hcju/proofwikiqa huggingface dataset:
Let $R = \Q \sqbrk X$ be the ring of polynomials in one variable $X$ over $\Q$. Let: :$J_1 = \set {a_0 + a_1 X + \cdots + a_n X^n \in R : a_0 = a_1 = 0}$ and :$J_2 = \set {a_0 + a_1 X + \cdots + a_n X^n \in R : a_0 = a_1 = a_3 = 0}$ First let us show that $J_1$ is an ideal of $R$. We establish the properties of the ideal test in order. $(1): \quad J_1 \ne \O$ This follows from the fact that $X^2 \in J_1$. $(2): \quad \forall P, Q \in J_1: P + \paren {-Q} \in J_1$ Let: :$\displaystyle P = \sum_{i \mathop = 0}^{+\infty} a_i X^i \in J_1$ :$\displaystyle Q = \sum_{i \mathop = 0}^{+\infty} b_i X^i \in J_1$ Then by the definition of addition of polynomials: :$\displaystyle P + \paren {-Q} = \sum_{i \mathop = 0}^{+\infty} c_i X^i$ where: :$c_i = a_i - b_i$ By assumption, $a_0 = a_1 = b_0 = b_1 = 0$. Therefore: :$c_0 = a_0 - b_0 = 0$ :$c_1 = a_1 - b_1 = 0$ Therefore $P + \left({-Q}\right) \in J_1$. $(3): \quad \forall P \in J_1, Q \in R: Q \cdot P \in J_1$ Let :$\displaystyle P = \sum_{i \mathop = 0}^{+\infty} a_i X^i \in J_1$ :$\displaystyle Q = \sum_{i \mathop = 0}^{+\infty} b_i X^i \in R$ By the definition of multiplication of polynomials: :$\displaystyle Q \cdot P = \sum_{i \mathop = 0}^{+\infty} c_i X^i$ where: :$c_i = \sum_{j + k \mathop = i} a_j b_k$ In particular, since $a_0 = 0$: :$c_0 = a_0 b_0 = 0$ Since $a_0 = a_1 = 0$: :$c_ 1 = a_0 b_1 + a_1 b_0 = 0$ Therefore $Q \cdot P \in J_1$. This shows that $J_1$ is an ideal of $R$. Next we show that $J$ is an ideal of $J_1$. Again, we verify the properties of the ideal test in turn. $(1): \quad J_2 \ne \O$ This follows from the fact that $X^2 \in J_2$. $(2): \quad \forall P, Q \in J_2: P + \paren {-Q} \in J_2$ Let: :$\displaystyle P = \sum_{i \mathop = 0}^{+\infty} a_i X^i \in J_2$ :$\displaystyle Q = \sum_{i \mathop = 0}^{+\infty} b_i X^i \in J_2$ Then by the definition of addition of polynomials: :$\displaystyle P + \paren {-Q} = \sum_{i \mathop = 0}^{+\infty} c_i X^i$ where: :$c_i = a_i - b_i$ By assumption: :$a_0 = a_1 = a_3 = b_0 = b_1 = b_3 = 0$ Therefore: :$c_0 = a_0 - b_0 = 0$ :$c_1 = a_1 - b_1 = 0$ :$c_3 = a_3 - b_3 = 0$ Therefore $P + \paren {-Q} \in J_2$. $(3): \quad \forall P \in J_2, Q \in J_1: Q \cdot P \in J_2$ Let :$\displaystyle P = \sum_{i \mathop = 0}^{+\infty} a_i X^i \in J_2$ :$\displaystyle Q = \sum_{i \mathop = 0}^{+\infty} b_i X^i \in J_1$ By the definition of multiplication of polynomials, :$\displaystyle Q \cdot P = \sum_{i \mathop = 0}^{+\infty} c_i X^i$ where: :$c_i = \sum_{j + k \mathop = i} a_j b_k$ Since $J_2 \subseteq J_1$, we have $P, Q \in J_1$. So we have already that $c_0 = c_1 = 0$. Moreover $a_0 = a_1 = b_0 = b_1 = 0$, so :$c_3 = a_0b_3 + a_1 b_2 + a_2 b_1 + a_3 b_0 = 0$ Therefore $Q \cdot P \in J_2$. This shows that $J_2$ is an ideal of $J_1$. Finally we wish to see that $J_2$ is not an ideal of $R$. We have that $X^2 \in J_2$ and $X \in R$. {{AimForCont}} $J_2$ were an ideal of $R$. This would imply that: :$X \cdot X^2 = X^3 \in J_2$ But by the definition of $J_2$, we must have that the coefficient of $X^3$ is $0$. This is a contradiction, so $J_2$ is not an ideal of $R$. {{qed}} Category:Ideal Theory i65nuorbt880ya12pvev69hzk3c32li
Let $b_1, b_2 \in \mathbb L_J$. Then from Set of Ideals forms Complete Lattice: :$(1): \quad b_1 + b_2 \in \mathbb L_J$ and is the supremum of $\set {b_1, b_2}$ :$(2): \quad b_1 \cap b_2 \in \mathbb L_J$ and is the infimum of $\set {b_1, b_2}$ Thus $\struct {\mathbb L_J, \subseteq}$ is a lattice. {{qed}}
