hcju/proofwikiqa huggingface dataset:
A field is by definition also a division subring. Thus $L$ is the intersection of the set of all division subrings of $F$ containing $S$. From Intersection of Division Subrings Containing Subset is Smallest, $L$ is the smallest division subring of $F$ containing $S$. From Intersection of Subfields is Subfield, $L$ is a subfield of $F$. As $L$ is the smallest division subring of $F$ containing $S$, and it is a subfield of $F$, there can be no smaller subfield of $F$ containing $S$ as it would then not be a division subring. So the intersection of the set of all subfields of $F$ containing $S$ is the smallest subfield of $F$ containing $S$.  {{qed}}
Let $L = \bigcap \mathbb K$. From Intersection of Subfields is Subfield, $\struct {L, +, \circ}$ is itself a subfield of $\struct {F, +, \circ}$. A field is by definition also a division subring. Thus $\struct {L, +, \circ}$ is the largest division subring of $F$ contained in each element of $\mathbb K$. But as $\struct {F, +, \circ}$ is a field, $\circ$ is commutative. So by Restriction of Commutative Operation is Commutative, $\struct {L, +, \circ}$ is also a field. Hence the result. {{qed}}
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be equivalent. By Norm is Power of Other Norm then: :$\exists \alpha \in \R_{\gt 0}: \forall q \in \Q: \norm q_1 = \norm q_2^\alpha$ In particular: :$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy: :$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$ By Norm of Negative then: :$\forall n \in \N: \norm {-n}_1 = \norm n_1 = \norm n_2^\alpha = \norm {-n}_2^\alpha$ Hence: :$\forall k \in \Z: \norm k_1 = \norm k_2^\alpha$ By Norm of Quotient then: :$\forall \dfrac a b \in \Q: \norm {\dfrac a b}_1 = \dfrac {\norm a_1} {\norm b_1} = \dfrac {\norm a_2^\alpha} {\norm b_2^\alpha} = \norm {\dfrac a b}_2^\alpha$ By Norm is Power of Other Norm then  $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equivalent.
From Null Ring is Ideal and Ring is Ideal of Itself, it is always the case that $\set {0_R}$ and $\struct {R, +, \circ}$ are ideals of $\struct {R, +, \circ}$. Let $a \in R^*$, where $R^* := R \setminus \set {0_R}$. Let $\ideal a$ be the principal ideal of $R$ generated by $a$. We have that $\ideal a$ is a non-null ideal and hence $\ideal a = R$. Thus $1_R \in \ideal a$. Thus $\exists x \in R: x \circ a = 1_R$ by the definition of principal ideal. Therefore $a$ is invertible. As $a$ is arbitrary, it follows that all such $a$ are invertible. Thus by definition $\struct {R, +, \circ}$ is a division ring such that $\circ$ is commutative. The result follows by definition of field. {{Qed}}
By definition, a field is a division ring. From Null Ring is Ideal and Ring is Ideal of Itself, it is always the case that $\set {0_F}$ and $\struct {F, +, \circ}$ are ideals of $\struct {F, +, \circ}$. From Ideals of Division Ring, it follows that the only ideals of $\struct {F, +, \circ}$ are $\struct {F, +, \circ}$ and $\set {0_F}$. {{qed}}
Let $S$ be the set of square matrices of order $2$ over the real numbers $\R$. $S$ is not a division ring, as for example: :$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ and so both $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ are proper zero divisors of $S$. Let $J$ be an ideal of $S$ containing a non-zero matrix $A$ whose $\tuple {r, s}$th entry is $\lambda \ne 0$ Let $E_{i j}$ be the matrix of order $2$ defined as: :$e_{a b} = \begin{cases} 1 & : a = i, b = j \\ 0 & : \text {otherwise} \end{cases}$ Thus for example: :$E_{1 2} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ Then for all $i, j \in \set {1, 2}$: {{begin-eqn}} {{eqn | l = \dfrac 1 \lambda E_{i r} A E_{s j}       | r = \dfrac 1 \lambda E_{i r} A E_{s j}       | c =  }} {{eqn | r = E_{i j}       | c =  }} {{end-eqn}} {{explain|Prove the above}} It follows that: :$\forall i, j \in \set {1, 2}: E_{i j} \in J$ It remains to be shown that $J$ is the whole of $S$.  {{finish}}.
