avewright/proofwiki-math huggingface dataset:
Theorem: Ring of Integers of Number Field is Dedekind Domain

Let $K$ be an algebraic number field.
Let $\OO_K$ be its ring of integers.
Then $\OO_K$ is a Dedekind domain.

Proof:
{{proof wanted}}
Category:Rings of Integers of Number Fields
Category:Dedekind Domains
Category:Algebraic Number Theory
o6i6oxc58m68oqpjj6hu6jlxxvuofok
Theorem: Sum of Reciprocals of Squares plus 1

:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2 + 1}  = \frac 1 2 \paren {\pi \coth \pi - 1}$

Proof:
Let $\map f x$ be the real function defined on $\hointl {-\pi} \pi$ as:
:$\map f x = e^x$
From Fourier Series: $e^x$ over $\openint {-\pi} \pi$, we have:
:$(1): \quad \ds \map f x \sim \map S x = \frac {\sinh \pi} \pi \paren {1 + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\cos n x} {1 + n^2} - \frac {n \sin n x} {1 + n^2} } }$
Let $x = \pi$.
By Fourier's Theorem, we have:
{{begin-eqn}}
{{eqn | l = \map S \pi
      | r = \frac 1 2 \paren {\lim_{x \mathop \to \pi^+} \map f x + \lim_{x \mathop \to \pi^-} \map f x}
      | c = 
}}
{{eqn | r = \dfrac {e^\pi + e^{-\pi} } 2
      | c = 
}}
{{eqn | r = \cosh \pi
      | c = {{Defof|Hyperbolic Cosine}}
}}
{{end-eqn}}
Thus setting $x = \pi$ in $(1)$, we have:
{{begin-eqn}}
{{eqn | l = \cosh \pi
      | r = \frac {\sinh \pi} \pi \paren {1 + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\cos \pi} {1 + n^2} - \frac {n \sin \pi} {1 + n^2} } }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\pi \cosh \pi} {\sinh \pi}
      | r = 1 + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\paren {-1}^n} {1 + n^2} - \frac {n \sin \pi} {1 + n^2} }
      | c = Cosine of Multiple of Pi and multiplying both sides by $\dfrac \pi {\sinh \pi}$
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\pi \coth \pi - 1} 2
      | r = \sum_{n \mathop = 1}^\infty \frac 1 {1 + n^2}
      | c = {{Defof|Hyperbolic Cotangent|index = 2}}, Sine of Multiple of Pi and simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Leading Coefficient of Product of Polynomials over Integral Domain

Let $R$ be an integral domain.
Let $f, g \in R \sqbrk x$ be polynomials.
Let $c$ and $d$ be their leading coefficients.
Then $f g$ has leading coefficient $c d$.

Proof:
Let $p = \deg f$ and $q = \deg g$ be their degrees.
By Degree of Product of Polynomials over Integral Domain, $\map \deg {f g} = p + q$.
For a natural number $n \ge 0$, let:
:$a_n$ be the coefficient of the monomial $x^n$ in $f$.
:$b_n$ be the coefficient of the monomial $x^n$ in $g$.
By Coefficients of Product of Two Polynomials, the coefficient of $x^{p + q}$ in $f g$ is the sum:
:$\ds \sum_{k \mathop = p}^p a_k b_{p + q - k}$
By Indexed Summation over Interval of Length One:
:$\ds \sum_{k \mathop = p}^p a_k b_{p + q - k} = a_p b_q = c d$
{{qed}}
Category:Polynomial Theory
i4jft09o5r7jccya4040z7whofeekm4
Theorem: Prime Ideal of Principal Ideal Domain is Maximal

Let $D$ be a principal ideal domain whose zero is $0_D$.
Let $J \subseteq D$ be a nonzero prime ideal.
Then $J$ is maximal.