Let $b \in \mathbb L_J$. From the way $\mathbb L_J$ is defined: : $J \subseteq b$ Thus by Preimage of Image of Subring under Ring Homomorphism: : $q_J^{-1} \left({q_J \left({b}\right)}\right) = b + J = b$ Let $c$ be an ideal of $R / J$. Then, by Image of Preimage of Subring under Ring Epimorphism: : $q_J \left({q_J^{-1} \left({c}\right)}\right) = c$ Thus by Bijection iff Left and Right Inverse, $\Phi_J$ is a bijection. Hence: : $\forall c \in \mathbb L \left({R / J}\right): q_J^{-1} \left({\Phi_J}\right) c = q_J^{-1} \left({c}\right)$ Now to show that $\Phi_J$ is an isomorphism. Let $b_1, b_2 \in \mathbb L_J$. Let $b_1 \subseteq b_2$. Then from Subset Maps to Subset: : $q_J \left({b_1}\right) \subseteq q_J \left({b_2}\right)$ Conversely, suppose $q_J \left({b_1}\right) \subseteq q_J \left({b_2}\right)$. By what we have just proved: : $b_1 = q_J^{-1} \left({q_J \left({b_1}\right)}\right) \subseteq  q_J^{-1} \left({q_J \left({b_2}\right)}\right) = b_2$ Thus $\Phi_J$ is an isomorphism. {{Qed}}
Let $J$ be an ideal of $\Z$. Then $J$ is a subring of $\Z$, and so $\left({J, +}\right)$ is a subgroup of $\left({\Z, +}\right)$. But by Integers under Addition form Infinite Cyclic Group, the group $\left({\Z, +}\right)$ is cyclic, generated by $1$. Thus by Subgroup of Cyclic Group is Cyclic, $\left({J, +}\right)$ is cyclic, generated by some $m \in \Z$. Therefore from the definition of principal ideal, $J = \left\{{k m: k \in \Z}\right\} = \left({m}\right)$, and is thus a principal ideal. {{qed}}
We have that Integers are Euclidean Domain. Then we have that Euclidean Domain is Principal Ideal Domain. Hence the result. {{qed}}
Let $U$ be an arbitrary ideal of $\Z$. Let $c$ be a non-zero element of $U$. Then both $c$ and $-c$ belong to $\ideal a$ and one of them is positive. Thus $U$ contains strictly positive elements. Let $b$ be the smallest strictly positive element of $U$. By the Set of Integers Bounded Below by Integer has Smallest Element, $b$ is guaranteed to exist. If $\ideal b$ denotes the ideal generated by $b$, then $\ideal b \subseteq U$ because $b\in U$ and $U$ is an ideal. Let $a \in U$. By the Division Theorem: :$\exists q, r \in \Z, 0 \le r < b: a = b q + r$ As $a, b \in U$ it follows that so does $r = a - b q$. By definition of $b$ it follows that $r = 0$. Thus: :$a = b q \in \ideal b$ and so: :$U \subseteq \ideal b$ From the above: :$U = \ideal b$ It follows by definition that $U$ is a principal ideal of $\Z$. Recall that $U$ was an arbitrary ideal of $\Z$. Hence by definition $\Z$ is a principal ideal domain. {{qed}}
It follows from Ring of Integers is Principal Ideal Domain that $J$ is a principal ideal. Let $c \in J, c \ne 0$. Then $-c \in J$ and by Natural Numbers are Non-Negative Integers, exactly one of them is strictly positive. Thus $J$ ''does'' actually contain strictly positive elements, so that's a start. Let $b$ be the smallest strictly positive element of $J$. This exists because Natural Numbers are Non-Negative Integers and the Well-Ordering Principle. By definition of a principal ideal, we have $\ideal b \subseteq J$ as $b \in J$. We need to show that $J \subseteq \ideal b$. So, let $a \in J$. By the Division Theorem, $\exists q, r: a = b q + r, 0 \le r < b$. As $a, b \in J$, then so does $r = a - b q$. So, by the definition of $b$, it follows that $r = 0$. Thus $a = b q \in \ideal b$. {{qed}}
