avewright/proofwiki-math huggingface dataset:
Theorem: Finite Multiplicative Subgroup of Field is Cyclic

Let $\struct {F, +, \times}$ be a field.
Let $\struct {F^*, \times}$ denote the multiplicative group of $F$.
Let $C$ be a finite subgroup of $\struct {F^*, \times}$.
Then $C$ is cyclic.

Proof:
We have {{afortiori}} that $\struct {F^*, \times}$ is an abelian group.
From Subgroup of Abelian Group is Abelian, $C$ is a finite abelian group.
From Fundamental Theorem of Finite Abelian Groups, $C$ is the internal group direct product of cyclic groups $H_1, H_2, \ldots, H_r$ whose orders are given as:
:$\order {H_i} = {p_i}^{e_i}$
for some prime numbers $p_1, p_2, \ldots, p_r$ and positive integers $e_1, e_2, \ldots, e_r$.
From Internal and External Group Direct Products are Isomorphic:
:$C \cong H_1 \times H_2 \times \cdots \times H_r$
where $\cong$ denotes (group) isomorphism.
As Powers of Coprime Numbers are Coprime, the orders of $H_1, \ldots, H_r$ are coprime.
By Group Direct Product of Cyclic Groups, $H_1 \times \ldots \times H_r$ is cyclic.
Hence $C$ is cyclic.
{{qed}}
Category:Finite Multiplicative Subgroup of Field is Cyclic
Category:Field Theory
Category:Cyclic Groups
ikzswvdi5kpdewnj93718b2rz2zltor
Theorem: Induced Homomorphism of Polynomial Forms

Let $R$ and $S$ be commutative rings with unity.
Let $\phi: R \to S$ be a ring homomorphism.
Let $R \sqbrk X$ and $S \sqbrk X$ be the rings of polynomial forms over $R$ and $S$ respectively in the indeterminate $X$.
Then the map $\overline \phi: R \sqbrk X \to S \sqbrk X$ given by:
:$\map {\overline \phi} {a_0 + a_1 X + \cdots + a_n X^n} = \map \phi {a_0} + \map \phi {a_1} X + \cdots + \map \phi {a_n} X^n$
is a ring homomorphism.

Proof:
Let:
:$f = a_0 + \cdots + a_n X^n$
:$g = b_0 + \cdots + b_m X^m \in R \sqbrk X$
We have:
{{begin-eqn}}
{{eqn | l = \map {\overline \phi} f \map {\overline \phi} g
      | r = \sqbrk {\map \phi {a_0} + \cdots + \map \phi {a_n} X^n} \sqbrk {\map \phi {b_0} + \cdots + \map \phi {b_m} X^m}
      | c = 
}}
{{eqn | r = \sum_{r \mathop = 0}^{m + n} \sum_{i \mathop + j \mathop = r} \map \phi {a_i} \map \phi {b_j} X^r
      | c = {{Defof|Multiplication of Polynomial Forms}}
}}
{{eqn | r = \sum_{r \mathop = 0}^{m + n} \map \phi {\sum_{i \mathop + j \mathop = r} a_i b_j} X^r
      | c = because $\phi$ is a homomorphism
}}
{{eqn | r = \map {\overline \phi} {f g}
      | c = 
}}
{{end-eqn}}
and
{{begin-eqn}}
{{eqn | l = \map {\overline \phi} f + \map {\overline \phi} g
      | r = \sqbrk {\map \phi {a_0} + \cdots + \map \phi {a_n} X^n} + \sqbrk {\map \phi {b_0} + \cdots + \map \phi {b_m} X^m}
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 0}^{\max \set {m, n} } \sqbrk {\map \phi {a_i} + \map \phi {b_i} } X^i
      | c = {{Defof|Addition of Polynomial Forms}}
}}
{{eqn | r = \sum_{i \mathop = 0}^{\max \set {m, n} } \map \phi {a_i + b_i} X^i
      | c = because $\phi$ is a homomorphism
}}
{{eqn | r = \map {\overline \phi} {f + g}
      | c = 
}}
{{end-eqn}}
Thus $\overline \phi$ is a ring homomorphism.
{{qed}}
Theorem: Structure of Simple Algebraic Field Extension