Proof:
As $D$ is a principal ideal domain, every ideal of $D$ is a principal ideal $\ideal r$ generated by some $r \in D$.
So, let $\ideal p$ be an arbitrary prime ideal of $D$ generated by $p$, where $p \ne 0_R$.
As $\ideal p$ is prime, $p$ is irreducible.
{{explain|Prove the above}}
The result follows from Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal.
{{qed}}
Theorem: Rodrigues' Formula for Legendre Polynomials

:$\map {P_n} x = \dfrac 1 {2^n n!} \dfrac {\d^n} {\d x^n} \paren {x^2 - 1}^n$
where:
:$n \in \N$ is a natural number
:$P_n$ is the $n$th Legendre polynomial.

Proof:
{{ProofWanted}}
{{Namedfor|Olinde Rodrigues|cat = Rodrigues}}
Theorem: Product of Coprime Ideals equals Intersection

Let $A$ be a commutative ring with unity.
Let $\mathfrak a, \mathfrak b \subseteq A$ be coprime ideals.
Then their product equals their intersection:
:$\mathfrak a \mathfrak b = \mathfrak a \cap \mathfrak b$

Proof:
By Intersection of Ideals of Ring contains Product:
:$\mathfrak a \mathfrak b \subseteq \mathfrak a \cap \mathfrak b$
It remains to show that $\mathfrak a \mathfrak b \supseteq \mathfrak a \cap \mathfrak b$.
Let $c \in \mathfrak a \cap \mathfrak b$.
Because $\mathfrak a$ and $\mathfrak b$ are coprime, there exist $x \in \mathfrak a$, $y \in \mathfrak b$ with $x + y = 1$.
Then:
:$c = c x + c y$
Because:
:$c \in \mathfrak b$ and $x \in \mathfrak a$
:$c \in \mathfrak a$ and $y \in \mathfrak b$
we have:
:$c \in \mathfrak a \mathfrak b$
{{qed}}
Theorem: Ideal Contained in Finite Union of Prime Ideals

Let $A$ be a commutative ring with unity.
Let $\mathfrak p_1, \ldots, \mathfrak p_n$ be prime ideals.
Let $\mathfrak a \subseteq \ds \bigcup_{i \mathop = 1}^n \mathfrak p_i$ be an ideal contained in their union.
Then $\mathfrak a \subseteq \mathfrak p_i$ for some $i \in \{1, \ldots, n\}$.

Proof:
The proof goes by induction on $n$.
For $n = 1$, the statement is trivial.
Let $n > 1$.
Suppose that the statement holds for $n - 1$ prime ideals.
It is to be shown that the statement holds for $n$ prime ideals.
{{AimForCont}} that $\mathfrak a \nsubseteq \mathfrak p_i$ for all $i \in \set {1, \ldots, n}$.
By the induction hypothesis, for all $i$ we have:
:$\mathfrak a \nsubseteq \ds \bigcup_{j \mathop \ne i} \mathfrak p_j$
For all $i$, let $x_i \in \mathfrak a \setminus \ds \bigcup_{j \mathop \ne i} \mathfrak p_j$.
Suppose there exists $i$ such that $x_i \notin \mathfrak p_i$.
Then:
:$x_i \in \mathfrak a \setminus \ds \bigcup_j \mathfrak p_j$
and we conclude that:
:$\mathfrak a \nsubseteq \ds \bigcup_j \mathfrak p_j$
which is a contradiction.
{{qed|lemma}}
The other possibility is that for all $i$:
:$x_i \in \mathfrak p_i$
Let:
:$y = \ds \sum_{i \mathop = 1}^n \prod_{\substack{j \mathop = 1 \\ j \mathop\neq i}}^n x_j$
be defined as a summation of products.
All terms except the $i$th are in $\mathfrak p_i$.
Since Sum of an Element in an Ideal and an Element not in the Ideal is not in the Ideal:
:$\forall i:\quad y \notin \mathfrak p_i$
Thus:
:$y \in \mathfrak a \setminus \bigcup_j \mathfrak p_j$
This also leads to a contradiction.
{{qed}}
Theorem: Modulus of Gamma Function of Imaginary Number

Let $t \in \R$ be a real number.
Then: 
:$\cmod {\map \Gamma {i t} } = \sqrt {\dfrac {\pi \csch \pi t} t}$
where:
:$\Gamma$ is the Gamma function
:$\csch$ is the hyperbolic cosecant function.

