hcju/proofwikiqa huggingface dataset:
=== Necessary condition === Let $a_0$ be a unit of $R$. For $i = 1, \ldots, n$, let $a_i$ be nilpotent in $R$. Because the nilradical is an ideal of $R$, it follows that: :$Q = -a_1 X + \dotsb + -a_n X^n$ is nilpotent. Moreover, multiplying through by $a_0^{-1}$ we may as well assume that $a_0 = 1_R$. Then: :$P = 1_R - Q$ and from Unity plus Negative of Nilpotent Ring Element is Unit $P$ is a unit of $R \sqbrk X$. {{qed|lemma}} === Sufficient condition === Let $\map P X$ be a unit of $R \sqbrk X$. That is, there exists: :$Q = b_0 + b_1 X + \dotsb + b_m X^m \in R \sqbrk X$ such that $P Q = 1$. By the definition of polynomial multiplication the degree zero term of $P Q$ is $a_0 b_0$. Therefore $a_0 b_0 = 1_R$. So $a_0$ is a unit of $R$. {{qed|lemma}} Next we show that $a_1, \dotsc, a_n$ are nilpotent. By Spectrum of Ring is Nonempty, $R$ has at least one prime ideal $\mathfrak p$. By Prime Ideal iff Quotient Ring is Integral Domain: :$R / \mathfrak p$ is an integral domain. By Ring of Polynomial Forms over Integral Domain is Integral Domain: :$R / \mathfrak p \sqbrk X$ is also an integral domain. For any polynomial $T \in R \sqbrk X$ let $\overline T$ denote the image of $T$ under the Induced Homomorphism of Polynomial Forms defined by the quotient mapping $R \to R / \mathfrak p$. Now we have: :$\overline P \cdot \overline Q = 1_{R / \mathfrak p}$ By Units of Ring of Polynomial Forms over Integral Domain this implies that $\overline P$ has degree zero. In particular for $i = 1, \dotsc, n$, the image of $a_i$ in $R / \mathfrak p$ is $0_{R / \mathfrak p}$. By definition, this means that $a_i \in \mathfrak p$. But this is true for every prime ideal $\mathfrak p$. Thus by definition: :$a_i \in \Nil R$ where $\Nil R$ denotes the nilradical of $R$. {{Qed}} Category:Polynomial Theory aewy1zp3boybd5gid8q2bag3v9j2elu
Let $p, q$ be polynomials in $x$ over $D$. We can express them as: : $\displaystyle p = \sum_{k \mathop = 0}^m a_k \circ x^k$ : $\displaystyle q = \sum_{k \mathop = 0}^n b_k \circ x^k$ where: : $(1): \quad a_k, b_k \in D$ for all $k$ : $(2): \quad m, n \in \Z_{\ge 0}$, that is, are non-negative integers. Suppose $m = n$. Then: : $\displaystyle p + q = \sum_{k \mathop = 0}^n a_k \circ x^k + \sum_{k \mathop = 0}^n b_k \circ x^k$ Because $\left({R, +, \circ}\right)$ is a commutative ring, it follows that: : $\displaystyle p + q = \sum_{k \mathop = 0}^n \left({a_k + b_k}\right) \circ x^k$ which is also a polynomials in $x$ over $D$. {{explain|Establish whether $R$ actually needs to be commutative for the above to hold.}} {{WLOG}}, suppose $m > n$. Then we can express $q$ as: : $\displaystyle \sum_{k \mathop = 0}^n b_k \circ x^k + \sum_{k \mathop = n \mathop + 1}^m 0_D \circ x^k$ Thus: : $\displaystyle p + q = \sum_{k \mathop = 0}^n \left({a_k + b_k}\right) \circ x^k + \sum_{k \mathop = n \mathop + 1}^m a_k \circ x^k$ which is also a polynomials in $x$ over $D$. Thus the sum of two polynomials in $x$ over $D$ is another polynomial in $x$ over $D$. Hence the result. {{qed}}
