tuple {y, z} \in A \land \tuple {x, y} \in B \implies \tuple {x, z} \in C}$
Using a different arrangement of variable names, statement $(2)$ can be written:
:$(2'): \quad \forall x, y, z \in S: \paren {\tuple {z, y} \in A^{-1} \land \tuple {x, z} \in \overline C \implies \tuple {x, y} \in \overline B}$
By the definition of inverse relation and the complement of a relation we can rewrite this as:
:$(2): \quad \forall x, y, z \in S: \paren {\tuple {y, z} \in A \land \tuple {x, z} \notin C) \implies \tuple {x, y} \notin B}$
We shall use the method of truth tables.
The two statements will be equivalent the columns under the main connectives, which is $\implies$ in each case, are identical.
Statement 1:
$\begin{array}{ccccc}
(\tuple {y, z} \in A & \land & \tuple {x, y} \in B) & \implies & \tuple {x, z} \in C \\
\hline
\T & \T & \T & \T & \T \\
\T & \T & \T & \F & \F \\
\T & \F & \F & \T & \T \\
\T & \F & \F & \T & \F \\
\F & \F & \T & \T & \T \\
\F & \F & \T & \T & \F \\
\F & \F & \F & \T & \T \\
\F & \F & \F & \T & \F \\
\end{array}$
Statement 2:
$\begin{array}{ccccc}
(\tuple {y, z} \in A & \land & \tuple {x, z} \notin C) & \implies & \tuple {x, y} \notin B \\
\hline
\T & \F & \F & \T & \F \\
\T & \T & \T & \F & \F \\
\T & \F & \F & \T & \T \\
\T & \T & \T & \T & \T \\
\F & \F & \F & \T & \F \\
\F & \F & \T & \T & \F \\
\F & \F & \F & \T & \T \\
\F & \F & \T & \T & \T \\
\end{array}$
=== $(2)$ iff $(3)$ ===
== Sources ==
* : $\S 4$ Proposition $4.4$
Category:Schröder Rule
Schröder Rule/Proof 2

== Theorem ==
== Proof ==
By the definition of relation composition and subset we have that statement $(1)$ may be written as:
:$(1') \quad \forall x, y, z \in S: \paren {\tuple {y, z} \in A \land \tuple {x, y} \in B \implies \tuple {x, z} \in C}$
Using a different arrangement of variable names, statement $(2)$ can be written:
:$(2') \quad \forall x, y, z \in S: \paren {\tuple {z, y} \in A^{-1} \land \tuple {x, z} \in \overline C \implies \tuple {x, y} \in \overline B}$
By the definition of the inverse and the complement of a relation we can rewrite this as:
:$(2) \quad \forall x, y, z \in S: \paren {\tuple {y, z} \in A \land \tuple {x, z} \notin C \implies \tuple {x, y} \notin B}$
Similarly, statement $(3)$ can be written:
:$(3') \quad \forall x, y, z \in S: \paren {\tuple {x, z} \in \overline C \land \tuple {y, x} \in B^{-1} \implies \tuple {y, z} \in \overline A}$
By the definition of the inverse and the complement of a relation we can rewrite this as:
:$(3) \quad \forall x, y, z \in S: \paren {\tuple {x, z} \notin C \land \tuple {x, y} \in B \implies \tuple {y, z} \notin A}$
So in all we have:
:$(1') \quad \forall x, y, z \in S: \paren {\tuple {y, z} \in A \land \tuple {x, y} \in B \implies \tuple {x, z} \in C}$
:$(2) \quad \forall x, y, z \in S: \paren {\tuple {y, z} \in A \land \tuple {x, z} \notin C \implies \tuple {x, y} \notin B}$
:$(3) \quad \forall x, y, z \in S: \paren {\tuple {x, z} \notin C \land \tuple {x, y} \in B \implies \tuple {y, z} \notin A}$
Category:Schröder Rule
Schur's Inequality