Suppose that $X = \C$. === Case 1 === If $n$ is even, then: {{begin-eqn}} {{eqn | l = \map f {x + n L}       | r = \map f {x + \paren {2 k} L}       | c = for some $k \in \Z$ }} {{eqn | r = \map f {x + \paren {k 2} L}       | c = Complex Multiplication is Commutative }} {{eqn | r = \map f {x + k \paren {2 L} }       | c = Complex Multiplication is Associative }} {{eqn | r = \map f x       | c = by Double of Antiperiodic Element is Periodic and the General Periodicity Property }} {{end-eqn}} {{qed|lemma}} === Case 2 === If $n$ is odd, then: {{begin-eqn}} {{eqn | l = \map f {x + n L}       | r = \map f {x + \paren {2 k + 1} L}       | c = for some $k \in \Z$ }} {{eqn | r = \map f {x + \paren {\paren {2 k} L + L} }       | c = Complex Multiplication Distributes over Addition }} {{eqn | r = \map f {\paren {x + \paren {2 k} L} + L}       | c = Complex Addition is Associative }} {{eqn | r = -\map f {x + \paren {2 k} L} }} {{eqn | r = -\map f x       | c = by '''Case 1''' }} {{end-eqn}} The proof for when $X = \R$ is nearly identical. {{qed}} Category:Antiperiodic Functions d6sgward9zyusbcx8pw9ly4hy5c3szv
{{AimForCont}} that $A \nmid L$. By the Division Theorem we have: :$\exists! q \in \Z, r \in \R: L = q A + r, 0 < r < A$ By Even and Odd Integers form Partition of Integers, it follows that $q$ must be either even or odd. === Case 1 === Suppose $q$ is even. Then: {{begin-eqn}} {{eqn | l = \map f {x + L}       | r = \map f {x + \paren {q A + r} } }} {{eqn | r = \map f {\paren {x + r} + q A} }} {{eqn | r = \map f {x + r}       | c = General Antiperiodicity Property }} {{eqn | r = -\map f x }} {{end-eqn}} And so $r$ is an anti-periodic element of $f$ that is less than $A$. But then $A$ cannot be the anti-period of $f$. Therefore by contradiction $q$ cannot be even. === Case 2 === Suppose $q$ is odd. Then: {{begin-eqn}} {{eqn | l = - \map f x       | r = \map f {x + L} }} {{eqn | r = \map f {x + \paren {q A + r} } }} {{eqn | r = \map f {\paren {x + r} + q A} }} {{eqn | r = - \map f {x + r}       | c = General Antiperiodicity Property }} {{end-eqn}} So: :$-\map f x = - \map f {x + r} \implies \map f x = \map f {x + r}$ It is seen that $r$ is a periodic element of $f$ such that $0 < r < A$. But consider $0 < A - r < A$: {{begin-eqn}} {{eqn | l = \map f {x + \paren {A - r} }       | r = \map f {\paren {x + A} - r} }} {{eqn | r = \map f {x + A} }} {{eqn | r = -\map f x }} {{end-eqn}} This contradicts the fact that $A$ is the anti-period of $f$. Hence the result. {{qed}} Category:Antiperiodic Functions 6lspg960b2ut2ajzetaiu2r0qfdbb9m
From Skewness in terms of Non-Central Moments:  :$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$  where $\mu$ is the mean of $X$, and $\sigma$ the standard deviation. We have, by Expectation of Poisson Distribution: :$\expect X = \lambda$ By Variance of Poisson Distribution: :$\var X = \sigma^2 = \lambda$ so: :$\sigma = \sqrt \lambda$ To now calculate $\gamma_1$, we must calculate $\expect {X^3}$. We find this using the moment generating function of $X$, $M_X$.  