Let $p, q$ be polynomials in $x$ over $S$. We can express them as: : $\displaystyle p = \sum_{k \mathop = 0}^m a_k \circ x^k$ : $\displaystyle q = \sum_{k \mathop = 0}^n b_k \circ x^k$ where: : $(1): \quad a_k, b_k \in S$ for all $k$ : $(2): \quad m, n \in \Z_{\ge 0}$, that is, are non-negative integers. Suppose $m = n$. Then: : $\displaystyle p + q = \sum_{k \mathop = 0}^n a_k \circ x^k + \sum_{k \mathop = 0}^n b_k \circ x^k$ Because $\left({R, +, \circ}\right)$ is a ring, it follows that: : $\displaystyle p + q = \sum_{k \mathop = 0}^n \left({a_k + b_k}\right) \circ x^k$ which is also a polynomial in $x$ over $S$. {{handwaving|What needs to be done here is to establish that each of the terms is an element of $R$ and that each of the coefficients is an element of $S$. Trivial, but important.}} Now suppose WLOG that $m > n$. Then we can express $q$ as: : $\displaystyle \sum_{k \mathop = 0}^n b_k \circ x^k + \sum_{k \mathop = n \mathop + 1}^m 0_D \circ x^k$ Thus: : $\displaystyle p + q = \sum_{k \mathop = 0}^n \left({a_k + b_k}\right) \circ x^k + \sum_{k \mathop = n \mathop + 1}^m a_k \circ x^k$ which is also a polynomial in $x$ over $S$. Thus the sum of two polynomials in $x$ over $S$ is another polynomial in $x$ over $S$. Hence the result. {{qed}} Category:Polynomial Theory i52w3woxyh57xvvzudjjhg7qfzd85ml
It  is immediate that $f \oplus g$ is a map from the free commutative monoid to $R$, so we need only prove  that $f \oplus g$ is nonzero on finitely many $\mathbf X^k$, $k \in Z$. Suppose that for some $k \in Z$, $a_k + b_k \ne 0$ This forces at least one of $a_k$ and $b_k$ to be non-zero. This can only be true for a finite number of terms because $f$ and $g$ are polynomials. The result follows. {{qed}} Category:Polynomial Theory {{proofread}} meubepouwyzytdegniyqmpwk5ne5cvx
Let $\textbf A = \left({A, \mu_A}\right)$ and $\textbf B = \left({B, \mu_B}\right)$ be fuzzy sets. === Proving Domain Equality === By the definition of fuzzy intersection the domain of $\textbf A \cap \textbf B$ is: :$A \cap B$ Similarly the domain of $\textbf B \cap \textbf A$ is: :$B \cap A$ By Intersection is Commutative: :$A \cap B = B \cap A$ Hence their domains are equal. {{qed|lemma}} === Proving Membership Function Equality === ==== Proving Form Equality ==== By the definition of fuzzy intersection the membership function of $\textbf A \cap \textbf B$ is of the form: :$\mu:A \cap B \to \left [{0 \,.\,.\, 1}\right]$ Similarly, the membership function of $\textbf B \cap \textbf A$ is of the form: :$\mu:B \cap A \to \left [{0 \,.\,.\, 1}\right]$ By Intersection is Commutative this is the same as: :$\mu:A \cap B \to \left [{0 \,.\,.\, 1}\right]$ Hence the membership functions are of the same form. ==== Proving Rule Equality ==== {{begin-eqn}} {{eqn | l=\forall x \in A \cap B: \mu_{A \cap B}(x) = \operatorname{min}\left({\mu_A(x), \mu_B(x)}\right)       | o=\iff       | r=\forall x \in B \cap A: \mu_{B \cap A}(x) = \operatorname{min}\left({\mu_A(x), \mu_B(x)}\right)       | c=Intersection is Commutative }} {{eqn | r=\forall x \in B \cap A: \mu_{B \cap A}(x) = \operatorname{min}\left({\mu_B(x), \mu_A(x)}\right)       | o=\iff       | c=Min Operation is Commutative }} {{end-eqn}} Hence the membership functions have the same rule. {{qed}} Category:Fuzzy Set Theory s2aicbj8uixr4w0uhz7a9168hepj7x5
A commutative ring with unity is a ring. An integral subdomain of a commutative ring with unity $R$ is a subring of $R$. The result then follows as a special case of Polynomials Closed under Addition: Polynomials over Ring {{qed}}