First we show that $\psi$ is injective. Suppose $0 < b < c$. Then $b \in \left({b}\right)$, but $b \notin \left({c}\right)$, because from Principal Ideals of Integers, $c$ is the smallest positive integer in $\left({c}\right)$. Thus $\left({b}\right) \ne \left({c}\right)$. It is also apparent that $b > 0 \implies \left({b}\right) \ne \left({0}\right)$ as $\left({0}\right) = \left\{{0}\right\}$. Thus $\psi$ is injective. Surjectivity follows from Principal Ideals of Integers: every integer is the smallest strictly positive element of a principal ideal of $\Z$. {{qed}}
Let $x, y \in \N_m$. By the Division Theorem: {{begin-eqn}} {{eqn | ll= \exists q, r \in \Z:       | l = x + y       | r = m q + r       | c = for $0 \le r < m$ }} {{eqn | ll= \exists p, s \in \Z:       | l = x y       | r = m p + s       | c = for $0 \le s < m$ }} {{end-eqn}} Then $x +_m y = r$ and $x \times_m y = s$, so: {{begin-eqn}} {{eqn | l = q_m \left({x +_m y}\right)       | r = q_m \left({r}\right)       | c =  }} {{eqn | r = q_m \left({m q}\right) + q_m \left({r}\right)       | c =  }} {{eqn | r = q_m \left({m q + r}\right)       | c =  }} {{eqn | r = q_m \left({x + y}\right)       | c =  }} {{eqn | r = q_m \left({x}\right) + q_m \left({y}\right)       | c =  }} {{end-eqn}} and similarly $q_m \left({x \times_m y}\right) = q_m \left({x y}\right) = q_m \left({x}\right) q_m \left({y}\right)$. So the restriction of $q_m$ to $\N_m$ is a homomorphism from $\left({\N_m, +_m, \times_m}\right)$ into $\left({\Z / \left({m}\right), +_{\left({m}\right)}, \times_{\left({m}\right)}}\right)$. Let $a \in \Z$. Then $\exists q, r \in \Z: a = q m + r: 0 \le r < m$, so $q_m \left({a}\right) = q_m \left({r}\right) \in q_m \left({\N_m}\right)$. Therefore $\Z / \left({m}\right) = q_m \left({\Z}\right) = q_m \left({\N_m}\right)$. Therefore the restriction of $q_m$ to $\N_m$ is surjective. If $0 < r < m$, then $r \notin \left({m}\right)$ and thus $q_m \left({r}\right) \ne 0$. Thus the kernel of the restriction of $q_m$ to $\N_m$ contains only zero. Therefore by the Quotient Theorem for Group Epimorphisms, the restriction of $q_m$ to $\N_m$ is an isomorphism from $\N_m$ to $\Z / \left({m}\right)$. {{wtd|Reference is made throughout to $\left({\N_m, +_m, \times_m}\right)$. It needs to be shown that this is the same (at least up to isomorphism) as the ring of integers modulo $m$ $\left({\Z, +_m, \times_m}\right)$.)}}
The ring of integers is a principal ideal domain. The result follows directly from Principal Ideals in Integral Domain. {{qed}}
From Element of Principal Ideal Domain is Finite Product of Irreducible Elements, each element which is neither $0$ nor a unit of a principal ideal domain has a factorization of irreducible elements. {{proof wanted|Need to prove that the factorization is unique}}
=== Maximal Ideal implies Quotient Ring is Field === {{:Maximal Ideal iff Quotient Ring is Field/Proof 1/Maximal Ideal implies Quotient Ring is Field}}{{qed|lemma}} === Quotient Ring is Field implies Ideal is Maximal === {{:Maximal Ideal iff Quotient Ring is Field/Proof 1/Quotient Ring is Field implies Ideal is Maximal}}{{qed}}