Proof:
By Euler's Reflection Formula:
:$\map \Gamma {i t} \, \map \Gamma {1 - i t} = \pi \, \map \csc {\pi i t}$
From Gamma Difference Equation: 
:$-i t \, \map \Gamma {i t} \, \map \Gamma {-i t} = \pi \, \map \csc {\pi i t}$
Then:
{{begin-eqn}}
{{eqn | l = \cmod {-i t} \cmod {\map \Gamma {i t} } \cmod {\map \Gamma {-i t} }
      | r = \cmod t \cmod {\map \Gamma {i t} } \cmod {\overline {\map \Gamma {i t} } }
      | c = Complex Conjugate of Gamma Function
}}
{{eqn | r = \cmod t \cmod {\map \Gamma {i t} }^2
      | c = $\cmod z = \cmod {\overline z}$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = \cmod {\pi \, \map \csc {\pi i t} }
      | r = \cmod {-i \pi \, \map \csch {\pi t} }
      | c = Hyperbolic Sine in terms of Sine
}}
{{eqn | r = \pi \, \map \csch {\pi \cmod t}
      | c = Hyperbolic Sine Function is Odd
}}
{{end-eqn}}
So:
:$\cmod {\map \Gamma {i t} }^2 = \dfrac {\pi \, \map \csch {\pi \cmod t} } {\cmod t}$
As $\cmod z \ge 0$ for all complex numbers $z$, we can take the non-negative square root and write:
:$\cmod {\map \Gamma {i t} } = \sqrt {\dfrac {\pi \, \map \csch {\pi \cmod t} } {\cmod t} }$
However, by Hyperbolic Sine Function is Odd: 
:$\dfrac {\pi \, \map \csch {-\pi t} } {-t} = \dfrac {-\pi \, \map \csch {\pi t} } {-t} = \dfrac {\pi \, \map \csch {\pi t} } t$
Hence we can remove the modulus and simply write: 
:$\cmod {\map \Gamma {i t} } = \sqrt {\dfrac {\pi \csch \pi t} t}$
{{qed}}
Theorem: Equivalence of Definitions of Unital Associative Commutative Algebra/Correspondence

Let $B$ be a algebra over $A$ that is unital, associative and commutative.
Let $\struct {C, f}$ be a ring under $A$.
{{TFAE}}
:$(1): \quad C$ is the underlying ring of $B$ and $f: A \to C$ is the canonical homomorphism to the unital algebra $B$.
:$(2): \quad B$ is the algebra defined by $f$.

Proof:
Let $\cdot: A \times B \to B$ the ring action of $B$.
Theorem: Existence of Homomorphism between Localizations of Ring

Let $A$ be a commutative ring with unity.
Let $S, T \subseteq A$ be multiplicatively closed subsets.
{{TFAE}}
:$(1): \quad$ There exists an $A$-algebra homomorphism $h : A_S \to A_T$ between localizations, the '''induced homomorphism'''.
:$(2): \quad S$ is a subset of the saturation of $T$.
:$(3): \quad$  The saturation of $S$ is a subset of the saturation of $T$.
:$(4): \quad$  Every prime ideal meeting $S$ also meets $T$.