Let $F / K$ be a field extension.
Let $\alpha \in F$ be algebraic over $K$.
Let $\mu_\alpha$ be the minimal polynomial of $\alpha$ over $K$.
Let $K \sqbrk X$ be the ring of polynomial functions.
Let $\gen {\mu_\alpha}$ be the ideal of $K \sqbrk X$ generated by $\mu_\alpha$.
Let $K \sqbrk \alpha$ be the subring of $F$ generated by $K \cup \set \alpha$.
Let $\map K \alpha$ be the subfield of $F$ generated by $K \cup \set \alpha$.
Then:
:$K \sqbrk \alpha = \map K \alpha \simeq K \sqbrk X / \gen {\mu_\alpha}$
Moreover:
:$n := \index {\map K \alpha} K = \deg \mu_\alpha$
and:
:$1, \alpha, \dotsc, \alpha^{n - 1}$ is a basis of $\map K \alpha$ over $K$.

Proof:
Define $\phi: K \sqbrk X \to K \sqbrk \alpha$ by:
:$\map \phi f = \map f \alpha$
We have
:$\map \phi f = 0 \iff \map f \alpha = 0 \iff \mu_\alpha \divides f$
where the last equivalence follows from $\mu_\alpha$ being the minimal polynomial of $\alpha$.
Thus:
:$\ker \phi = \set {f \in K \sqbrk X: \mu_\alpha \divides f} =: \gen {\mu_\alpha}$
By {{Corollary|Field Adjoined Set}}, $\phi$ is surjective.
So by the First Ring Isomorphism Theorem:
:$K \sqbrk X / \gen {\mu_\alpha} \simeq K \sqbrk \alpha$
By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal:
:$\gen {\mu_\alpha}$ is maximal.
So $K \sqbrk \alpha$ is a field by Maximal Ideal iff Quotient Ring is Field.
Also $K \sqbrk \alpha$ is the smallest ring containing $K \cup \set \alpha$.
So because a field is a ring it is also the smallest field containing $K \cup \set \alpha$.
This shows that $K \sqbrk \alpha = \map K \alpha$.
By Field Adjoined Set:
:$K \sqbrk \alpha = \set {\map f \alpha: f \in K \sqbrk X}$
where $K \sqbrk X$ is the ring of polynomial functions over $K$.
From Division Theorem for Polynomial Forms over Field, for each $f \in K \sqbrk X$ there are $q, r \in K \sqbrk X$ such that:
:$f = q \mu_\alpha + r$
and:
:$\deg r < d =: \deg \mu_\alpha$
Therefore, since $\map {\mu_\alpha} \alpha = 0$, we have 
{{begin-eqn}}
{{eqn | l = K \sqbrk \alpha
      | r = \set {\map r \alpha: r \in K \sqbrk X, \ \deg r < \deg \mu_\alpha}
}}
{{eqn | r = \set {a_0 + a_1 \alpha + \dotsb + a_{d - 1} \alpha^{d - 1}: \alpha_0, \dotsc, \alpha_d \in K}
}}
{{end-eqn}}
Therefore $1, \dotsc, \alpha^{d - 1}$ span $K \sqbrk \alpha / K$ as a vector space.
Moreover if $s \in K \sqbrk X$, then $\map s \alpha = 0$ and $\deg s < \deg \mu_\alpha$ implies that $s = 0$.
Thus $1, \dotsc, \alpha^{d - 1}$ are linearly independent.
Therefore no non-zero $K$-combination of $1, \dotsc, \alpha^{d - 1}$ is zero.
Therefore $\set {1, \dotsc, \alpha^{d - 1} }$ is a basis of $K \sqbrk \alpha / K$.
The degree of $K \sqbrk \alpha / K$ is by definition the number of elements of a basis for $K \sqbrk \alpha / K$.
Therefore:
:$d = \deg \mu_\alpha = \index {K \sqbrk \alpha} K$
{{qed}}
Category:Field Extensions
kthcd9ji6cr59rz76letoipwq9e0437
Theorem: Subset of Particular Point Space is either Open or Closed

Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $H \subseteq S$ be any subset of $T$.
Then $H$ is either open or closed in $T$.
The only sets which are both closed and open in $T$ are $S$ and $\O$.