By Moment Generating Function of Poisson Distribution, this is given by:  :$\displaystyle \map {M_X} t = e^{\lambda \paren {e^t - 1} }$ From Moment in terms of Moment Generating Function: :$\expect {X^3} = \map {M_X'''} 0$ In Variance of Poisson Distribution: Proof 3, it is shown that: :$\displaystyle \map {M_X''} t = \lambda \paren {\lambda e^t + 1} e^{\lambda \paren {e^t - 1} + t}$ So: {{begin-eqn}} {{eqn	| l = \map {M_X'''} t 	| r = \lambda^2 e^{\lambda \paren {e^t - 1} + 2t} + \lambda \paren {\lambda e^t + 1} \frac \d {\d t} \paren {\lambda \paren {e^t - 1} + t} \frac \d {\d \paren {\lambda \paren {e^t - 1} + t} } \paren {e^{\lambda \paren {e^t - 1} + t} } 	| c = Product Rule, Exponential of Sum, Chain Rule for Derivatives }} {{eqn	| r = \lambda^2 e^{\lambda \paren {e^t - 1} + 2t} + \lambda \paren {\lambda e^t + 1}^2 e^{\lambda \paren {e^t - 1} + t} 	| c = Derivative of Exponential Function, Derivative of Power }} {{end-eqn}} Setting $t = 0$: {{begin-eqn}} {{eqn	| l = \expect {X^3} 	| r = \lambda^2 e^{\lambda \paren {e^0 - 1} + 0} + \lambda \paren {\lambda e^0 + 1}^2 e^{\lambda \paren {e^0 - 1} + 0} }} {{eqn	| r = \lambda^2 + \lambda \paren {\lambda + 1}^2 	| c = Exponential of Zero }} {{eqn	| r = \lambda^3 + 3 \lambda^2 + \lambda 	| c = Square of Sum }} {{end-eqn}} So: {{begin-eqn}} {{eqn	| l = \gamma_1 	| r = \frac {\lambda^3 + 3 \lambda^2 + \lambda - 3 \lambda^2 - \lambda^3} {\lambda^{3/2} } }} {{eqn	| r = \frac \lambda {\lambda^{3/2} } }} {{eqn	| r = \frac 1 {\sqrt \lambda} }} {{end-eqn}} {{qed}} Category:Skewness Category:Poisson Distribution oml15vhprg2hwwjnhtgtfbkhko1uqex
Consider $A + L$: {{begin-eqn}} {{eqn | l = \map f {x + \paren {A + L} }       | r = \map f {\paren {x + A} + L} }} {{eqn | r = \map f {x + A} }} {{eqn | r = -\map f x }} {{end-eqn}} Hence $A + L$ is an anti-periodic element of $f$. Combining Antiperiodic Element is Multiple of Antiperiod, Divides is Reflexive, and Common Divisor Divides Difference yields:  :$A \divides \paren {A + L} \land A \divides A \implies A \divides L$ Hence the result. {{qed}} Category:Antiperiodic Functions hywff4b2rkdxz178xdjt5efld55qywm
From the definition of skewness:  :$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$ where:  :$\mu$ is the mean of $X$ :$\sigma$ is the standard deviation of $X$.  From the definition of the continuous uniform distribution, $X$ has probability density function: :$\map {f_X} x = \dfrac 1 {b - a}$ So, from Expectation of Function of Continuous Random Variable: :$\displaystyle \gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3} = \frac 1 {\paren {b - a} \sigma^3} \int_a^b \paren {x - \mu}^3 \rd x$ Then:  {{begin-eqn}} {{eqn	| l = \frac 1 {\paren {b - a} \sigma^3} \int_a^b \paren {x - \mu}^3 \rd x 	| r = \frac 1 {\paren {b - a} \sigma^3} \int_a^b \paren {x - \frac {a + b} 2}^3 \rd x 	| c = Expectation of Continuous Uniform Distribution }} {{eqn	| r = \frac 1 {\paren {b - a} \sigma^3} \int_{a - \frac {a + b} 2}^{b - \frac {a + b} 2} u^3 \rd u 	| c = substituting $u = x - \dfrac {a + b} 2$ }} {{eqn	| r = \frac 1 {\paren {b - a} \sigma^3} \int_{-\frac {b - a} 2}^{\frac {b - a} 2} u^3 \rd u }} {{end-eqn}} We have:  :$\paren {-u}^3 = -u^3$ so we can see the integrand is odd.  So, by Definite Integral of Odd Function:  :$\displaystyle \gamma_1 = \frac 1 {\paren {b - a} \sigma^3} \int_{-\frac {b - a} 2}^{\frac {b - a} 2} u^3 \rd u = 0$ {{qed}} Category:Skewness Category:Continuous Uniform Distribution 2a2owjs7yj6v2pzrxxl73wkltthj1ya