Proof:
Let $i : A \to A_S$ and $j : A \to A_T$ be the localization homomorphisms.
Theorem: Half-Range Fourier Cosine Series/Identity Function/0 to Pi

Let $\map f x$ be the real function defined on $\openint 0 \pi$ as:
:400pxthumbright$\map f x$ and its $4$th approximation
:$\map f x = x$
Then its half-range Fourier cosine series can be expressed as:
{{begin-eqn}}
{{eqn | l = x
      | o = \sim
      | r = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\cos \paren {2 n - 1} x} {\paren {2 n - 1}^2}
      | c = 
}}
{{eqn | r = \frac \pi 2 - \frac 4 \pi \paren {\cos x + \frac {\cos 3 x} {3^2} + \frac {\cos 5 x} {5^2} + \cdots}
      | c = 
}}
{{end-eqn}}

Proof:
Let $\map f x: \openint 0 \lambda \to \R$ be the identity function on the open real interval $\openint 0 \lambda$:
:$\forall x \in \openint 0 \lambda: \map f x = x$
From Half-Range Fourier Cosine Series for Identity Function, the half-range Fourier cosine series for $\map f x$ can be expressed as:
{{begin-eqn}}
{{eqn | l = \map f x
      | o = \sim
      | r = \frac \lambda 2 - \frac {4 \lambda} {\pi^2} \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} \cos \dfrac {\paren {2 n + 1} \pi x} \lambda
      | c = 
}}
{{eqn | r = \frac \lambda 2 - \frac {4 \lambda} {\pi^2} \paren {\cos \dfrac {\pi x} \lambda + \frac 1 {3^2} \cos \dfrac {3 \pi x} \lambda + \frac 1 {5^2} \cos \dfrac {5 \pi x} \lambda + \dotsb}
      | c = 
}}
{{end-eqn}}
The result follows by setting $\lambda = \pi$.
{{qed}}
Theorem: Existence of Homomorphism between Localizations of Ring at Elements

Let $A$ be a commutative ring with unity.
Let $f, g \in A$.
{{TFAE}}
:$(1): \quad$ There exists an $A$-algebra homomorphism $h : A_f \to A_g$ between localizations, the '''induced homomorphism'''.
:$(2): \quad f$ divides some power of $g$.
:$(3): \quad$ There is an inclusion of vanishing sets: $\map V f \subseteq \map V g$. That is, every prime ideal containing $f$ also contains $g$.
:$(4): \quad$ There is an inclusion of principal open subsets: $\map D f \supseteq \map D g$

Proof:
=== 1 implies 2 ===
Let there exist an $A$-algebra homomorphism $A_f \to A_g$.
By Existence of Homomorphism between Localizations of Ring, $f$ is contained in the saturation of the set of powers $\set {g^n : n \in \N}$.
That is, $f$ divides some power of $g$.
{{qed|lemma}}
Theorem: Hilbert's Basis Theorem for Finitely Generated Algebras

Let $A$ be a Noetherian ring.
Let $B$ be a finitely generated algebra over $A$.
Then $B$ is a Noetherian ring.

Proof:
Let $\set {b_1, \ldots , b_n} \subseteq B$ be a generator of $B$.
Let $A \sqbrk {X_1, \ldots, X_n}$ be the ring of polynomial forms over $A$ in $\sequence {X_1, \ldots ,X_n}$.
By Hilbert's basis theorem for polynomial rings, $A \sqbrk {X_1, \ldots, X_n}$ is Noetherian.
Let $\phi : A \sqbrk {X_1, \ldots, X_n} \to B$ be the homomorphism defined by:
:$\map f {X_1, \ldots , X_n} \mapsto \map f {b_1, \ldots , b_n}$
Then, by {{Defof|Generator of Algebra}}:
:$\Img \phi = B$
Thus by First Ring Isomorphism Theorem:
:$B \cong A \sqbrk {X_1, \ldots, X_n} / \map \ker \phi$
Now the {{RHS}} is Noetherian, since Quotient Ring of Noetherian Ring is Noetherian.
Thus $B$ is Noetherian.
{{qed}}
{{Namedfor|David Hilbert|cat = Hilbert}}
Category:Noetherian Rings
Category:Commutative Algebra
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Theorem: Zariski's Lemma

Let $L / k$ be a field extension.
Let $L$ be finitely generated as an algebra over $k$.
Then $L / k$ is a finite field extension.