Let $\mathbb L_J$ be the set of all ideals of $R$ which contain $J$. Let the ordered set $\struct {\map {\mathbb L} {R / J}, \subseteq}$ be the set of all ideals of $R / J$. Let the mapping $\Phi_J: \struct {\mathbb L_J, \subseteq} \to \struct {\map {\mathbb L} {R / J}, \subseteq}$ be defined as: :$\forall a \in \mathbb L_J: \map {\Phi_J} a = \map {q_J} a$ where $q_J: a \to a / J$ is the quotient epimorphism from $a$ to $a / J$. Then from Ideals Containing Ideal Isomorphic to Quotient Ring, $\Phi_J$ is an isomorphism. Now from Quotient Ring Defined by Ring Itself is Null Ring, $\map {q_J} J$ is the null ideal of $R / J$. At the same time, $\map {q_J} R$ is the entire ring $R / J$. Let $R / J$ not be the null ring. Then by Epimorphism Preserves Rings and Epimorphism Preserves Commutativity, $R / J$ is a commutative ring with unity. By definition, $J$ is a maximal ideal of $R$ {{iff}} $\mathbb L_J = \set {J, R}$ and $J$ is a proper ideal of $R$. By Ideals of Field, $R / J$ is a field {{iff}}: :$\map {\mathbb L} {R / J} = \set {\map {q_J} J, \map {q_J} R}$ and the null ideal $\map {q_J} J$ is a proper ideal of $R / J$. As $\Phi_J: \mathbb L_J \to \map {\mathbb L} {R / J}$ is an isomorphism, $J$ is a maximal ideal {{iff}} $J$ is a field. {{qed}}
Let $J$ be a maximal ideal. We have by definition of quotient ring that $J$ is the zero element of $R / J$. Let $A \in R / J$ be a non-zero element of $R / J$. Let $x \in A$. Since $A \ne J$, we have that $x \notin J$. Let the ideal $K = J + A$ of $R$ be formed. This contains all the elements of the form $j + r a$, with $j \in J$ and $r \in R$. As $J$ is maximal and $J \subsetneq K$, it follows that: :$K = R$ and so: :$1_R \in K$ That is: :$\exists j \in J, r \in R: j + r a = 1_R$ Thus: :$\paren {r + J} \paren {a + J} = \paren {1 - u} J = 1_R + J$ and so $\paren {r + J}$ is the product inverse of $\paren {a + J}$. So every non-zero element of $R / J$ has a product inverse. That is, $R / J$ is a field. {{qed|lemma}} Let $R / J$ be a field. Let $K$ be an ideal of $R$ such that: :$J \subsetneq K \subseteq R$ Let $a \in K$ such that $a \notin J$. Then: :$J + \ideal a \subsetneq K$ But as $a \notin J$, we have that $a + J$ is a non-zero element of $R / J$. Thus as $R / J$ is a field, $a + J$ has a product inverse $r + J$: :$\paren {r + J} \paren {a + J} = 1_R + J$ So: :$\exists r \in R, j \in J: r a + \paren {-1_R} = j$ That is: :$r a + \paren {-1_R} \in J$ So: :$j + r a = 1_R$ and from Ideal of Unit is Whole Ring: Corollary this implies: :$J + \ideal a = R$ So: :$K = R$ and it follows by definition that $J$ is maximal. {{qed}}
First, note that $\Z$ is a principal ideal domain, so all ideals are principal. Suppose $\ideal p$ is prime. From Integer Divisor is Equivalent to Subset of Ideal, $m \divides n \iff \ideal n \subseteq \ideal m$. But as $p$ is prime, the only divisors of $p$ are $1$ and $p$ itself. By Natural Numbers Set Equivalent to Ideals of Integers, it follows that if $p$ is prime, then $\ideal p$ must be a maximal ideal. {{qed|lemma}} Conversely, let $p \in \Z_{>0}$ such that $\ideal p$ is maximal. Then if $\ideal p \subseteq \ideal q$ for some $q \in \Z_{>0}$, we have: :$\ideal q = \ideal p \implies q = p$ or :$\ideal q = \ideal 1 \implies q = 1$ Hence, as $p \in \ideal q$, we have: :$p \in q \iff q = 1 \text { or } q = p$ and: :$p \in \ideal q \implies q \divides p$ Hence $p$ is prime, so $\ideal p$ is a prime ideal. {{qed}}