Let $L_f$ be the set of all periodic elements of $f$. By Periodic Element is Multiple of Antiperiod and Absolute Value of Real Number is not less than Divisors: :$\forall p \in L_f: A \divides p \land A \le p$ Suppose there is a $p \in L_f$ such that $p = A$.  Then:  {{begin-eqn}} {{eqn | l = \map f x       | r = \map f {x + A}       | c =  }} {{eqn | r = -\map f x       | c =  }} {{eqn | ll= \leadsto       | l = \map f x       | r = 0       | c =  }} {{end-eqn}} which contradicts Constant Function has no Period. Therefore $\forall p \in L_f: A \divides p \land A < p$. The smallest number $r$ such that $A < r \land A \divides r$ is $2A$. But by Double of Antiperiodic Element is Periodic, this is a periodic element of $f$. Hence the result. {{qed}} Category:Antiperiodic Functions g6ijhsitxemjy4cby6j0aqeniqikh1o
We have: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \ideal {a^{n + 1} }       | c =  }} {{eqn | ll= \leadsto       | lo= \exists r \in D:       | l = x       | r = r \circ a^{n + 1}       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \paren {r \circ a} \circ a^n       | c =  }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = \ideal {a^n}       | c =  }} {{eqn | ll= \leadsto       | l = \ideal {a^{n + 1} }       | o = \subseteq       | r = \ideal {a^n}       | c =  }} {{end-eqn}} It remains to be shown that $\ideal {a^{n + 1} } \ne \ideal {a^n}$. {{AimForCont}} $\ideal {a^{n + 1} } = \ideal {a^n}$. Then: {{begin-eqn}} {{eqn | l = \ideal {a^{n + 1} }       | r = \ideal {a^n}       | c =  }} {{eqn | ll= \leadsto       | l = a^n       | o = \in       | r = \ideal {a^{n + 1} }       | c =  }} {{eqn | ll= \leadsto       | lo= \exists x \in D:       | l = a^n       | r = x \circ a^{n + 1}        | c =  }} {{eqn | ll= \leadsto       | l = 1_D       | r = x \circ a        | c = Cancellation Law of Ring Product of Integral Domain }} {{end-eqn}} That is, $a$ is a unit of $D$. This contradicts the assertion that $a$ is a proper element of $D$. The result follows by Proof by Contradiction. {{qed}}
Let $a \in D$ be a proper element of $D$. Because $\struct {D, +, \circ}$ is not a field, such an element is known to exist. From Principal Ideal in Integral Domain generated by Power Plus One is Subset of Principal Ideal generated by Power: :$\forall n \in \Z_{\ge 0}: \ideal {a^{n + 1} } \subsetneq \ideal {a_n}$ where $\ideal x$ denotes the principal ideal of $D$ generated by $x$. Hence the set: :$S = \set {\ideal {1_D}, \ideal a, \ideal {a^2}, \ideal {a^3}, \dotsc}$ is infinite. {{qed}}
{{AimForCont}} $\map \phi {1_R} = 0_D$. Let $x \in R$ be arbitrary. Then: {{begin-eqn}} {{eqn | l = \map \phi x       | r = \map \phi {x \circ_R 1_R}       | c = {{Defof|Unity of Ring}} }} {{eqn | r = \map \phi x \circ_D \map \phi {1_R}       | c = {{Defof|Ring Homomorphism}} }} {{eqn | r = \map \phi x \circ_D 0_D       | c = by hypothesis }} {{eqn | r = 0_D       | c = {{Defof|Ring Zero}} }} {{end-eqn}} But this contradicts the assertion that $\map \ker \phi \ne R$. It follows that $\map \phi {1_R} \ne 0_D$. Then we have: {{begin-eqn}} {{eqn | l = 1_D \circ_D \map \phi {1_R}       | r = \map \phi {1_R}       | c = {{Defof|Unity of Ring}} }} {{eqn | r = \map \phi {1_R \circ_R 1_R}       | c = {{Defof|Unity of Ring}} }} {{eqn | r = \map \phi {1_R} \circ_D \map \phi {1_R}       | c = {{Defof|Ring Homomorphism}} }} {{eqn | ll= \leadsto       | l = 1_D       | r = \map \phi {1_R}       | c = Cancellation Law of Ring Product of Integral Domain }} {{end-eqn}} Hence the result. {{qed}}