First note that when $R$ is the null ring the result is false. This is because although $0_R$ is nilpotent element in the null ring, it is not actually a zero divisor. Hence in this case $0_R$ is both nilpotent and a zero divisor. So, let $R$ be a non-null ring. By hypothesis, there exists $n \in \Z_{>0}$ such that $x^n = 0_R$. If $n = 1$, then $x = 0_R$. By hypothesis, $R$ is not the null ring, so we may choose $y \in R \setminus \set 0$. By Ring Product with Zero: :$y \circ x = y \circ 0_R = 0_R$ Therefore $x$ is a zero divisor in $R$. If $n \ge 2$, define $y = x^{n - 1}$. Then: :$y \circ x = x^{n - 1} \circ x = x^n = 0_R$ so $x$ is a zero divisor in $R$. {{Qed}} Category:Nilpotent Ring Elements Category:Zero Divisors 820we46dktq2afef9pcvxgsk5f4sprg
Let $x \in D$ be a nilpotent element. Then by Nilpotent Element is Zero Divisor, $x$ is a zero divisor in $D$. By the definition of an integral domain, this means that $x = 0$. Therefore the only nilpotent element of $D$ is $0$. That is, $D$ is reduced. {{Qed}} Category:Ring Theory fm7xo02i9ov8o6hcq03u0qp2633wo8v
It is immediate that a unit of $D$ is also a unit of $D \sqbrk X$. Let $P$ be a unit of $D \sqbrk X$. Then there exists $Q \in D \sqbrk X$ such that $P Q = 1$. By Corollary 2 to Degree of Product of Polynomials over Ring we have: :$0 = \map \deg 1 = \map \deg P + \map \deg Q$ Therefore: :$\map \deg P = \map \deg Q = 0$ That is, $P \in R$ and $Q \in R$. Moreover $P Q = 1$ in $R$, so it follows that $P$ is a unit of $R$. {{Qed}} Category:Polynomial Theory 4dkr1neieh4qu20hakri4h3i80a93f5
Let $x, y \in \phi^{-1} \left({e}\right)$. It is to be shown that: :$x * y \in \phi^{-1} \left({e}\right)$ Thus: {{begin-eqn}} {{eqn | l = x, y \in \phi^{-1} \left({e}\right)       | o = \iff       | r = \left({\phi \left({x}\right) = e}\right) \land \left({\phi \left({y}\right) = e}\right)       | c = Definition of Kernel }} {{eqn | o = \iff       | r = \phi \left({x}\right) \circ \phi \left({y}\right) = e       | c = Definition of Identity }} {{eqn | o = \iff       | r = \phi \left({x * y}\right) = e       | c = Definition of Homomorphism }} {{eqn | o = \iff       | r = x*y \in \phi^{-1} \left({e}\right)       | c = Definition of Preimage }} {{end-eqn}} Hence the result. {{qed}} Category:Abstract Algebra 677e0zsz2ea0h4jfe8kvauldn584i1e
=== Necessary Condition === Let $P$ be irreducible. Suppose further that $ \lambda P$ has a non-trivial factorization: :$\displaystyle \lambda P = Q_1 Q_2$ that is, such that $Q_1$ and $Q_2$ are not units of $K \left[{X}\right]$. By Units of Ring of Polynomial Forms over Field it follows that $\deg Q_1 \ge 1$ and $\deg Q_2 \ge 1$. Let $Q_1' = \lambda^{-1} Q_1$. This implies that: :$P = Q_1' Q_2$ with $\deg Q_1' = \deg Q_1 \ge 1$. But this is a non-trivial factorization of $P$ in $K \left[{X}\right]$. This contradicts our supposition that $P$ is irreducible. Therefore $\lambda P$ has no non-trivial factorization, that is, $\lambda P$ is irreducible. {{qed|lemma}} === Sufficient Condition === Let $\lambda P$ be irreducible. Let $Q = \lambda P$. From the necessary condition, we know that any scalar multiple of $Q$ is irreducible. In particular: :$\lambda^{-1}Q = \lambda^{-1}\lambda P = P$ is irreducible, the required result. {{Qed}} Category:Polynomial Theory nw6yg0yol9cx0ajlkik5il3g22753ym