By Prime Number iff Generates Principal Maximal Ideal and Maximal Ideal iff Quotient Ring is Field, $(1)$ implies $(3)$, and from Field is Integral Domain, $(3)$ implies $(2)$. By the definition of Integral Domain, $\Z_p$ is an integral domain iff $\left({\Z_p^*, \times_p}\right)$ is a semigroup. Let $\left({p}\right)$ be the principal ideal of $\left({\Z, +, \times}\right)$ generated by $p$. From the Canonical Epimorphism from Integers by Principal Ideal, $\left({\Z_p, +_p, \times_p}\right)$ is isomorphic to $\left({\Z, +, \times}\right) / \left({p}\right)$. So, we can let $q_p \left({m}\right): \Z \to \Z_p$ be the quotient mapping from $\left({\Z, +, \times}\right)$ to $\left({\Z_p, +_p, \times_p}\right)$. Let $0_p$ denote the zero of $\Z_p$. Suppose $p = m n$ where $1 < m < p, 1 < n < p$. Then in the ring $\Z_p$ we have $q_p \left({m}\right) \ne 0_p, q_p \left({n}\right) \ne 0_p$. But as $q_p$ is an epimorphism and therefore obeys the morphism property, $q_p \left({m}\right) \times_p q_p \left({n}\right) = q_p \left({m n}\right) = q_p \left({p}\right) = 0_p$. But by definition, $0_p \notin \Z_p^*$. Thus if $p = m n$, then $\left({\Z_p^*, \times_p}\right)$ is not a semigroup. {{Qed}}
{{begin-eqn}} {{eqn | l = \ideal 0       | r = \set {\sum^n_{i \mathop = 1} r_i \times 0 \times s_i: n \in \N; r_i, s_i \in \Z}       | c = {{Defof|Principal Ideal of Ring}} }} {{eqn | r = \set {\sum^n_{i \mathop = 1} 0: n \in \N}       | c = $0$ is the zero under integer multiplication }} {{eqn | r = \set 0       | c = Integer Addition Identity is Zero }} {{end-eqn}} {{finish}} Category:Quotient Rings Category:Integers 0pghfflaz3nxk2mqcj13qe7xh1y0fpw
=== Necessary Condition === {{:Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Forward Implication}} === Sufficient Condition === {{:Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Reverse Implication}} Category:Principal Ideals Category:Principal Ideal Domains Category:Factorization Category:Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal Category:Maximal Ideals of Rings sqqhg4vygfe3be9qkn9ckj88k6xwfri
By the Index Law for Sum of Indices and Powers of Ring Elements, we have $\left({n 1_R}\right) \left({m 1_R}\right) = n \left({m 1_R}\right) = \left({n m}\right) 1_R$. Thus $g$ is an epimorphism from $\Z$ onto $S$. Assume that $R$ has no proper zero divisors. By Kernel of Ring Epimorphism is Ideal, the kernel of $g$ is an ideal of $\Z$. By Ring of Integers is Principal Ideal Domain, the kernel of $g$ is $\left({p}\right)$ for some $p \in \Z_{>0}$. By Kernel of Ring Epimorphism is Ideal (don't think this is the correct reference - check it), $S$ is isomorphic to $\Z_p$ and also has no proper zero divisors. So from Integral Domain of Prime Order is Field either $p = 0$ or $p$ is prime. Now we need to show that $g$ is unique. Let $h$ be  a non-zero (ring) homomorphism from $\Z$ into $R$. As $h \left({1}\right) = h \left({1^2}\right) = \left({h \left({1}\right)}\right)^2$, either $h \left({1}\right) = 1_R$ or $h \left({1}\right) = 0_R$ by Idempotent Elements of Ring with No Proper Zero Divisors. But, by Homomorphism of Powers: Integers, $\forall n \in \Z: h \left({n}\right) = h \left({n 1}\right) = n h \left({1}\right)$ So if $h \left({1}\right) = 0_R$, then $\forall n \in \Z: h \left({n}\right) = n 0_R = 0_R$. Hence $h$ would be a zero homomorphism, which contradicts our stipulation that it is not. So $h \left({1}\right) = 1_R$, and thus $\forall n \in \Z: h \left({n}\right) = n 1 = g \left({n}\right)$. {{qed}} {{Proofread}}
From Null Ring iff Zero and Unity Coincide, $1_R \ne 0_R$ except when $R = \left\{{0_R}\right\}$. {{qed}} Category:Ring Theory 6oqldbeeezzek7hrj6fbojap6vj42gf