== Theorem ==
Let $x, y, z \in \R_{\ge 0}$ be positive real numbers.
Let $t \in \R, t > 0$ be a (strictly) positive real number.
Then:
:$x^t \paren {x - y} \paren {x - z} + y^t \paren {y - z} \paren {y - x} + z^t \paren {z - x} \paren {z - y} \ge 0$
The equality holds either:
: $x = y = z$
: Two of them are equal and the other is zero.
When $t$ is a positive even integer, the inequality holds for all real numbers $x, y, z$.
== Proof ==
We note that the inequality, as stated, is symmetrical in $x, y$ and $z$.
, we can assume that $x \ge y \ge z \ge 0$.
Consider the expression:
:$\paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z}} + z^t \paren {x - z} \paren {y - z}$
We see that every term in the above is non-negative. So, directly:
:$(1): \quad \paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z}} + z^t \paren {x - z} \paren {y - z} \ge 0$
If $x = y = z$, all of $x - y$, $x - z$ and $y - z$ are $0$.
Thus equality holds.
Inspection on a case-by-case basis provides evidence for the other conditions for equality.
To show these are the only cases, we suppose $x, y, z$ are not equal.
Then $x > y > z \ge 0$.
We thus have $x - z > y - z$.
Hence:
Now we suppose two numbers are equal, but the other is neither the same number or $0$.
If $x = y > z > 0$:
If $x > y = z \ge 0$:
This shows the equality conditions is .
$(1)$ can then be rearranged to Schur's inequality.
Now, let $t$ be a positive even integer.
, we can assume that $x \ge y \ge z$.
By Pigeonhole Principle, at least $2$ of them have the same sign.
Suppose $x, y$ are positive.
Once again we consider the expression:
:$\paren {x - y} \paren {x^t \paren {x - z} - y^t \paren {y - z}} + z^t \paren {x - z} \paren {y - z}$
The first term is still non-negative.
The second term is non-negative, since:
:$z^t \ge 0 \quad$ Even Power is Non-Negative
:$x \ge z$, $y \ge z$
Thus we can still conclude $(1)$, which can then be rearranged to Schur's inequality.
Suppose $y, z$ are negative.
Then $-z, -y$ are positive, and $-z \le -y \le -x$.
Substituting $x, y, z$ for $-z, -y, -x$ in the above, the result follows.
Category:Algebra
Schur's Lemma (Representation Theory)

== Theorem ==
Let $\struct {G, \cdot}$ be a finite group.
Let $V$ and $V'$ be two irreducible $G$-modules.
Let $f: V \to V'$ be a homomorphism of $G$-modules.
Then either:
:$\map f v = 0$ for all $v \in V$
or:
:$f$ is an isomorphism.
=== Corollary ===
== Proof ==
From Kernel is G-Module, $\map \ker f$ is a $G$-submodule of $V$.
From Image is G-Module, $\Img f$ is a $G$-submodule of $V'$.
By the definition of irreducible:
:$\map \ker f = \set 0$
or:
:$\map \ker f = V$
If $\map \ker f = V$ then by definition:
:$\map f v = 0$ for all $v \in V$
Let $\map \ker f = \set 0$.
Then from Linear Transformation is Injective iff Kernel Contains Only Zero:
:$f$ is injective.
It also follows that:
:$\Img f = V'$
Thus $f$ is surjective and injective.
Thus by definition $f$ is a bijection and thence an isomorphism.
Category:Representation Theory
Schur's Lemma (Representation Theory)/Corollary

== Corollary to Schur's Lemma ==
Let $\struct {G, \cdot}$ be a finite group.
Let $\struct {V, \phi}$ be a $G$-module.
Let the underlying field $k$ of $V$ be algebraically closed.
Let:
:$\map {\mathrm {End}_G} V := \leftset {f: V \to V: f}$ is a homomorphism of $G$-modules$\rightset {}$
Then:
:$\map {\mathrm {End}_G} V$
is a field, with the same structure as $k$.
== Proof ==
Denote the identity mapping on $V$ as $I_V: V \to V$.
If $f = 0$, since $0\in k$ it can be written $f = 0 I_V$.
Let $f$ be an automorphism.
We have that $k$ is algebraically closed.
Therefore the characteristic polynomial of $f$ is complete reducible in $k \sqbrk x$.
Hence $f$ has all eigenvalue in $k$.
Let $\lambda \in k$ be an eigenvalue of $f$.
Consider the endomorphism:
:$f - \lambda I_V: V \to V$
Because $\lambda$ is an eigenvalue:
:$\map \ker {f - \lambda I_V} \ne \set 0$
From Schur's Lemma:
:$f = \lambda I_V$
:$\paren {\lambda I_V} \circ \paren {\mu I_V} = \paren {\lambda \mu} I_V$
:$\lambda I_V + \paren {-\mu I_V} = \paren {\lambda - \mu} I_V$
From Subring Test:
:$\map {\mathrm {End}_G} V$ is a subring of the ring endomorphisms of $V$ as an abelian group.
Let $\phi: \map {\mathrm {End}_G} V \to k$ be defined as:
:$\map \phi {\lambda I_V} = \lambda$
Then:
:$\map \phi {\lambda I_V + \mu I_V} = \lambda + \mu = \map \phi {\lambda I_V} + \map \phi {\mu I_V}$
:$\map \phi {\paren {\lambda I_V} \circ \paren {\mu I_V} } = \lambda \mu = \map \phi {\lambda I_V} \map \phi {\mu I_V}$
Hence $\phi$ is a ring isomorphism.
But since $k$ is a field it is a field isomorphism.
 