{{BeginTableau|\left({p \lor q}\right) \land \left({r \lor s}\right) \vdash \left({p \lor r}\right) \lor \left({q \land s}\right)}} {{Premise|1|\left({p \lor q}\right) \land \left({r \lor s}\right)}} {{SequentIntro|2|1|\left({p \land \left({r \lor s}\right)}\right) \lor \left({q \land \left({r \lor s}\right)}\right)|1|Conjunction Distributes over Disjunction}} {{TheoremIntro|3|p \land \left({r \lor s}\right) \implies p|Simplification}} {{TheoremIntro|4|q \land \left({r \lor s}\right) \implies \left({q \land r}\right) \lor \left({q \land s}\right)|Conjunction Distributes over Disjunction}} {{TheoremIntro|5|q \land r \implies r|Simplification}} {{TheoremIntro|6|q \land s \implies q \land s|Law of Identity/Formulation 2}} {{SequentIntro|7||\left({q \land r}\right) \lor \left({q \land s}\right) \implies r \lor \left({q \land s}\right)|5,6|Constructive Dilemma}} {{SequentIntro|8||q \land \left({r \lor s}\right) \implies r \lor \left({q \land s}\right)|4,7|Hypothetical Syllogism}} {{SequentIntro|9||\left({p \lor q}\right) \land \left({r \lor s}\right) \implies p \lor \left({r \lor \left({q \land s}\right)}\right)|3,8|Constructive Dilemma}} {{ModusPonens|10|1|p \lor \left({r \lor \left({q \land s}\right)}\right)|9|1}} {{SequentIntro|11|1|\left({p \lor r}\right) \lor \left({q \land s}\right)|10|Rule of Association}} {{EndTableau}} {{qed}} Category:Conjunction Category:Disjunction 75311jdpussf09y7sbnft5t38dm2bjf
{{finish|The following is an outline only}} Suppose that $D \sqbrk X$ exists. Let $\displaystyle \map P X = \sum_{k \mathop = 0}^n a_k X^k$, where $a_n \ne 0_D$, be an arbitrary element of $D \sqbrk X$. Then $\displaystyle \map P X$ corresponds to, and is completely described by, the ordered tuple of coefficients $\tuple {a_0, a_1, \dotsc, a_n, 0_D, 0_D, 0_D, \dotsc}$. Consider the set $S$ of infinite sequences of elements of $D$ which are eventually $0_D$. That is, whose elements are of the form $\tuple {b_0, b_1, \dotsc, b_n, 0_D, 0_D, 0_D, \dotsc}$ where $b_0, \ldots, b_n \in D$. Consider the polynomial ring over $S$ by defining the operations: {{begin-axiom}} {{axiom | n = 1         | lc= '''Ring Addition:'''         | ml= \sequence {r_0, r_1, r_2, \ldots} + \sequence {s_0, s_1, s_2, \ldots}         | mo= =         | mr= \sequence {r_0 + s_0, r_1 + s_1, r_2 + s_2, \ldots}         | c =  }} {{axiom | n = 2         | lc= '''Ring Negative:'''         | ml= -sequence {r_0, r_1, r_2, \ldots}         | mo= =         | mr= \sequence {-r_0, -r_1, -r_2, \ldots}         | c =  }} {{axiom | n = 3         | lc= '''Ring Product:'''         | ml= \sequence {r_0, r_1, r_2, \ldots} \circ \sequence {s_0, s_1, s_2, \ldots}         | mo= =         | mr= \sequence {t_0, t_1, t_2, \ldots}         | rc= where $\displaystyle t_i = \sum_{j \mathop + k \mathop = i} r_j s_k$ }} {{end-axiom}} From Polynomial Ring of Sequences is Ring we have that $\struct {S, +, \circ}$ is a ring. {{Finish|To be proved: a) that the sequences $\tuple {a_0, 0_D, 0_d}$ form a subring $D'$ of $\struct {S, +, \circ}$ isomorphic to $D$, b) the sequence $\tuple {0_D, 1_D, 0_D, 0_D, \dotsc}$ is transcendental over $D'$, and c) that $D' \sqbrk X$ is the whole of $R$. Thus we have constructed $D' \simeq D$. If we now ignore the difference between $a_0 \in D$ and $\tuple {a_0, 0_D, 0_d} \in D'$ so that $D'$ is identified with $D$ the ring $D \sqbrk X$ has been constructed as required.}}