Proof:
Let $H \subseteq S$.
There are two cases to consider:
:$p \in H$
:$p \notin H$
If $p \in H$ then by definition of a particular point topology, $H$ is open.
If $p \notin H$, then $p \in \relcomp S H$, where $\relcomp S H$ is the relative complement of $H$ in $S$.
So $\relcomp S H$ is open by definition of a particular point topology.
From the definition of a closed set, $\relcomp S {\relcomp S H}$ is closed in $T$.
From Relative Complement of Relative Complement we have that $\relcomp S {\relcomp S H} = H$ and so $H$ is closed in $T$.
Now suppose $H \subseteq T$ is both closed and open in $T$.
From Open and Closed Sets in Topological Space, if $H = \O$ or $H = S$ then H is both closed and open in $T$.
So, suppose $H \ne \O$ or $H \ne S$.
As $H \subseteq T$ is both closed and open in $T$, $\relcomp S H$ is also both closed and open in $T$.
From Boundary is Intersection of Closure with Closure of Complement we have:
:$\partial H = H^- \cap \paren {\relcomp S H}^-$
where $H^-$ is the closure of $H$.
By Closure of Open Set of Particular Point Space we have that $H^- = S$, and of course $\paren {\relcomp S H}^- = S$.
So:
:$\partial H = H^- \cap \paren {\relcomp S H}^- = S$
However, from Set Clopen iff Boundary is Empty we have that $\partial H = \O$.
Hence if $H$ is both closed and open in $T$, then $H = \O$ or $H = S$.
{{qed}}
Category:Particular Point Topologies
Category:Open Sets
Category:Closed Sets
Category:Clopen Sets
i8o1wvls9jj3k3ytrshzdxggmxac1yv
Theorem: Particular Point Space is Weakly Locally Compact

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is weakly locally compact.

Proof:
By Compact Space in Particular Point Space, $\set p$ is compact in $T$.
But by definition $\set p$ is open in $T$.
So $p$ is contained in a compact neighborhood, that is, $\set p$.
Now let $x \in S: x \ne p$.
Then $\set {x, p}$ is open in $T$.
We have that $\set {x, p}$ trivially has an open cover, that is, $\set {\set {x, p} }$ itself.
Hence any open cover of $\set {x, p}$ has a finite subcover: any one set that contains $x$ and $p$ is a cover for $\set {x, p}$.
So $\set {x, p}$ is a neighborhood of $x$ which is compact in $T$.
So $x$ is contained in a compact neighborhood.
Hence the result, by definition of weakly locally compact.
{{qed}}
Theorem: Infinite Particular Point Space is not Strongly Locally Compact

Let $T = \struct {S, \tau_p}$ be an infinite particular point space.
Then $T$ is not strongly locally compact.

Proof:
By definition, $T$ is strongly locally compact {{iff}} every point of $S$ is contained in an open set whose closure is compact.
Let $x \in S: x \ne p$.
Let $x \in U$ where $U$ is open in $T$.
From Closure of Open Set of Particular Point Space we have:
:$U^- = S$
where $U^-$ is the closure of $U$.
But from Infinite Particular Point Space is not Compact we have that $S$ is not compact.
Hence the result, by definition of strongly locally compact.
{{qed}}
Theorem: Infinite Particular Point Space is not Compact

Let $T = \struct {S, \tau_p}$ be an infinite particular point space.
Then $T$ is not compact.

Proof:
Consider the open cover of $T$:
:$\CC = \set {\set {x, p}: x \in S, x \ne p}$
As $S$ is infinite, then so is $\CC$, as we can set up a bijection from $\phi: S \setminus \set p \leftrightarrow \CC$:
:$\forall x \in S \setminus \set p: \map \phi x = \set {x, p}$
Hence $\CC$ has no finite subcover.
The result follows by definition of compactness.
{{qed}}
Theorem: Uncountable Particular Point Space is not Lindelöf

Let $T = \struct {S, \tau_p}$ be an uncountable particular point space.
Then $T$ is not a Lindelöf space.