Proof:
{{MissingLinks}}
By Noether Normalization Lemma, we find a finite and injective morphism:
:$\alpha: k \sqbrk {x_1, \dotsc, x_n} \to L$
If we can prove that $n = 0$, the proof is complete.
{{AimForCont}} $n > 0$.
Then:
:$x_1 \in k \sqbrk {x_1, \dotsc, x_n}$
and:
:$\map \alpha {x_1} \ne 0$
We have that $\map \alpha {x_1}^{-1}$ is integral over $k \sqbrk {x_1, \dotsc, x_n}$.
Thus there exists a $m \in \N$ and $a_0, \dotsc, a_{m - 1} \in k \sqbrk {x_1, \dotsc, x_n}$ such that:
:$\ds \map \alpha {x_1}^{-m} + \sum_{i \mathop = 0}^{m - 1} \map \alpha {a_i} \map \alpha {x_1}^{-i} = 0$
Multiplying by $\map \alpha {x_1}^m$:
{{begin-eqn}}
{{eqn | l = 0
      | r = 1 + \sum_{i \mathop = 0}^{m - 1} \map \alpha {a_i} \map \alpha {x_1}^{m - i}
}}
{{eqn | r = \map \alpha {1 +  x_1 \paren {\sum_{i \mathop = 0}^{m - 1} a_i x_1^{m - i - 1} } }
}}
{{end-eqn}}
and thus, since $\alpha$ is injective, we find that:
:$\ds 1 = x_1 \paren {-\sum_{i \mathop = 0}^{m - 1} a_i x_1^{m - i - 1} }$
which means that $x_1$ is invertible. 
This contradiction shows that $n = 0$.
{{qed}}
{{Namedfor|Oscar Zariski|cat = Zariski}}
Category:Commutative Algebra
Category:Field Extensions
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Theorem: Cayley-Hamilton Theorem/Matrix

Let $A$ be a commutative ring with unity.
Let $\mathbf N = \sqbrk {a_{i j} }$ be an $n \times n$ matrix with entries in $A$.
Let $\mathbf I_n$ denote the $n \times n$ unit matrix.
Let $\map {p_{\mathbf N} } x$ be the determinant $\map \det {x \cdot \mathbf I_n - \mathbf N}$.
Then:
:$\map {p_{\mathbf N} } {\mathbf N} = \mathbf 0$
as an $n \times n$ zero matrix.
That is:
:$\mathbf N^n + b_{n - 1} \mathbf N^{n - 1} + \cdots + b_1 \mathbf N + b_0 = \mathbf 0$
where the $b_i$ are the coefficients of $\map {p_{\mathbf N} } x$.

Proof:
Taking $\phi = \mathbf N$ in the proof of Cayley-Hamilton Theorem for Finitely Generated Modules we see that $\mathbf N$ satisfies:
:$\map {p_{\mathbf N} } x = \map \det {x \cdot \mathbf I_n - \mathbf N} = 0$
Take $\mathfrak a$ to be the ideal generated by the entries of $\mathbf N$.
{{qed}}
{{explain}}
{{Namedfor|Arthur Cayley|name2 = William Rowan Hamilton}}
Theorem: Cayley-Hamilton Theorem/Finitely Generated Module

Let $A$ be a commutative ring with unity.
Let $M$ be a finitely generated $A$-module.
Let $\mathfrak a$ be an ideal of $A$.
Let $\phi$ be an endomorphism of $M$ such that $\phi \sqbrk M \subseteq \mathfrak a M$.
Then $\phi$ satisfies an equation of the form:
:$\phi^n + a_{n - 1} \phi^{n-1} + \cdots + a_1 \phi + a_0 = 0$
with the $a_i \in \mathfrak a$.