Category:Representation Theory
Schur's Theorem (Ramsey Theory)

== Theorem ==
Let $r$ be a positive integer.
Then there exists a positive integer $S$ such that:
:for every partition of the integers $\set {1, \ldots, S}$ into $r$ parts, one of the parts contains integers $x$, $y$ and $z$ such that:
::$x + y = z$
== Proof ==
Let:
:$n = \map R {3, \ldots, 3}$
where $\map R {3, \ldots, 3}$ denotes the Ramsey number on $r$ colors.
Take $S$ to be $n$.
partition the integers $\set {1, \ldots, n}$ into $r$ parts, which we denote by colors.
That is:
:the integers in the first part are said to be colored $c_1$
:the integers in the second part are said to be colored $c_2$ 
and so on till color $c_r$.
Thus $\set {1, \ldots, S}$ has been $r$-colored.
(This terminology is common in Ramsey theory.)
Now consider the complete graph $K_n$.
Now color the edges of $K_n$ as follows:
:An edge $xy$ is given color $c$ if $\size {x - y}$ was colored $c$ in the partitioning.
From the definition of $\map R {3, \ldots, 3}$ and Ramsey's Theorem, $K_n$ will definitely contain a monochromatic triangle, say built out of the vertices $i > j > k$.
Let the triangle be colored $c_m$.
Now $i - j$, $i - k$ and $j - k$ will also be colored $c_m$.
That is, $i - j$, $i - k$ and $j - k$ will belong to the same part in the partition.
It only remains to take $x = i - j$, $y = j - k$ and $z = i - k$ to complete the proof.
== An extension ==
The above proving technique allows to obtain a variety of similar and further going results. Here is just a sample:
THEOREM 1
Let $r$ be a positive integer.
Then there exists a positive integer $S$ such that:
:for every partition of the integers $\set {1, \ldots, S}$ into $r$ parts, one of the parts contains integers $x$, $y$ and $z$ such that:
::$x + y = z$
:and:
::$x \ne y$
This theorem follows from the following one:
THEOREM 2
Let $r$ be a positive integer.
Then there exists a positive integer $S$ such that:
for every partition of the integers $\set {1, \ldots, S}$ into $r$ parts, one of the parts contains integers $a$, $b$, $a + b$, $c$, $b + c$, and $d$ such that:
::$a + b + c = d$
PROOF
The proof is nearly the same as of the original Schur's theorem above, except that one uses $\map R {4, \ldots, 4}$.
Category:Combinatorics
Category:Ramsey Theory
Schur-Zassenhaus Theorem