Proof:
Consider the open cover of $T$:
:$\CC = \set {\set {x, p}: x \in S, x \ne p}$
As $S$ is uncountable, then so is $\CC$, as we can set up a bijection $\phi: S \setminus \set p \leftrightarrow \CC$:
:$\forall x \in S \setminus \set p: \map \phi x = \set {x, p}$
Hence $\CC$ has no countable subcover.
The result follows by definition of Lindelöf space.
{{qed}}
Theorem: Countable Particular Point Space is Lindelöf

Let $T = \struct {S, \tau_p}$ be a countable particular point space.
Then $T$ is a Lindelöf apace.

Proof:
Consider the open cover of $T$:
:$\CC = \set {\set {x, p}: x \in S, x \ne p}$
As $S$ is countable, then so is $\CC$, as we can set up a bijection from $\phi: S \setminus \set p \leftrightarrow \CC$:
:$\forall x \in S \setminus \set p: \map \phi x = \set {x, p}$
Hence $\CC$ is its own countable subcover.
The result follows by definition of Lindelöf apace.
{{qed}}
Theorem: Particular Point Space is Separable

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is separable.

Proof:
By definition, $T$ is separable {{iff}} there exists a countable subset of $S$ which is everywhere dense in $T$.
Consider $U := \set p \subseteq S$.
By definition, $U$ is open in $T$.
From Closure of Open Set of Particular Point Space we have that $U^- = S$, where $U^-$ is the closure of $U$.
By definition, $U$ is everywhere dense in $T$.
$U$ is (trivially) countable.
Hence the result, by definition of a separable space.
{{qed}}
Theorem: Field Adjoined Set

Let $F$ be a field.
Let $S \subseteq F$ be a subset of $F$.
Let $K \le F$ be a subfield of $F$.
The subring $K \sqbrk S$ of $F$ generated by $K \cup S$ is the set of all finite linear combinations of powers of elements of $S$ with coefficients in $K$.
The subfield $\map K S$ of $F$ generated by $K \cup S$ is the set of all $x y^{-1} \in F$ with $a, b \in K \sqbrk S$, $b \ne 0$.
$\map K S$ is isomorphic to the field of quotients $Q$ of $K \sqbrk S$.

Proof:
{{explain|Most of this needs to be explained in more detail. Suffers from coherence issues.}}
Let $\set {X_s: s \in S}$ be a family of indeterminates indexed by $S$.
Let $\phi$ be the Evaluation Homomorphism such that $\phi \sqbrk {X_s} = s$.
{{explain|Can the above Evaluation Homomorphism link be replaced by Definition:Polynomial Evaluation Homomorphism?}}
By Ring Homomorphism Preserves Subrings, $\Img \phi$ is a subring of $F$
This subring contains $\phi \sqbrk {X_s} = s$ for each $s \in S$ and $\map \phi k = k$ for all $k \in K$.
Moreover since it is obtained by evaluating polynomials on $S$ it is the set of all finite linear combinations of powers elements of $S$ with coefficients in $K$ as claimed.
All these linear combinations must belong to any subring of $F$ that contains $K$ and $S$ (otherwise it is not closed), so $\Img \phi$ is the smallest such subring.
By Universal Property for Field of Quotients, the inclusion $K \sqbrk S \to F$ extends uniquely to a homomorphism $\psi : Q \to F$, given by $\map \psi {a / b} = a b^{-1}$.
We have:
:$\Img \psi = \set {a b^{-1} \in F: a, b \in K \sqbrk S, \ b \ne 0} \simeq Q$
The isomorphism comes from the fact that a field homomorphism is injective and the First Ring Isomorphism Theorem.
Thus $\Img \phi$ is a subfield of $F$ containing $K \sqbrk S$ and $K \cup S$.
Any subfield of $F$ containing $K$ and $S$ must contain $K \cup S$, $K \sqbrk S$ and all $a b^{-1}$ with $a, b \in K \sqbrk S$ (otherwise it would not be closed).
Therefore $Q \simeq \Img \phi$ is the smallest such subfield.
{{qed}}
Category:Field Extensions
2njf4e7mslqjt9fcdptfh8kouoney9q
Theorem: Separability in Uncountable Particular Point Space

Let $T = \struct {S, \tau_p}$ be an uncountable particular point space.
Let $H = S \setminus \set p$ where $\setminus$ denotes set difference.
Then $H$ is not separable.