Suppose $\Char R = n$ where $n$ is composite. Let $n = r s$, where $r, s \in \Z, r > 1, s > 1$. First note that: {{begin-eqn}} {{eqn | l = \paren {r \cdot 1_R} \circ \paren {s \cdot 1_R}       | r = \paren {r s} \paren {1_R \circ 1_R}       | c = Powers of Ring Elements }} {{eqn | r = \paren {r s} 1_R       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \paren {r \cdot 1_R} \circ \paren {s \cdot 1_R}       | r = n \cdot 1_R       | c =  }} {{eqn | r = 0_R       | c =  }} {{eqn | ll= \leadsto       | l = r \cdot 1_R = 0_R       | o = \lor       | r = s \cdot 1_R = 0_R       | c =  }} {{end-eqn}} as $R$ has no proper zero divisors. But both $r$ and $s$ are less than $n$ which contradicting the minimality of $n$. So if $\Char R = n$ it follows that $n$ must be prime. Now let $x \in R^*$. Then by Characteristic times Ring Element is Ring Zero, $n \cdot x = 0_R$. It follows from Element to Power of Multiple of Order is Identity that: :$\order x \divides n$ Since $n$ is prime, either $\order x = 1$ or $\order x = n$. It cannot be $1$, from Null Ring iff Characteristic is One, so the result follows. {{qed}}
Let $\struct {D, +, \circ}$ be an integral domain, whose zero is $0_D$ and whose unity is $1_D$, such that $\Char D = 0$. Let $x \in D, x \ne 0_D$. Then: {{begin-eqn}} {{eqn | ll= \forall n \in \Z_{>0}:       | l = n \cdot x       | r = n \cdot \paren {x \circ 1_D}       | c =  }} {{eqn | r = \paren {n \circ 1_D} \cdot x       | c = Powers of Ring Elements }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = x       | o = \ne       | r = 0_D       | c =  }} {{eqn | ll= \leadsto       | l = n \cdot 1_D       | o = \ne       | r = 0_D       | c =  }} {{eqn | ll= \leadsto       | l = n \cdot x       | o = \ne       | r = 0_D       | c = {{Defof|Integral Domain}} }} {{end-eqn}} That is, $x$ has infinite order in $\struct {D, +}$. {{qed}}
From the definition, a field is a ring with no zero divisors. So by Characteristic of Finite Ring with No Zero Divisors, if $\Char F \ne 0$ then it is prime. {{Qed}}
Follows directly from: * Subring Generated by Unity of Ring with Unity * Quotient Theorem for Monomorphisms {{qed}}
Let $\struct {F, +, \times}$ be a field whose unity is $1_F$ such that $\Char F = p$. Let $P$ be a prime subfield of $F$. From Field has Prime Subfield, this has been shown to exist. We can consistently define a mapping $\phi: \Z_p \to F$ by: :$\forall n \in \Z_p: \map \phi {\eqclass n p} = n \cdot 1_F$ Suppose $a, b \in \eqclass n p$. Then: :$a = n + k_1 p, b = n + k_2 p$ So: {{begin-eqn}} {{eqn | l = \map \phi a       | r = \map \phi {n + k_1 p}       | c =  }} {{eqn | r = \paren {n + k_1 p} \cdot 1_F       | c =  }} {{eqn | r = n \cdot 1_F + k_1 p \cdot 1_F       | c =  }} {{eqn | r = n \cdot 1_F       | c =  }} {{end-eqn}} and similarly for $b$, showing that $\phi$ is well-defined. Let $C_a, C_b \in \Z_p$. Let $a \in C_a, b \in C_b$ such that $a = a' + k_a p, b = b' + k_b p$. Then: {{begin-eqn}} {{eqn | l = \map \phi {C_a} + \map \phi {C_b}       | r = \map \phi {a' + k_a p} + \map \phi {b' + k_b p}       | c =  }} {{eqn | r = \paren {a' + k_a p} \cdot 1_F + \paren {b' + k_b p} \cdot 1_F       | c =  }} {{eqn | r = \paren {a' + k_a p + b' + k_b p} \cdot 1_F       | c = Integral Multiple Distributes over Ring Addition }} {{eqn | r = \paren {a' + b'} \cdot 1_F + \paren {\paren {k_a p + k_b p} \cdot 1_F}       | c =  }} {{eqn | r = \paren {a' + b'} \cdot 1_F       | c =  }} {{eqn | r = \map \phi {C_a +_p C_b}       | c =  }} {{end-eqn}} Similarly for $\map \phi {C_a} \times \map \phi {C_b}$. So $\phi$ is a ring homomorphism. From Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, it