By definition of nilpotent element: :$x^n = 0_R$ for some $n \in \Z_{>0}$. From Difference of Two Powers: {{begin-eqn}} {{eqn | l = a^n - b^n       | r = \paren {a - b} \circ \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} \circ b^j       | c =  }} {{eqn | r = \paren {a - b} \circ \paren {a^{n - 1} + a^{n - 2} \circ b + a^{n - 3} \circ b^2 + \dotsb + a \circ b^{n - 2} + b^{n - 1} }       | c =  }} {{end-eqn}} for $a, b \in R$. Putting $a = 1_R$ and $b = x$, we have: {{begin-eqn}} {{eqn | l = \paren {1_R}^n - x^n       | r = \paren {1_R - x} \circ \paren {1_R^{n - 1} + 1_R^{1_R - 2} \circ x + 1_R^{n - 3} \circ x^2 + \dotsb + 1_R \circ x^{n - 2} + x^{n - 1} }       | c =  }} {{eqn | ll= \leadsto       | l = 1_R - 0_R       | r = \paren {1_R - x} \circ \paren {1_R + 1_R \circ x + 1_R \circ x^2 + \dotsb + 1_R \circ x^{n - 2} + x^{n - 1} }       | c = as $x^n = 0_R$, $1_R^k = 1_R$ }} {{eqn | ll= \leadsto       | l = 1_R       | r = \paren {1_R - x} \circ \paren {1_R + x + x^2 + \dotsb + x^{n - 2} + x^{n - 1} }       | c =  }} {{end-eqn}} Thus by definition $1_R - x$ has a product inverse $1_R + x + x^2 + \dotsb + x^{n - 2} + x^{n - 1}$. Hence by definition $1_R - x$ is a unit of $R$. {{qed}}
From Nilpotent Elements of Commutative Ring form Ideal, $\struct {N, +, \circ}$ is an ideal of $\struct {R, +, \circ}$. Hence the quotient ring $R / N$ is defined. By definition of the ideal of $\struct {R, +, \circ}$, $N$ is the zero of $R / N$. Let $\paren {x + N}^n \in N$. Then: :$x^n \in N$ and so: :$\paren {x^n}^m = 0_R$ for some $m \in \Z_{>0}$. Hence: :$x \in N$ and so: :$x + N = N$ That is, if a (left) coset of $N$ is nilpotent in $R / N$ that (left) coset is $N$ itself. The result follows. {{qed}}
Consider the set $S$ defined as: :$S = U_R \setminus \set {1_D, -1_D}$ If $S$ has even cardinality, it can be partitioned into doubletons of the form $\set {u, u^{-1} }$. Each of these doubletons has a product of $1_D$. The product of all these with $1_D$ and $-1_D$ is $-1_D$. It remains to be shown that $S$ cannot be of odd cardinality. {{AimForCont}} $S$ has an odd number of elements. Then after pairing off each $u \in D$ with its product inverse $u^{-1}$, we are left with one element of $S$ which is self-inverse. But from Self-Inverse Element of Integral Domain is Unity or its Negative, this self-inverse element is either $1_D$ or $-1_D$. These have already been counted. Hence there cannot be an odd number of elements of $S$. {{qed}}
Let $p$ be prime. Consider $\struct {\Z_p, +, \times}$, the ring of integers modulo $m$. From Ring of Integers Modulo Prime is Field, $\struct {\Z_p, +, \times}$ is a field. Hence, apart from $\eqclass 0 p$, all elements of $\struct {\Z_p, +, \times}$ are units As $\struct {\Z_p, +, \times}$ is a field, it is also by definition an integral domain, we can apply: From Product of Units of Integral Domain with Finite Number of Units, the product of all elements of $\struct {\Z_p, +, \times}$ is $-1$. But the product of all elements of $\struct {\Z_p, +, \times}$ is $\paren {p - 1}!$ The result follows. {{qed}}
From Prime Ideal iff Quotient Ring is Integral Domain: :$J$ is a prime ideal of $\Z$ {{iff}} $\Z / J$ is an integral domain. From Quotient Ring of Integers and Zero: :$\Z / \set 0 \cong \Z$ As $\Z$ is an integral domain, it follows that $\set 0$ is a prime ideal of $\Z$. From Quotient Ring of Integers with Principal Ideal: :$\struct {\Z, +, \times} / \ideal p$ is isomorphic to $\struct {\Z_p, +_p, \times_p}$, the ring of integers modulo $p$. From Ring of Integers Modulo Prime is Integral Domain: :$\struct {\Z_n, +_n, \times_p}$ is an integral domain {{iff}} $p$ is prime. The result follows. {{qed}}