{{BeginTableau|\vdash \paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} } }} {{Assumption|1|p \land \paren {q \lor r} }} {{SequentIntro|2|1|\paren {\paren {p \land q} \lor \paren {p \land r} }|1|Conjunction is Left Distributive over Disjunction: Formulation 1}} {{Implication|3||\paren {p \land \paren {q \lor r} } \implies \paren {\paren {p \land q} \lor \paren {p \land r} }|1|2}} {{Assumption|4|\paren {p \land q} \lor \paren {p \land r} }} {{SequentIntro|5|4|p \land \paren {q \lor r}|4|Conjunction is Left Distributive over Disjunction: Formulation 1}} {{Implication|6||\paren {\paren {p \land q} \lor \paren {p \land r} } \implies \paren {p \land \paren {q \lor r} }|4|5}} {{BiconditionalIntro|7||\paren {p \land \paren {q \lor r} } \iff \paren {\paren {p \land q} \lor \paren {p \land r} }|3|6}} {{EndTableau}} {{qed}}
{{BeginTableau|\vdash \left({p \land \left({q \lor r}\right)}\right) \iff \left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right)}} {{TheoremIntro|1|\left({p \land \left({q \lor r}\right)}\right) \implies \left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right)|Conjunction Distributes over Disjunction: Forward Implication}} {{TheoremIntro|2|\left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right) \implies \left({p \land \left({q \lor r}\right)}\right)|Conjunction Distributes over Disjunction: Reverse Implication}} {{BiconditionalIntro|3||\left({p \land \left({q \lor r}\right)}\right) \iff \left({\left({p \land q}\right) \lor \left({p \land r}\right)}\right)|1|2}} {{EndTableau}} {{qed}}
By Eisenstein Integers form Subring of Complex Numbers we know that $\struct {\Z \sqbrk \omega, +, \times}$ is a subring of the complex numbers $\C$. Let $1_\C$ be the unity of $\C$. Let $1_\omega$ be the unity of $\Z \sqbrk \omega$. By the Subdomain Test it suffices to show that $1_\C = 1_\omega$. By Unity of Ring is Unique it suffices to show that $1_\C$ is a unity of $\Z \sqbrk \omega$. First we note that: :$\Z \sqbrk \omega = \set {a + b\omega: a, b \in \Z}$ In particular: :$1_\C \in \Z \sqbrk \omega$ Moreover, by definition, $\Z \sqbrk \omega$ inherits its ring product from $\C$. For any $\alpha \in \Z \sqbrk \omega$: :$1_\C \alpha = \alpha 1_\C = \alpha$ in $\C$. Therefore this identity holds in $\Z \sqbrk \omega$ as well. {{Qed}} Category:Number Theory Category:Integral Domains Category:Eisenstein Integers 5sqorqvi9uyhayy87nukjuax98mo1k8
We will use the Subring Test. This is valid, as the set of complex numbers $\C$ forms a field, which is by definition itself a ring. We note that $\Z \sqbrk \omega$ is not empty, as (for example) $0 + 0 \omega \in \Z \sqbrk \omega$. Let $a + b \omega, c + d \omega \in \Z \sqbrk \omega$. Then we have $-\paren {c + d \omega} = -c - d \omega$, and so: {{begin-eqn}} {{eqn | l = \paren {a + b \omega} + \paren {-\paren {c + d \omega} }       | r = \paren {a + b \omega} + \paren {-c - d \omega}       | c =  }} {{eqn | r = \paren {a + \paren {-c} } + \paren {b + \paren {-d} } \omega       | c =  }} {{eqn | r = \paren {a - c} + \paren {b - d} \omega       | c =  }} {{end-eqn}} We have that $a, b, c, d \in \Z$ and $\Z$ is an integral