follows that $\phi$ is a ring monomorphism. Thus it follows that $P = \Img \phi$ is a subfield of $F$ such that $P \cong \Z_p$. Let $K$ be a subfield of $F$. let $P = \Img \phi$ as defined above. We know that $1_F \in K$. It follows that $1_F \in K \implies P \subseteq K$. Thus $K$ contains a subfield $P$ such that $P$ is isomorphic to $\Z_p$. The uniqueness of $P$ follows from the fact that if $P_1$ and $P_2$ are both minimal subfields of $F$, then $P_1 \subseteq P_2$ and $P_2 \subseteq P_1$, thus $P_1 = P_2$. {{qed}}
By Intersection of Division Subrings is Division Subring, the intersection $P$ of the set of all division subrings of $K$ is a division ring. Let $\map Z K$ be the center of $K$. From Center of Ring is Commutative Subring, $\map Z K$ is a commutative subring of $K$. Therefore $\map Z K$ is a commutative division ring Thus $\map Z K$ is a subfield of $K$. But as $P$ is contained in $\map Z K$, it is itself commutative. By its definition, $P$ contains no proper subfield and hence is a prime field. Also, $P$ is contained in every other subfield of $K$. Therefore $P$ is therefore the ''only'' prime subfield of $K$. {{qed}} Category:Subfields Category:Division Subrings daxy7sy883yl680few7ik2tj3tnrzis
* By Properties of Ordered Ring $(5)$: :$\forall n \in \Z_{>0}: n \cdot 1_D > 0$ Thus $\operatorname{Char} \left({D}\right) \ne p$ for any $p > 0$. Hence $\operatorname{Char} \left({D}\right) = 0$ and so $g$ is a monomorphism from $\Z$ into $D$. Also, if $m < p$, then $p - m \in \Z_+$, so $p \cdot 1_D - m \cdot 1_D > 0_D$. Hence $g \left({m}\right) < g \left({p}\right)$ and thus by Monomorphism from Total Ordering, $g$ is a monomorphism from $\Z$ into $D$. {{qed}} {{Proofread}} {{Improve|We have a new and more axiomatic definition of an ordered integral domain which we may want to bring this into line with. Ultimately it boils down to the fact that an ordering on an integral domain is a lot more rigid than for a ring, such that totality is a consequence of the trichotomy law.}}
Follows from: : Characteristic of Ordered Integral Domain is Zero : Order Embedding between Quotient Fields is Unique. {{Qed}}
By application of the Subring Test: As $D$ is an integral domain, it has a unity $1_D$ and so $x = 1_D x$. Hence $x \in D \sqbrk x$ and so $D \sqbrk x \ne \O$. Let $p, q \in D \sqbrk x$. Then let: :$\displaystyle p = \sum_{k \mathop = 0}^m a_k \circ x^k, q = \sum_{k \mathop = 0}^n b_k \circ x^k$ Thus: :$\displaystyle -q = -\sum_{k \mathop = 0}^n b_k \circ x^k = \sum_{k \mathop = 0}^n \paren {-b_k} \circ x^k$ and so: :$q \in D \sqbrk x$ Thus as Polynomials Closed under Addition, it follows that: : $p + \paren {-q} \in D \sqbrk x$ Finally, from Polynomials Closed under Ring Product, we have that $p \circ q \in D \sqbrk x$. All the criteria of the Subring Test are satisfied. Hence the result. {{qed}}
{{handwaving}} It is immediate that $f \otimes g$ is a map from the free commutative monoid to $R$, so we need only prove that $f \otimes g$ is nonzero on finitely many $\mathbf X^k$, $k \in Z$. Suppose that for some $k \in Z$: :$\displaystyle \sum_{\substack{p + q = k \\ p, q \mathop \in Z}} a_p b_q \ne 0$ Therefore if $c_k \ne 0$ there exist $p, q \in Z$ such that $p + q = k$ and $a_p$, $b_q$ are both nonzero. Since $f$ and $g$ are polynomials, the set $\left\{{p + q: a_p \ne 0, \ b_q \ne 0}\right\}$ is finite, and we are done. {{qed}} {{proofread}} Category:Polynomial Theory 9e0x49comm43jlp1wrmcmg69fsco1f5