Proof:
Recall that the set of endomorphisms $\map {\End_A} M$ is a unitary $A$-module.
Let $R \subseteq \map {\End_A} M$ be the $A$-submudole generated by $\set {\phi^n : n \in \N}$.
That is:
:$R = \set {a_0 + a_1 \phi + \cdots + a_k \phi^k : k \in \N, a_1, \ldots, a_k \in A}$
Note that $R$ is a commutative ring with unity.
Recall also the ring of $n$-matrices $\map {\MM_R} n$.
Let $\set {m_1, \ldots, m_n}$ be a generator for $M$.
Then for each $i \in \set {1, \ldots, n}$, as $\map \phi {m_i} \in \mathfrak a M$, we can write:
:$\ds \map \phi {m_i} = \sum_{j \mathop = 1}^n a_j m_j$
where $a_j \in A$ for $j \in \set {1, \ldots, n}$.
Thus for each $i \in \set {1, \ldots, n}$:
:$(1): \quad \ds \sum_{j \mathop = 1}^n \sqbrk {\delta_{ij} \phi - a_{ij} \mathbf 1_R} m_i = 0$
where:
:$\delta_{i j}$ is the Kronecker delta with values in $A$
Now let $\Delta \in \map {\MM_R} n$ defined as:
:$\Delta := \paren {\phi \delta_{i j} - a_{i j} \mathbf 1_R }$
Then $\paren 1$ can be written as:
:$\Delta \cdot \begin{bmatrix} m_1 \\ \vdots \\ m_n \end{bmatrix} = \mathbf 0_{R^n}$
Let $\adj \Delta \in \map {\MM_R} n$ be the adjugate matrix of $\Delta$.
By Matrix Product with Adjugate Matrix:
{{begin-eqn}}
{{eqn | l = \adj \Delta \cdot \Delta
      | r = \Delta \cdot \adj \Delta
      | c = 
}}
{{eqn | r = \map \det \Delta \mathbf I_n
      | c = 
}}
{{eqn | r = <nowiki>\begin{bmatrix}
\map \det \Delta & 0 & \cdots & 0 \\
0 & \map \det \Delta & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \map \det \Delta \\
\end{bmatrix}</nowiki>
}}
{{end-eqn}}
where $\map \det \Delta \in R$ denotes the determinant of $\Delta$.
Multiplying the above equality by $\adj \Delta$ and applying Cramer's Rule:
:$\ds \map \det \Delta m_i = 0$
for each $i \in \set {1, \ldots , n}$.
Therefore:
:$\map \det \Delta = \mathbf 0_R$
On the other hand, $\map \det \Delta$ is a monic polynomial in $\phi$, since:
{{begin-eqn}}
{{eqn | l = \map \det \Delta
      | r = \map \det {\delta_{i j} \phi - a_{i j} \mathbf 1_R}
}}
{{eqn | r = \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n \paren {\delta_{k \map \lambda k} \phi - a_{k \map \lambda k} \mathbf 1_R} }
      | c = {{Defof|Determinant|subdef = Matrix|index = 1}}
}}
{{eqn | r = \phi^n + a_{n - 1} \phi^{n-1} + \cdots + a_1 \phi + a_0 \mathbf 1_R
}}
{{end-eqn}}
where:
:$\lambda$ goes over all permutations of $\set {1, 2, \ldots, n}$
:$\map \sgn \lambda$ is the sign of the permutation $\lambda$
:$a_0, a_1, \ldots, a_{n-1} \in \mathfrak a$
As $R \subseteq \map {\End_A} M$, this means:
:$\phi^n + a_{n - 1} \phi^{n-1} + \cdots + a_1 \phi + a_0 = 0$
{{qed}}
{{Namedfor|Arthur Cayley|name2 = William Rowan Hamilton}}
Category:Cayley-Hamilton Theorem
Category:Commutative Algebra
jl6kdhd2nhc321y50wel7tcg5m5yk9h
Theorem: Weak Nullstellensatz

Let $K$ be an algebraically closed field.
Let $n \ge 0$ be an natural number.
Let $K \sqbrk {x_1, \ldots, x_n}$ be the polynomial ring in $n$ variables over $k$.
Let $I \subseteq K \sqbrk {x_1,\ldots, x_n}$ be an ideal.
{{TFAE}}
# $I$ is the unit ideal: $I = (1)$.
# Its zero-locus is empty set: $\map V I = \O$.