domain. Therefore by definition $\Z$ is a ring. So it follows that $a - c \in \Z$ and $b - d \in \Z$. Hence $\paren {a - c} + \paren {b - d} \omega \in \Z \sqbrk \omega$. Now consider $\paren {a + b \omega} \paren {c + d \omega}$. By the definition of complex multiplication, we have: :$\paren {a + b \omega} \paren {c + d \omega} = \paren {a c - b d} + \paren {a d + b c} \omega$ As $a, b, c, d \in \Z$ and $\Z$ is a ring, it follows that $a c - b d \in \Z$ and $ad + bc \in \Z$. Hence: :$\paren {a + b \omega} \paren {c + d \omega} \in \Z \sqbrk \omega$ So by the Subring Test, $\Z \sqbrk \omega$ is a subring of $\C$. {{qed}}
We find that: {{begin-eqn}} {{eqn | l = \cmod \alpha^2       | r = \alpha \overline \alpha       | c = Modulus in Terms of Conjugate }} {{eqn | r = \paren {a + b \omega} \paren {\overline {a + b \omega} }       | c = Modulus in Terms of Conjugate }} {{eqn | r = \paren {a + b \omega} \paren {\overline a + \overline b \overline \omega}       | c = Sum of Complex Conjugates and Product of Complex Conjugates }} {{eqn | r = \paren {a + b \omega} \paren {a + b \overline \omega}       | c = Complex Number equals Conjugate iff Wholly Real }} {{eqn | r = a^2 + \paren {\omega + \overline \omega} a b + \omega \overline \omega b^2       | c =  }} {{end-eqn}} By the definition of the polar form of a complex number: :$\omega = \exp \paren {\dfrac {2 \pi i} 3} = \map \cos {\dfrac {2 \pi} 3} + i \, \map \sin {\dfrac {2 \pi} 3} = -\dfrac 1 2 + i \dfrac {\sqrt 3} 2$ Thus by Sum of Complex Number with Conjugate: :$\omega + \overline \omega = 2 \cdot \paren {-\dfrac 1 2} = -1$ Also: {{begin-eqn}} {{eqn | l = \omega \overline \omega       | r = \map \exp {\dfrac {2 \pi i} 3} \, \overline {\map \exp {\dfrac {2 \pi i} 3} }       | c =  }} {{eqn | r = \map \exp {\dfrac {2 \pi i} 3} \, \map \exp {-\dfrac {2 \pi i} 3}       | c = Polar Form of Complex Conjugate }} {{eqn | r = \map \exp {\dfrac {2 \pi i} 3 - \dfrac {2 \pi i} 3}       | c = Exponential of Sum }} {{eqn | r = \map \exp 0       | c =  }} {{eqn | r = 1       | c = Exponential of Zero }} {{end-eqn}} Therefore: :$\cmod \alpha^2 = a^2 + \paren {\omega + \overline \omega} a b + \omega \overline \omega b^2 = a^2 - a b + b^2$ as required. {{qed}} Category:Algebraic Number Theory 033fhr3hho1o0vaercxao9w8ucefa8h
As $\struct {F, +, \circ}$ is a field, it is also by definition a ring. Thus from Polynomial Ring of Sequences is Ring we have that $\struct {\GF, \oplus, \otimes}$ is a ring. {{explain|Use an analogous result to Ring of Polynomial Forms is Commutative Ring with Unity to get the CRU bit done}} From Field is Integral Domain, a field is also by definition an integral domain. Let $f, g \in \GF$ such that neither $f$ nor $g$ are the null polynomial. Let: :$\deg f = m, \deg g = n$ where $\deg$ denotes the degree of $f$ and $g$ respectively. By Degree of Product of Polynomials over Integral Domain, the degree of $f \times g$ is $m + n$. Then by definition of polynomial multiplication, its leading coefficient is $a_m \circ b_n$. As by definition an integral domain has no proper zero divisors: :$a_m \circ b_n \ne 0_F$. So, by definition, $f \otimes g$ has a leading coefficient which is not $0_F$. That is, $f \otimes g$ is not the null polynomial The result follows by definition of integral domain.