== Theorem ==
Let $G$ be a finite group and $N$ be a normal subgroup in $G$.
Let $N$ be a Hall subgroup of $G$.
Then there exists $H$, a complement of $N$, such that $G$ is the semidirect product of $N$ and $H$.
== Proof ==
The proof proceeds by induction.
By definition, $N$ is a Hall subgroup the index and order of $N$ in $G$ are relatively prime numbers.
Let $G$ be a group whose identity is $e$.
We induct on $\order G$, where $\order G$ is the order of $G$.
We may assume that $N \ne \set e$.
Let $p$ be a prime number dividing $\order N$.
Let $\Syl p N$ be the set of Sylow $p$-subgroups of $N$.
By the First Sylow Theorem:
:$\Syl p N \ne \O$
Let:
: $P \in \Syl p N$
: $G_0$ be the normalizer in $G$ of $P$
: $N_0 = N \cap G_0$.
By Frattini's Argument:
:$G = G_0 N$
By the Second Isomorphism Theorem for Groups and thence Lagrange's Theorem (Group Theory), it follows that:
: $N_0$ is a Hall subgroup of $G_0$
: $\index {G_0} {N_0} = \index G H$
Suppose $G_0 < G$.
Then by induction applied to $N_0$ in $G_0$, we find that $G_0$ contains a complement $H \in N_0$.
We have that:
:$\order H = \index {G_0} {N_0}$
and so $H$ is also a complement to $N$ in $G$.
So we may assume that $P$ is normal in $G$ (that is: $G_0 < G$).
Let $Z \paren P$ be the center of $P$.
By:
:Center is Characteristic Subgroup
:$P$ is normal in $G$
:Characteristic Subgroup of Normal Subgroup is Normal
$Z \paren P$ is also normal in $G$.
Let $Z \paren P = N$.
Then there exists a long exact sequence of cohomology groups:
:$0 \to H^1 \paren {G / N, P^N} \to H^1 \paren {G, P} \to H^1 \paren {N, P} \to H^2 \paren {G / N, P} \to H^2 \paren {G, P}$
which splits as desired.
Otherwise:
:$Z \paren P \ne N$
In this case $N / Z \paren P$ is a normal (Hall) subgroup of $G / Z \paren P$.
By induction:
:$N / Z \paren P$ has a complement $H / Z \paren P$ in $E // Z \paren P$.
Let $G_1$ be the preimage of $H // Z \paren P$ in $G$ (under the equivalence relation).
Then:
:$\order {G_1} = \order {K / Z \paren P} \times \order {Z \paren P} = \order {G / N} \times \order {Z \paren P}$
Therefore, $Z \paren P$ is normal Hall subgroup of $G_1$.
By induction, $Z \paren P$ has a complement in $G_1$ and is also a complement of $N$ in $G$.
== Also known as ==
Some sources refer to this theorem as Schur's theorem, but that name is also used for an unrelated result in Ramsey theory.
Category:Group Theory
Category:Hall Subgroups
Schwarz's Lemma

== Theorem ==
Let $D$ be the unit disk centred at $0$.
Let $f: D \to \C$ be a holomorphic function.
Let $\map f 0 = 0$ and $\cmod {\map f z} \le 1$ for all $z \in D$. 
Then $\cmod {\map {f'} 0} \le 1$, and $\cmod {\map f z} \le \cmod z$ for all $z \in D$.
=== Corollary ===
== Proof ==
First a lemma:
=== Lemma ===
Let $g : D \to \C$ be a complex function such that:
:$\map g z = \begin {cases} \frac {\map f z} z & z \ne 0 \\ \map {f'} 0 & z = 0\end {cases}$
By the lemma, $g$ is holomorphic on $D$. 
Let $r$ be a real number with $0 < r < 1$. 
Let $C_r$ be the closed disk of radius $r$ centered at $0$.
Since $C_r$ is compact, by Continuous Function on Compact Space is Bounded, $g$ attains its maximum on $C_r$.
Let $\alpha \in C_r$ be a point where this maximum is attained.
If $g$ is constant, then we can pick $\alpha$ to be any point on $C_r$. 
If $g$ is non-constant, then by the maximum modulus principle, $\alpha$ necessarily lies on the boundary of $C_r$.
That is, $\cmod \alpha = r$ if $g$ is non-constant. 
If $g$ is constant, pick $\alpha$ such that $\cmod \alpha = r$ to avoid splitting into cases.
With that, for all $z \in C_r$ we have:
:$\cmod {\map g z} \le \cmod {\map g \alpha}$
Since $0$ lies in the interior of $C_r$, we must have $\alpha \ne 0$, so:
So, for all $z \in C_r$ we have: 
:$\cmod {\map g z} \le \dfrac 1 r$
 there exists $\omega \in D$ with: 
:$\cmod \omega < 1$
:$\cmod {\map g \omega} = 1 + \varepsilon > 1$
Note that we then have: 
:$0 < \dfrac 1 {1 + \varepsilon} < 1$
Note that on $C_R$ with $R > \map \max {\dfrac 1 {1 + \varepsilon}, \cmod \omega}$, we should have: 
:$\cmod {\map g z} \le \dfrac 1 R < 1 + \varepsilon$
However, we have $\omega \in C_R$ and: 
:$\cmod {\map g \omega} = 1 + \varepsilon$
a contradiction. 
So, it must be the case that: 
:$\cmod {\map g z} \le 1$
for all $z \in D$.
We have: 
:$\cmod {\map {f'} 0} = \cmod {\map g 0} \le 1$
Note that: 
:$\cmod {\map f 0} = 0$
so certainly: 
:$\cmod {\map f z} \le \cmod z$
for $z = 0$. 
Take $z \in D \setminus \set 0$, from: 
:$\cmod {\map g z} \le 1$
we have: 
:$\cmod {\dfrac {\map f z} z} \le 1$
that is: 
:$\cmod {\map f z} \le \cmod z$
hence the result.
Category:Schwarz's Lemma
Category:Complex Analysis
Schwarz's Lemma/Corollary