Proof:
By definition, $H$ is separable {{iff}} there exists a countable subset of $S$ which is everywhere dense in $T$.
Let $V \subseteq H$ where $V$ is countable.
$V$ is not open in $T$ as it does not contain $p$.
From Subset of Particular Point Space is either Open or Closed it follows that $V$ is closed.
From Closed Set Equals its Closure, $V^- = V$.
But $V^- \ne H$ as $V$ is countable and $H$ is uncountable.
So whatever $V$ is, if it is countable it is not everywhere dense.
The result follows from definition of separable.
{{qed}}
Theorem: Uncountable Particular Point Space is not Second-Countable

Let $T = \struct {S, \tau_p}$ be an uncountable particular point space.
Then $T$ is not second-countable.

Proof:
Let $H = S \setminus \set p$ where $\setminus$ denotes set difference.
Every subset $V \subseteq H$ is a closed set from Subset of Particular Point Space is either Open or Closed.
Thus we can consider $H$ as an uncountable discrete space.
The result follows from Uncountable Discrete Space is not Second-Countable.
{{qed}}
Theorem: Particular Point Space is First-Countable

Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $T$ is first-countable.

Proof:
Let $x \in S: x \ne p$.
Consider the set $U_x = \set {x, p} \subseteq S$.
Now let $V \in \tau_p$ be an open set in $S$ such that $x \in V$.
So $x \in V$, by definition of $V$, and $p \in V$ as $V$ is open.
It follows directly that $U_x \subseteq V$.
So $\set {U_x}$ is a local basis at $x$ which is (trivially) countable.
If $x = p$ then $U_x = \set p$ and the same argument still applies.
So every element of $T$ has a countable local basis.
Hence $T$ is first-countable by definition.
{{qed}}
Theorem: Kronecker’s Theorem

Let $K$ be a field.
Let $f$ be a polynomial over $K$ of degree $n \ge 1$.
Then there exists a finite extension $L / K$ of $K$ such that $f$ has at least one root in $L$.
Moreover, we can choose $L$ such that the degree $\index L K$ of $L / K$ satisfies $\index L K \le n$.
{{explain|Work out exactly which definitions of Polynomial and Degree are appropriate here.}}

Proof:
Let $K \sqbrk X$ be the ring of polynomial forms over $K$.
By Polynomial Forms over Field form Unique Factorization Domain, $K \sqbrk X$ is a unique factorisation domain.
Therefore, we can write $f = u g_1 \cdots g_r$, where $u$ a unit of $K \sqbrk X$ and $g_i$ is irreducible for $i = 1, \ldots, r$.
Clearly it is sufficient to find an extension of $K$ in which the irreducible factor $g_1$ of $f$ has a root.
Let $L = K \sqbrk X / \gen {g_1}$, where $\gen {g_1}$ is the ideal generated by $g_1$.
By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, $\gen {g_1}$ is maximal.
Therefore by Maximal Ideal iff Quotient Ring is Field, $L$ is a field.
Moreover, writing the quotient ring epimorphism $\overline {\map p X} = \map p X + \gen {g_1}$ for $\map p X \in K \sqbrk X$:
{{begin-eqn}}
{{eqn | l = \map {g_1} {\overline X}
      | r = \overline {\map {g_1} X}
      | c = 
}}
{{eqn | r = \overline {0_{K \sqbrk X} }
      | c = 
}}
{{eqn | r = 0_L
      | c = 
}}
{{end-eqn}}
So $\overline X$ is a root of $g_1$ in $L$.
It remains to show that $\index L K \le n$.
By the Division Theorem for Polynomial Forms over Field, every polynomial $\map p X \in K \sqbrk X$ can be written as:
:$\map p X = \map q X \map {g_1} X + \map r X$
with $\map \deg {\map r X} < \map \deg {\map {g_1} X} \le \map \deg {\map f X} = n$.
Now we have by the definition of the quotient ring:
{{begin-eqn}}
{{eqn | l = \overline {\map p X}
      | r = \overline {\map q X \map {g_1} X + \map r X}
      | c = 
}}
{{eqn | r = \overline {\map r X}
      | c = 
}}
{{end-eqn}}
So if $\map r X = r_0 + r_1 X + \cdots + r_{n - 1} X^{n - 1}$, we have:
:$\overline {\map p X} = \overline {r_0 + r_1 X + \cdots + r_{n - 1} X^{n - 1} }$
Our choice of $p$ was arbitrary, so every polynomial can be written in this form.
In particular the set $\set {\overline 1, \overline X, \ldots, \overline {X^{n - 1} } }$ spans $L$ over $K$.
{{explain|At this stage, how is it established that $L$ is a vector space?}}
Thus by Generator of Vector Space Contains Basis, a basis of $L$ has at most $n$ elements.
That is, $\index L K \le n$.
{{qed}}
{{Namedfor|Leopold Kronecker|cat = Kronecker}}
Category:Field Extensions
5266qki92kdstp8h61e9gzc2j0gxdh0
Theorem: Homeomorphic Non-Comparable Particular Point Topologies