First we check that the operations of ring product and ring addition are closed in $A$. Let $Z$ be the set of all multiindices indexed by $J$. Let: :$\displaystyle f = \sum_{k \mathop \in Z} a_k \mathbf X^k, \ g = \sum_{k \mathop \in Z} b_k \mathbf X^k \in R \sqbrk {\set {X_j: j \in J} }$. Under the evaluation homomorphism, $f$ and $g$ map to: :$\displaystyle A \owns \hat f: \forall x \in R^J: \map {\hat f} x = \sum_{k \mathop \in Z} a_k x^k$ :$\displaystyle A \owns \hat g: \forall x \in R^J: \map {\hat g} x = \sum_{k \mathop \in Z} b_k x^k$ {{explain|Clarification needed on the above link: exactly what the "evaluation homomorphism" is needs to be established.}} Then the induced pointwise sum of $\hat f$ and $\hat g$ is: {{begin-eqn}} {{eqn | l = \map {\hat f} x + \map {\hat g} x       | r = \sum_{k \mathop \in Z} a_k x^k + \sum_{k \mathop \in Z} b_k x^k }} {{eqn | r = \sum_{k \mathop \in Z} \paren {a_k + b_k} x^k }} {{eqn | r = \map {\widehat {f + g} } x       | c = {{Defof|Addition of Polynomial Forms}} }} {{end-eqn}} Thus polynomial functions are closed under ring addition. The induced pointwise product of $\hat f$ and $\hat g$ is: {{begin-eqn}} {{eqn | l = \map {\hat f} x \circ \map {\hat g} x       | r = \paren {\sum_{k \mathop \in Z} a_k x^k} \circ \paren {\sum_{k \mathop \in Z} a_k x^k} }} {{eqn | r = \sum_{k \mathop \in Z} \paren {\sum_{p + q \mathop = k} a_p b_q} \mathbf X^k }} {{eqn | r = \map {\widehat {f \circ g} } x       | c = {{Defof|Multiplication of Polynomial Forms}} }} {{end-eqn}} Thus polynomial functions are closed under ring product. Finally, we invoke Structure Induced by Ring Operations is Ring, which shows that $\struct {A, +, \circ}$ is a commutative ring with unity. {{qed}} Category:Polynomial Theory d0rhzr8jw0klq6yvjr2jn9d9ky9hnhe
Let $P = \left\{{x \in L: x \preceq f \left({x}\right)}\right\}$. Let $p = \bigvee P$, the supremum of $P$. Let $x \in P$. Then by the definition of supremum: : $x \preceq p$ Since $f$ is increasing: : $f \left({x}\right) \preceq f \left({p}\right)$ By the definition of $P$: : $x \preceq f \left({x}\right)$ Thus because $\preceq$ is an ordering, and therefore transitive: : $x \preceq f \left({p}\right)$ As this holds for all $x \in P$, $f \left({p}\right)$ is an upper bound of $P$. By the definition of supremum: : $p \preceq f \left({p}\right)$ As $f$ is increasing: : $f \left({p}\right) \preceq f \left({f \left({p}\right)}\right)$ Thus by the definition of $P$: : $f \left({p}\right) \in P$ Since $p$ is the supremum of $P$: : $f \left({p}\right) \preceq p$ Since we already know that $p \preceq f \left({p}\right)$: : $f \left({p}\right) = p$ because $\preceq$ is an ordering and therefore antisymmetric. Thus $p$ is a fixed point of $f$. We have that $\preceq$ is an ordering, and therefore reflexive. Thus every fixed point of $f$ is in $P$. So $p$ is the greatest fixed point of $f$. Now note that $f$ is also increasing in the dual ordering. Thus $f$ also has a greatest fixed point in the dual ordering. That is, it has a least fixed point in the original ordering. {{qed}} {{Namedfor|Bronisław Knaster|name2 = Alfred Tarski|cat = Knaster|cat2 = Tarski}}
Let $S \subseteq F$. Let $s = \bigvee S$ be the supremum of $S$. We wish to show that there is an element of $F$ that succeeds all elementd of $S$ and is the smallest element of $F$ to do so. By the definition of supremum, an element succeeds all elements of $S$ {{iff}} it succeeds $s$. Let $U = s^\succeq$ be the upper closure of $s$. Thus we seek the smallest fixed point of $f$ that lies in $U$. Note that $U = \left[{s \,.\,.\, \top}\right]$, the closed interval between $s$ and the top element of $L$. First we show that $U$ is closed under $f$. We have that: :$\forall a \in S: a \preceq s$ so: $a = f \left({a}\right) \preceq f \left({s}\right)$ Thus $f \left({s}\right)$ is an upper bound of $S$, so by the definition of supremum, $s \preceq f \left({s}\right)$. Let $x \in U$. Then $s \preceq x$. So: : $f \left({s}\right) \preceq f \left({x}\right)$ Since $s \preceq f \left({s}\right)$, it follows that: : $s \preceq f \left({x}\right)$ so: : $f \left({x}\right) \in U$ Thus the restriction of $f$ to $U$ is an increasing mapping from $U$ to $U$. By Interval in Complete Lattice is Complete Lattice, $\left({U, \preceq}\right)$ is a complete lattice. Thus by Knaster-Tarski Lemma, $f$ has a smallest fixed point in $U$. Thus $S$ has a supremum in $F$. A precisely similar argument shows that $S$ has an infimum in $F$. Since this holds for all $S \subseteq F$, it follows that $\left({F, \preceq}\right)$ is a complete lattice. {{qed}}