Proof:
{{proof wanted}}
Category:Algebraic Geometry
ia15k477bwcg0dqffpy95zox0hid4sq
Theorem: Finite-Dimensional Integral Domain over Field is Field

Let $k$ be a field.
Let $R$ be a $k$-algebra of finite dimension which is an integral domain.
Then $R$ is a field.

Proof:
{{tidy}}
We will prove this using the Rank - Nullity Theorem applied to $A$ viewing it as a finite dimensional vector space over $K$. Pick a non-zero element $y \in A$ and consider the $K$ - linear transformation $$\begin{eqnarray*} T&\colon&A \longrightarrow A \\
&& x\mapsto yx.\end{eqnarray*} $$
Now the kernel of this linear transformation is trivial because $A$ is an integral domain, and hence by rank-nullity we have that 
$$\begin{eqnarray*} \dim_K A &=& \dim_K \ker T + \dim_K \operatorname{im} T \\
&=& 0 + \dim_K \operatorname{im} T\\
&=& \dim_K \operatorname{im} T \end{eqnarray*}$$
from which it follows that $T$ is surjective. In particular there exists $z \in A$ such that $zy = 1$, so that since $y$ was arbitrary we have shown that every element in $A$ is invertible, i.e. $A$ is a field.
{{qed}}
Theorem: Preimage of Maximal Ideal of Finitely Generated Algebra is Maximal

Let $k$ be a field.
Let $A$ and $B$ be $k$-algebras.
Let $f: A \to B$ be a $k$-algebra homomorphism.
Let $B$ be finitely generated over $k$.
Let $\mathfrak m$ be a maximal ideal of $B$.
Then its preimage $\map {f^{-1} } {\mathfrak m}$ is a maximal ideal of $A$.

Proof:
We have an injective morphism:
:$\dfrac A {\map {f^{-1} } {\mathfrak m} } \to \dfrac B {\mathfrak m}$
{{explain|The exact meaning of the above line needs to be defined.}}
We have that $\dfrac B {\mathfrak m}$ is a field extension of $k$ which is finitely generated.
Thus, by Zariski's Lemma, $\dfrac B {\mathfrak m}$ is a finite field extension. 
By Unital Subalgebra of Finite Field Extension is Field, $\dfrac A {\map {f^{-1} } {\mathfrak m} }$ is a field.
Thus $\map {f^{-1} } {\mathfrak m}$ is a maximal ideal.
{{qed}}
Theorem: Unital Subalgebra of Algebraic Field Extension is Field

Let $E / F$ be an algebraic field extension.
Let $A \subseteq E$ be a unital subalgebra over $F$.
Then $A$ is a field.

Proof:
By Integral Ring Extension is Integral over Intermediate Ring, $E$ is integral over $A$.
Let $a \in A$ be nonzero.
Because $E$ is a field, $a$ is a unit of $E$.
By Ring Element is Unit iff Unit in Integral Extension, $a$ is a unit of $A$.
Thus $A$ is a field.
{{qed}}
Theorem: Unital Subalgebra of Finite Field Extension is Field

Let $E / F$ be a finite field extension.
Let $A \subseteq E$ be a unital subalgebra over $F$.
Then $A$ is a field.

Proof:
By Finite Field Extension is Algebraic, $E/F$ is algebraic.
The result follows from Unital Subalgebra of Algebraic Field Extension is Field
{{qed}}
Theorem: Piecewise Continuous Function with Improper Integrals may not be Bounded

Let $f$ be a real function defined on a closed interval $\closedint a b$, $a < b$. 
Let $f$ be a piecewise continuous function with improper integrals.
Then $f$ may not be piecewise continuous and bounded on $\closedint a b$.