Let $S$ be a set with at least two elements.
Let $p, q \in S: p \ne q$.
Let $\tau_p$ and $\tau_q$ be the particular point topologies on $S$ by $p$ and $q$ respectively.
Then the topological spaces $T_p = \struct {S, \tau_p}$ and $T_q = \struct {S, \tau_q}$ are homeomorphic.
However, $\tau_p$ and $\tau_q$ are not comparable.

Proof:
We can set up the mapping $\phi: S \to S$:
:<nowiki>$\forall x \in S: \map \phi x = \begin {cases}
q & : x = p \\
p & : x = q \\
x & : \text {otherwise} \end {cases}$</nowiki>
It is straightforward to show that $\phi$ is a homeomorphism.
However, we have, for example, that $\set q \notin \tau_p$ and $\set p \notin \tau_q$.
So neither $\tau_p$ nor $\tau_q$ is a subset of the other.
Hence by definition $\tau_p$ and $\tau_q$ are not comparable. 
{{qed}}
Theorem: Finite Field Extension is Algebraic

Let $L / K$ be a finite field extension.
Then $L / K$ is algebraic.

Proof:
Let $x \in L$ be arbitrary.
Let $n = \index L K$ be the degree of $L$ over $K$.
From Size of Linearly Independent Subset is at Most Size of Finite Generator, there is a $K$-linear combination of $\set {1, \ldots, x^n}$ equal to $0$.
Say $a_n x^n + \cdots + a_1 x + a_0 = 0$, $a_i \in K$, $i = 0, \ldots, n$.
Therefore $x$ satisfies a polynomial with coefficients in $K$.
{{explain|Clarify whether the above link is what it is supposed to be}}
That is, $x$ is algebraic.
Since $x \in L$ was chosen arbitrarily, $L / K$ is algebraic.
{{qed}}
Category:Field Extensions
a8tbknuuwb412ec7o6pcnvbduym0pww
Theorem: Finitely Generated Algebraic Extension is Finite

Let $L / K$ be a field extension.
Let $\alpha_1, \ldots, \alpha_n \in L$ be algebraic over $K$.
Let $K \sqbrk {\alpha_1, \ldots, \alpha_n}$ be the field generated by $\set {\alpha_1, \ldots, \alpha_n}$.
Then $K \sqbrk {\alpha_1, \ldots, \alpha_n} / K$ is a finite field extension.

Proof:
Let $S = \set {\alpha_1, \ldots, \alpha_n}$.
We show by induction on $n$ that $K \sqbrk S / K$ is finite.
Clearly $K$ is finite over itself, so the result holds when $n = 0$.
Now suppose that for all sets $T \subseteq L$ with $\card T \le n - 1$ and each element of $T$ algebraic over $K$, $K \sqbrk T / K$ is finite.
We have:
:$K \sqbrk S = K \sqbrk {\alpha_1, \ldots, \alpha_{n - 1} } \sqbrk {\alpha_n}$
By the induction hypothesis $K \sqbrk {\alpha_1, \ldots, \alpha_{n - 1} } / K$ is finite.
By the Structure of Simple Algebraic Field Extension $K \sqbrk S / K \sqbrk {\alpha_1, \ldots, \alpha_{n - 1} }$ is finite.
Therefore, by the Tower Law, $K \sqbrk S / K$ is finite.
{{qed}}
Category:Field Extensions
Category:Proofs by Induction
7mzefnsxf3mxrqi82p6sq21tqcmgm8l
Theorem: Transitivity of Algebraic Extensions

Let $E / F / K$ be a tower of field extensions.
Let $E$ be algebraic over $F$.
Let $F$ be algebraic over $K$.
Then $E$ is algebraic over $K$.