== Corollary to Schwarz's Lemma ==
Let $D$ be the unit disk centred at $0$.
Let $f : D \to \C$ be a holomorphic function.
If:
:$\cmod {\map f \omega} = \cmod \omega$ for some $\omega \in D \setminus \set 0$
or:
:$\cmod {\map {f'} 0} = 1$
then $\map f z = a z$ for all $z \in D$ for some $a \in \C$ with $\cmod a = 1$.
== Proof ==
Let $g : D \to \C$ be a complex function with: 
:$\map g z = \begin{cases}\frac {\map f z} z & z \ne 0 \\ \map {f'} 0 & z = 0\end{cases}$
By Schwarz's Lemma: Lemma, $g$ is holomorphic on $D$.
By Schwarz's Lemma, we have: 
:$\cmod {\map g z} \le 1$
for all $z \in D$. 
Suppose that:
:$\cmod {\map f \omega} = \cmod \omega$ for some $\omega \in D \setminus \set 0$.
Then, for this $\omega$, we have: 
:$\cmod {\map g \omega} = 1$
Hence $g$ attains a maximum at $\omega \in D \setminus \set 0$.
As $D$ is a domain, by the Maximum Modulus Principle $g$ is therefore constant.
Suppose instead that: 
:$\cmod {\map {f'} 0} = 1$
Then: 
:$\cmod {\map g 0} = 1$
In this case, $g$ attains a maximum at $0 \in D$, again implying that $g$ is constant.
That is, there exists some $a \in \C$ such that: 
:$\map g z = a$
Note that we then have: 
:$\cmod {\map g 0} = \cmod a = 1$
For $z \in D \setminus \set 0$, we therefore have: 
:$\dfrac {\map f z} z = a$
That is: 
:$\map f z = a z$
As $\map f 0 = 0$, we have: 
:$\map f z = a z$
for all $z \in D$, for some $a \in \C$ with $\cmod a = 1$.
Category:Complex Analysis
Category:Schwarz's Lemma
Schwarz's Lemma/Lemma

