Ring Theory - A Rosetta Stone
Ch. 7 Introduction to Rings
Following the enumeration of results found in Dummit and Foote's Abstract Algebra, chapters 7 through 9, we relate the results of ring theory in Mizar.
§7.1 Basic Definitions and Examples
- Definition. A Ring \(R\) is a non-empty set with two binary \(+\) and
\(\times\) (called addition and multiplication) such that (
vectsp_1)- addition is commutative: for all \(x,y\in R\), we have \(x+y=y+x\) (
Abelian) - associativity of addition \((x+y)+z=x+(y+z)\) for all \(x,y,z\in R\)
(
add-associative) - additive inverse: for each \(x\in R\), there exists a \(y\in R\) such
that \(x+y=0\) (i.e., every element is
right_complementable) - additive identity \(x+0=x\) (
right_zeroed) and by commutativity \(0+x=x\) for all \(x\in R\) - multiplication is associative: for all \(x,y,z\in R\) we have \((xy)z=x(yz)\)
- it is well-unital: for all \(x\in R\), we have \(x\cdot 1=x\) and \(1\cdot x=x\)
- multiplication is distributive over addition: for all \(x,y,z\in R\) we have \(x(y+z)=xy+xz\) and \((x+y)z=xz+yz\)
- addition is commutative: for all \(x,y\in R\), we have \(x+y=y+x\) (
- Remark. This is technically a little "overfitted": Mathematicians can work with nonassociative rings, and do so all the time.
- Definition. A ring \(R\) is Commutative if multiplication is commutative.
- Definition. A ring \(R\) is said to have an Identity
(i.e., is Unital) if there is an element \(1\in R\) such
that for all \(a\in R\) we have \(1\times a=a\times 1=a\).
- Remark. Sadly, Mizar's ring is defined to always be unital. Therefore we will tend to drop the "unital" prefix.
- Implicit Definition. A ring \(R\) is called Degenerate if \(1=0\). Otherwise, \(R\) is said to be Non-degenerate.
- Definition. A nondegenerate ring \(R\) is called a Division Ring
(or a Skew-Field) if every nonzero \(a\in R\) has a
multiplicative inverse (
vectsp_1). - Definition. A Field is a commutative Skew-Field (
vectsp_1). - Examples.
- The integers form a ring, which Mizar calls
INT.Ring - The real numbers form a field which Mizar calls
F_Real - The rational numbers form a field which Mizar calls
F_Rat - The complex numbers form a field which Mizar calls
F_Complex - The integers modulo \(n\) forms a commutative ring which Mizar
calls
INT.Ring n - The quaternions form a ring \(\mathbb{H}\) which Mizar calls
R_Quaternion - If \(A\) is a ring and \(X\) is a nonempty set, then the set \(R\) of all functions \(X\to A\) forms a ring Missing from Mizar
- The integers form a ring, which Mizar calls
- Proposition 1. Let \(R\) be a ring. Then
- For all \(a\in R\) we have \(0a=a0=0\) (
vectsp_1:7andvectsp_1:14) - For all \(a,b\in R\) we have
- (
vectsp_1:8) \(a(-b)=-(ab)\) - (
vectsp_1:9) \((-a)b=-(ab)\)
- (
- For all \(a,b\in R\) we have \((-a)(-b)=ab\) (
vectsp_1:10) - For all \(a\in R\) we have \(-a=(-1)a\) (
vectsp_1:14)
- For all \(a\in R\) we have \(0a=a0=0\) (
- Definition. Let \(R\) be a ring.
- We say \(x\in R\) is a Zero Divisor if there is a \(y\in R\)
such that \(y\neq0\) and \(xy=0\). Mizar requires \(R\) to be
commutative. (
ringfrac) - If \(R\) is nondegenerate (i.e., \(1\neq0\)), we say \(x\in R\) is
Unital if there exists some \(y\in R\) such that \(xy=1\)
and \(yx=1\) (
gcd_1:def 2)
- We say \(x\in R\) is a Zero Divisor if there is a \(y\in R\)
such that \(y\neq0\) and \(xy=0\). Mizar requires \(R\) to be
commutative. (
- Examples.