Let $x, y \in \GF$. Then: {{begin-eqn}} {{eqn | l = \map \sigma {x y}       | r = \paren {x y}^p       | c = Definition of $\sigma$ }} {{eqn | r = x^p y^p       | c = Power of Product of Commutative Elements in Group }} {{eqn | r = \map \sigma x \, \map \sigma y       | c = Definition of $\sigma$ }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \map \sigma {x + y}       | r = \paren {x + y}^p       | c = Definition of $\sigma$ }} {{eqn | r = \sum_{k \mathop = 0}^p \dbinom p k x^k y^{p - k}       | c = Binomial Theorem }} {{eqn | o = \equiv       | r = x^p + y^p       | rr= \pmod p       | c = Power of Sum Modulo Prime }} {{eqn | r = \map \sigma x + \map \sigma y       | c = Definition of $\sigma$ }} {{end-eqn}} Thus it has been demonstrated that $\sigma$ is a homomorphism. Then we have: {{begin-eqn}} {{eqn | l = \map \ker \sigma       | r = \set {x \in \GF: \map \sigma x = 0_\GF}       | c = {{Defof|Kernel of Ring Homomorphism}} }} {{eqn | r = \set {x \in \GF: x^p = 0_\GF}       | c = Definition of $\sigma$ }} {{eqn | r = \set {x \in \GF: x = 0_\GF}       | c = Congruence of Powers }} {{eqn | r = \set {0_\GF}       | c = Congruence of Powers }} {{end-eqn}} From Kernel is Trivial iff Monomorphism, $\sigma$ is a monomorphism. That is, $\sigma$ is an injection. Then from Injection from Finite Set to Itself is Surjection, $\sigma$ is a surjection. Thus $\sigma$ is a bijective homomorphism to itself. The result follows by definition of automorphism. {{qed}}
{{AimForCont}} $\phi: \struct {F_{\ne 0_F}, \times} \to \struct {F, +}$ is an isomorphism. By definition: :$0_F$ is the identity of $\struct {F, +}$ and :$1_F$ is the identity of $\struct {F_{\ne 0_F}, \times}$. We have that: {{begin-eqn}} {{eqn | l = 0_F       | r = \map \phi {1_F}       | c = Epimorphism Preserves Identity }} {{eqn | r = \map \phi {\paren {-1_F} \times \paren {-1_F} }       | c =  }} {{eqn | r = \map \phi {-1_F} + \map \phi {-1_F}       | c = {{Defof|Group Isomorphism}} }} {{end-eqn}} and so by definition $F$ has characteristic $2$. Let $x \in \struct {F_{\ne 0_F}, \times}$. Then: {{begin-eqn}} {{eqn | l = \map \phi {x^2}       | r = \map \phi x + \map \phi x       | c = {{Defof|Group Isomorphism}} }} {{eqn | r = 0_F       | c = as $F$ has characteristic $2$ }} {{end-eqn}} As $\phi$ is an isomorphism, it is also a monoomorphism. From Kernel is Trivial iff Monomorphism :$\map \ker \phi = \set {1_F}$ and so $x^2 = 1$. Thus $x = 1$ and so $\order F = 2$. Thus $\order {F_{\ne 0_F} } = 1$. So $\phi$ is a bijection from a set of cardinality $1$ to a set of cardinality $2$. So $\phi$ cannot be a bijection and so cannot be an isomorphism. Hence the result by Proof by Contradiction. {{qed}}
Let $K = \displaystyle \bigcup_{n \mathop \in \N} J_n$. Then from Increasing Union of Sequence of Ideals is Ideal, $K$ is an ideal of $D$. We have that $D$ is a principal ideal domain. Hence there exists $a \in D$ such that: :$K = \ideal a$ where $\ideal a$ is the principal ideal of $D$ generated by $a$. But $a \in J_m$ for some $m \in \N$. Thus $K \subseteq J_m$ Thus it follows that $J_{m + 1} \subseteq J_m$ which contradicts our initial assertion that: :$\forall n \in \N: J_n \subsetneq j_{n + 1}$ Hence the result. {{qed}}