== Theorem ==
Let $D$ be the unit disk centred at $0$.
Let $g : D \to \C$ be a complex function such that: 
:$\map g z = \begin {cases} \dfrac {\map f z} z & z \ne 0 \\ \map {f'} 0 & z = 0\end {cases}$
Then $g$ is holomorphic on $D$.
== Proof ==
By Differentiable Function is Continuous, $f$ is continuous.
So by Quotient Rule for Continuous Complex Functions:
:$g$ is continuous on $D \setminus \set 0$.
We aim to show that $f$ is continuous on $D$. 
Note that since $f$ is holomorphic on $D$ and $0 \in D$ we have, by the definition of the complex derivative: 
:$\ds \lim_{z \mathop \to 0} \frac {\map f z - \map f 0} z = \map {f'} 0 \in \C$
Since $\map f 0 = 0$, we furthermore have: 
:$\ds \map {f'} 0 = \lim_{z \mathop \to 0} \frac {\map f z} z$
That is: 
:$\ds \map g 0 = \lim_{z \mathop \to 0} \map g z$
so $g$ is continuous at $0$.
Since $f$ is holomorphic on $D$, by the Quotient Rule for Continuous Complex Functions:
:$g$ is differentiable on $D \setminus \set 0$.
It remains to show that $g$ is differentiable at $0$. 
Take $z \ne 0$ and consider: 
:$\dfrac {\map g z - \map g 0} z$
We have: 
Since $f$ is holomorphic on $D$, by Holomorphic Function is Analytic, there exists a positive real number $R$ such that the series: 
:$\ds \sum_{n \mathop = 0}^\infty \frac {\map {f^{\paren n} } 0} {n!} z^n$
converges to $\map f z$ on $\cmod z < R$.
Note that since $\map f 0 = 0$, the first term of this series is zero.
With that, we have: 
Taking $z \to 0$ we have: 
:$\ds \lim_{z \mathop \to 0} \frac {\map g z - \map g 0} z = \frac 1 2 \map {f} 0$
so $g$ is indeed differentiable at $0$.
Hence by definition $g$ is holomorphic on $D$.
Category:Complex Analysis
Category:Schwarz's Lemma
Schönemann-Eisenstein Theorem

== Theorem ==
Let $\map f x = a_d x^d + a_{d - 1} x^{d - 1} + \dotsb + a_0 \in \Z \sqbrk x$ be a polynomial over the ring of integers $\Z$.
Let $p$ be a prime such that:
:$(1): \quad p \divides a_i \iff i \ne d$
:$(2): \quad p^2 \nmid a_0$
where $p \divides a_i$ signifies that $p$ is a divisor of $a_i$.
Then $f$ is irreducible in $\Q \sqbrk x$.
== Proof ==
By Gauss's Lemma on Irreducible Polynomials, it suffices to show that $f$ is irreducible in $\Z \sqbrk x$.
 that $f = g h$ where $g, h \in \Z \sqbrk x$ are both non-constant.
Let:
:$\map g x = b_e x^e + b_{e - 1} x^{e - 1} + \dotsb + b_0$
:$\map h x = c_f x^f + c_{f - 1} x^{f - 1} + \dotsb + c_0$
Then we have for each $i$:
:$\ds a_i = \sum_{j + k \mathop = i} {b_j c_k}$
In particular, it follows that:
:$a_0 = b_0 c_0$
Possibly after exchanging $g$ and $h$, we may arrange that:
:$p \nmid c_0$
by condition $(2)$.
From condition $(1)$, it follows that then necessarily:
:$p \divides b_0$
We also have:
:$a_d = b_e c_f$
and by condition $(1)$:
:$p \nmid a_d$
and hence:
:$p \nmid b_e$
It follows that there exists a smallest positive $i$ such that:
:$p \nmid b_i$
Naturally, $i \le e$. 
By assumption, both $g$ and $h$ are non-constant.
Hence by Degree of Product of Polynomials over Integral Domain:
:$i < d$
Consider:
:$a_i = b_0 c_i + b_1 c_{i - 1} + \dotsb + b_i c_0$
with the convention that $c_j = 0$ if $j > f$.
By the minimality of $i$, it follows that:
:$p \divides b_k$
for $0 \le k < i$.
Also, since neither $c_0$ nor $b_i$ is divisible by $p$, the last term $b_i c_0$ is not divisible by $p$.
Thus, we conclude that:
:$p \nmid a_i$
which contradicts condition $(1)$.
Therefore, $f$ is irreducible.
=== Warning ===
== Examples ==
== Also known as ==
== Sources ==
* 
* 
* 
Category:Schönemann-Eisenstein Theorem
Category:Algebraic Number Theory
Schönemann-Eisenstein Theorem/Examples

== Examples of Use of Schönemann-Eisenstein Theorem ==
=== Example: $x^3 + 2 x + 2$ ===
=== Example: $x^3 + 2 x + 4$ ===
=== Example: $x^3 - 2 x + 4$ ===
Category:Examples of Use of Schönemann-Eisenstein Theorem
Schönemann-Eisenstein Theorem/Examples/x^3 + 2x + 2