Proof:
Consider the function:
:$\map f x = \begin{cases}
0 & : x = a \\
\dfrac 1 {\sqrt{x - a} } & : x \in \hointl a b
\end{cases}$
Since $\dfrac 1 {\sqrt{x - a} }$ is continuous on $\openint a b$, $f$ is continuous on $\openint a b$.
Therefore, $f$ satisfies $(1)$ in the requirements of a piecewise continuous function with improper integrals for the subdivision $\set {a, b}$ of $\closedint a b$.
Also: 
{{begin-eqn}}
{{eqn | l = \int_{a+}^{b-} \map f x \rd x
      | r = \int_{a+}^{b-} \dfrac 1 {\sqrt{x - a} } \rd x
}}
{{eqn | r = \bigintlimits {2 \sqrt {x - a} } {a+} {b-}
}}
{{eqn | r = \lim_{x \mathop \to b-} 2 \sqrt{x - a} - \lim_{x \mathop \to a+} 2 \sqrt{x - a}
}}
{{eqn | r = 2 \sqrt {b - a} - 2 \sqrt {a - a}
      | c = as $2 \sqrt{x - a}$ is left-continuous at $b$ and right-continuous at $a$
}}
{{eqn | r = 2 \sqrt{b - a}
}}
{{end-eqn}}
Hence $\ds \int_{a+}^{b-} \map f x \rd x$ exists.
Thus $f$ is a piecewise continuous function with improper integrals.
However, we have that $\map f x$ approaches $\infty$ as $x$ approaches $a$ from above.
Thus $f$ is not bounded.
Therefore $f$ is not piecewise continuous and bounded.
Hence the result.
{{qed}}
Theorem: Weierstrass's Elliptic Function is Even in z

Let $\omega_1$ and $\omega_2$ be non-zero complex constants with $\dfrac {\omega_1} {\omega_2}$ having a positive imaginary part. 
For $z \in \C \setminus \set {2 m \omega_1 + 2 n \omega_2: \tuple {n, m} \in \Z^2}$:
:$\ds \map \wp {-z; \omega_1, \omega_2} = \map \wp {z; \omega_1, \omega_2}$
That is, Weierstrass's elliptic function is even in $z$.

Proof:
{{begin-eqn}}
{{eqn	| l = \map \wp {-z; \omega_1, \omega_2}
	| r = \frac 1 {\paren {-z}^2} + \sum_{\tuple {n, m} \mathop \in \Z^2 \setminus \tuple {0, 0} } \paren {\frac 1 {\paren {-z - 2 m \omega_1 - 2 n \omega_2}^2} - \frac 1 {\paren {2 m \omega_1 + 2 n \omega_2}^2} }
	| c = {{Defof|Weierstrass's Elliptic Function}}
}}
{{eqn	| r = \frac 1 {z^2} + \paren {\sum_{\tuple {n, m} \mathop \in \Z_{\ge 0}^2 \setminus \tuple {0, 0} } + \sum_{\tuple {n, m} \mathop \in \Z_{<0}^2} } \paren {\frac 1 {\paren {z - 2 m \omega_1 - 2 n \omega_2}^2} - \frac 1 {\paren {-2 m \omega_1 - 2 n \omega_2}^2} }
}}
{{eqn	| r = \frac 1 {z^2} + \paren {\sum_{\tuple {n, m} \mathop \in \Z_{\le 0}^2 \setminus \tuple {0, 0} } } + \sum_{\tuple {n, m} \mathop \in \Z_{> 0}^2} \paren {\frac 1 {\paren {z - 2 m \omega_1 - 2 n \omega_2}^2} - \frac 1 {\paren {2 m \omega_1 + 2 n \omega_2}^2} }
	| c = letting $\tuple {n, m} \to \tuple {-n, -m}$
}}
{{eqn	| r = \frac 1 {z^2} + {\sum_{\tuple {n, m} \mathop \in \Z^2 \setminus \tuple {0, 0} } } \paren {\frac 1 {\paren {z - 2 m \omega_1 - 2 n \omega_2}^2} - \frac 1 {\paren {2 m \omega_1 + 2 n \omega_2}^2} }
}}
{{eqn	| r = \map \wp {z; \omega_1, \omega_2}
	| c = {{Defof|Weierstrass's Elliptic Function}}
}}
{{end-eqn}}
{{qed}}