First it is shown that $\map N z = 0 \iff z = 0$. {{begin-eqn}} {{eqn | l = z       | r = 0       | c =  }} {{eqn | r = 0 + 0 i       | c =  }} {{eqn | ll= \leadsto       | l = \map N z       | r = 0^2 + 0^2       | c = Definition of $N$ }} {{eqn | r = 0       | c =  }} {{end-eqn}} Let $z = x + i y$. {{begin-eqn}} {{eqn | l = \map N z       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \map N {x + i y}       | r = 0       | c = Definition of $z$ }} {{eqn | ll= \leadsto       | l = x^2 + y^2       | r = 0       | c = Definition of $N$ }} {{eqn | ll= \leadsto       | l = a       | r = 0       | c = Square of Real Number is Non-Negative }} {{eqn | l = b       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = z       | r = 0       | c = Definition of $z$ }} {{end-eqn}} Then we have: {{begin-eqn}} {{eqn | l = \map N z       | r = \map N {x + i y}       | c = Definition of $z$ }} {{eqn | r = x^2 + y^2       | c = Definition of $N$ }} {{eqn | 0 = \ge       | r = 0       | c = Square of Real Number is Non-Negative }} {{end-eqn}} Hence the result by definition of positive definite. {{qed}} Category:Field Norm of Complex Number fgrcmu847s49n7be08imferkgspbgaq
{{begin-eqn}} {{eqn | l = \map N {z_1 z_2}       | r = \cmod {z_1 z_2}^2       | c = Definition of $N$ }} {{eqn | r = \paren {\cmod {z_1} \cmod {z_2} }^2       | c = Complex Modulus of Product of Complex Numbers }} {{eqn | r = \cmod {z_1}^2 \cmod {z_2}^2       | c =  }} {{eqn | r = \map N {z_1} \map N {z_2}       | c = Definition of $N$ }} {{end-eqn}} So $N$ is a multiplicative function by definition. {{qed}}
Proof by Counterexample: Let $z_1 = z_2 = 1$. Then: {{begin-eqn}} {{eqn | l = \map N {z_1 + z_2}       | r = \cmod {z_1 + z_2}^2       | c = Definition of $N$ }} {{eqn | r = 2^2       | c =  }} {{eqn | r = 4       | c =  }} {{end-eqn}} But: {{begin-eqn}} {{eqn | l = \map N {z_1} + \map N {z_2}       | r = \cmod {z_1}^2 + \cmod {z_2}^2       | c = Definition of $N$ }} {{eqn | r = 1^2 + 1^2       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} So we have that for these instances of $z_1$ and $z_2$: :$\map N {z_1 + z_2} > \map N {z_1} + \map N {z_2}$ and so the triangle inequality is not satisfied. Hence by definition $N$ is not a norm on $\C$. {{qed}} Category:Field Norm of Complex Number dofpceh1eri25d4elz5bhvmm0obtoo4
Let $\map N z$ denote the field norm of $z \in \Z \sqbrk {i \sqrt 5}$. Let $z_1 \in \Z \sqbrk {i \sqrt 5}$ be a unit of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$. Thus by definition: :$\exists z_2 \in \Z \sqbrk {i \sqrt 5}: z_1 \times z_2 = 1$ Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$. Then: {{begin-eqn}} {{eqn | l = 1       | r = \map N 1       | c = Field Norm on 5th Cyclotomic Ring }} {{eqn | r = \map N {z_1 \times z_2}       | c =  }} {{eqn | r = \map N {z_1} \times \map N {z_2}       | c =  }} {{eqn | r = \paren { {x_1}^2 + 5 {y_1}^2} \paren { {x_2}^2 + 5 {y_2}^2}       | c = Field Norm on 5th Cyclotomic Ring }} {{eqn | ll= \leadsto       | l = {x_1}^2       | r = 1       | c =  }} {{eqn | lo= \land       | l = {x_2}^2       | r = 1       | c =  }} {{eqn | lo= \land       | l = y_1       | r = y_2 = 0       | c =  }} {{end-eqn}} The result follows. {{qed}}