== Example of Use of Schönemann-Eisenstein Theorem ==
Consider the polynomial:
:$\map P x = x^3 + 2 x + 2$
By the Schönemann-Eisenstein Theorem, $\map P x$ is irreducible over $\Q \sqbrk x$.
== Proof ==
We note that the prime number $2$:
:is a divisor of the coefficient of $x^1$, that is, $2$
:is not a divisor of the degree of $\map P x$, that is, $3$
and that:
:$2^2$ is not a divisor of the coefficient of $x_0$, that is, $2$.
Hence, by the Schönemann-Eisenstein Theorem, $\map P x$ is irreducible over $\Q \sqbrk x$.
== Sources ==
* 
* 
Category:Examples of Use of Schönemann-Eisenstein Theorem
Schönemann-Eisenstein Theorem/Examples/x^3 + 2x + 4

== Example of Use of Schönemann-Eisenstein Theorem ==
Consider the polynomial:
:$\map P x = x^3 + 2 x + 4$
By the Schönemann-Eisenstein Theorem, it is not possible to tell whether or not $\map P x$ is irreducible over $\Q \sqbrk x$.
In fact, in this case $\map P x$ is indeed irreducible.
== Proof ==
We note that the prime number $2$:
:is a divisor of the coefficient of $x^1$, that is, $2$
:is not a divisor of the degree of $\map P x$, that is, $3$
and that:
:$2^2$ is a divisor of the coefficient of $x_0$, that is, $4$.
Hence, by the Schönemann-Eisenstein Theorem, it is not necessarily the case that $\map P x$ is not irreducible over $\Q \sqbrk x$.
== Sources ==
* 
* 
Category:Examples of Use of Schönemann-Eisenstein Theorem
Schönemann-Eisenstein Theorem/Examples/x^3 - 2x + 4

== Example of Use of Schönemann-Eisenstein Theorem ==
Consider the polynomial:
:$\map P x = x^3 - 2 x + 4$
By the Schönemann-Eisenstein Theorem, it is not possible to tell whether or not $\map P x$ is irreducible over $\Q \sqbrk x$.
In fact, in this case $\map P x$ is not irreducible over $\Q \sqbrk x$.
== Proof ==
We note that the prime number $2$:
:is a divisor of the coefficient of $x^1$, that is, $-2$
:is not a divisor of the degree of $\map P x$, that is, $3$
and that:
:$2^2$ is a divisor of the coefficient of $x_0$, that is, $4$.
Hence, by the Schönemann-Eisenstein Theorem, it is not necessarily the case that $\map P x$ is not irreducible over $\Q \sqbrk x$.
However, we note that:
:$x^3 - 2 x + 4 = \paren {x + 2} \paren {x^2 - 2 x + 2}$
demonstrating that $\map P x$ is indeed not irreducible over $\Q \sqbrk x$.
== Sources ==
* 
* 
Category:Examples of Use of Schönemann-Eisenstein Theorem
Schönemann-Eisenstein Theorem/Warning

== Schönemann-Eisenstein Theorem: Warning ==
The converse of the Schönemann-Eisenstein Theorem does not hold.
That is, if a polynomial over $\Z$ is irreducible in $\Q \sqbrk x$, it is not necessarily the case that the criteria:
:$(1): \quad p \divides a_i \iff i \ne d$
:$(2): \quad p^2 \nmid a_0$
where:
:$p$ is a prime
:$p \divides a_i$ signifies that $p$ is a divisor of $a_i$
both hold.
See Schönemann-Eisenstein Theorem: $x^3 + 2 x + 4$ for a counterexample.
Category:Schönemann-Eisenstein Theorem
Scientific Notation/Examples/Arbitrary Examples

== Examples of Scientific Notation ==
Various numbers are specified in scientific notation as follows:
=== Example 1 ===
=== Example 2 ===
=== Example 3 ===
=== Example 4 ===
=== Example 5 ===
=== Example 6 ===
=== Example 7 ===
=== Example 8 ===
Category:Examples of Scientific Notation
Scientific Notation/Examples/Division/Arbitrary Example

== Example of Multiplication and Division using Scientific Notation ==
:$\dfrac {\paren {0 \cdotp 006} \paren {80 \, 000} } {0 \cdotp 04} = 12 \, 000$
== Proof ==
== Sources ==
* : Chapter $1$: Scientific Notation: Example 3.
Category:Examples of Scientific Notation