Proof:
Let $x \in E$.
There are $a_0, \ldots, a_n \in F$ such that $a_0 + \cdots + a_n x^n = 0$.
Let $L = \map K {a_0, \ldots, a_n}$.
We have that $L / K$ is finitely generated and algebraic.
Therefore by Finitely Generated Algebraic Extension is Finite this extension is finite.
We have that $\map L x / L$ is simple and algebraic.
So by Structure of Simple Algebraic Field Extension, this extension is also finite.
Therefore, by the Tower Law:
:$\map L x / K$ is finite.
That is, $x$ is contained in a finite extension of $K$.
Therefore because a Finite Field Extension is Algebraic, it follows that $x$ is algebraic over $K$, as was to be proved.
{{qed}}
Theorem: Maximal Algebraic Extension is Subfield

Let $L / K$ be a field extension.
Let $K^a$ be the maximal algebraic extension of $K$ contained in $L$.
Then $K^a$ is a subfield of $L$.

Proof:
Let $\alpha, \beta \in K^a$.
By Field Adjoined Algebraic Elements is Algebraic, $\map K {\alpha, \beta} / K$ is algebraic.
By definition, $\map K {\alpha, \beta}$ is a field.
Therefore $\alpha \beta$, $\alpha^{-1}$ and $\alpha - \beta$ all lie in $\map K {\alpha, \beta}$.
Hence all are algebraic over $K$.
Also:
$K \subseteq K^a$
so:
:$K^a \ne \O$
By the Subfield Test:
:$K^a \le L$ is a subfield of $L$.
{{qed}}
Category:Field Extensions
c5a0ij0kr7yf609wr0mmvevthoo5u4x
Theorem: Equivalence of Definitions of Algebraically Closed Field

Let $K$ be a field. 
{{TFAE|def = Algebraically Closed Field}}

Proof:
=== Definition $(1)$ implies Definition $(2)$ ===
Let $K$ be algebraically closed by definition 1.
Let $f$ be an irreducible polynomial over $K$.
By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, the ideal $\gen f$ generated by $f$ is maximal.
So by Maximal Ideal iff Quotient Ring is Field:
:$L = K \sqbrk X / \gen f$ is a field
where $L$ is a field extension over $K \sqbrk X$.
Now:
:$L = \set {g + \gen f: g \in K \sqbrk X}$
From Division Theorem for Polynomial Forms over Field:
:$\forall g \in K \sqbrk X: \exists q, r \in K \sqbrk X: g = q f + r, \deg r < \deg f =: n$
Therefore:
:$L = \set {r + \gen f: r \in K \sqbrk X, \deg r < n}$
By Basis for Quotient of Polynomial Ring, this has basis:
:$1 + \gen f, \ldots, X^{n - 1} + \gen f$ span $L$.
Thus $L$ is finite.
By Finite Field Extension is Algebraic, $L$ is algebraic.
Also $K \subseteq L$.
So by hypothesis $K = L$.
This implies:
:$\index L K = 1$
where $\index L K$ is the degree of $L$ over $K$.
Hence:
:$n = \deg f = 1$
Thus $K$ is algebraically closed by definition 2.
{{qed|lemma}}
Theorem: Closed Set in Particular Point Space has no Limit Points

Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $H \subsetneq S$ be closed in $T$.
Then $H$ has no limit points.

Proof:
Let $H$ be closed in $T$.
Then by definition $p \notin H$.
Let $x \in H$.
By definition, $x$ is a limit point of $H$ if every open set $U \in \tau$ such that $x \in U$ contains some point of $H$ other than $x$.
Consider the set $U_x := \set {x, p} \subseteq S$.
As $p \in U_x$ we have that $U_x$ is open in $T$.
But there is no $y \in U_x: y \in H, y \ne x$ and so $x$ is not a limit point of $H$.
So $H$ has no limit points.
{{qed}}