- The example \(\mathbb{Q}(\sqrt{D})\) where \(D^{2}\notin\mathbb{Q}\)
is a specialized example of adjoining the squareroot of a
nonsquare element of a field. In Mizar, this is done by
FieldAdjunction(F,{sqrt a})(see, e.g., Lemmadegainfield_9)
- The example \(\mathbb{Q}(\sqrt{D})\) where \(D^{2}\notin\mathbb{Q}\)
is a specialized example of adjoining the squareroot of a
nonsquare element of a field. In Mizar, this is done by
- Definition. A commutative nondegenerate ring is called an
Integral Domain (
vectsp_2) if it has no zero divisors, i.e.,domRing-like: for any \(x,y\in R\), if \(xy=0\) then \(x=0\) or \(y=0\)
- Proposition 2. Let \(R\) be a ring, let \(a,b,c\in R\). If \(a\) is not a zero-divisor and \(ab=ac\), then \(b=c\). Missing from Mizar
- Corollary 3. Any finite integral domain is a field. Missing from Mizar
- Definition. Let \(R\) be a ring. A Subring of \(R\) is a
subgroup of \(R\) that is closed under multiplication. (
c0sp1:def 3) - Examples.
- \(\mathbb{Z}\) is a subring of \(\mathbb{Q}\) (
ring_3:47) - \(\mathbb{Q}\) is a subring of \(\mathbb{R}\) (
liouvil2:4) - \(\mathbb{R}\) is a subring of \(\mathbb{C}\) (
liouvil2:3) - The integral quaternions \(S=\mathbb{Z}+i\mathbb{Z}+j\mathbb{Z}+k\mathbb{Z}\) is a subring of the quaternions Missing from Mizar
- If \(R\) is a subring of a field \(F\) that contains the identity, then \(R\) is an integral domain Missing from Mizar
- \(\mathbb{Z}\) is a subring of \(\mathbb{Q}\) (
§7.2 Examples: Polynomial rings, Matrix rings, and group rings
Polynomial rings
Polynomial rings are…well…tricky. After all, it's given an entire chapter in Dummit and Foote, but it could easily be given its own book.
- Proposition 4. Let \(R\) be an integral domain, let \(p(x)\) and
\(q(x)\) be nonzero elements of \(R[x]\). Then
- \(\deg(p(x)q(x))=\deg(p(x))+\deg(q(x))\)
- the units of \(R[x]\) are just the units of \(R\)
- \(R[x]\) is an integral domain
Matrix Rings
This has been formalized in the submitted-but-not-yet-accepted
matgrp_0 article.
Group Rings
Missing from Mizar
§7.3 Ring Homomorphisms and Quotient Rings
- Definition. (
quofield) Let \(R\) and \(S\) be rings. A Ring Homomorphism is a map \(\varphi\colon R\to S\) such that- additive: \(\varphi(a+b)=\varphi(a)+\varphi(b)\) for all \(a,b\in R\)
- multiplicative: \(\varphi(ab)=\varphi(a)\varphi(b)\) for all \(a,b\in R\)
- unity-preserving: \(\varphi(1)=1\)
- Implicit Definition. Let \(R\) and \(S\) be rings. We say \(S\) is
$R$-homomorphic if there exists a ring homomorphism
\(\varphi\colon R\to S\). (
ring_2:def 4) This is needed because types in Mizar can't be empty, and we want to defineHomomorphism of R,Sring_2. - Definition. Let \(R\) and \(S\) be rings. The Kernel of a
ring homomorphism \(\varphi\), denoted \(\ker(\varphi)\), is the set of
elements of \(R\) that map to \(0\) in \(S\). (
vectsp10:def 9) - Definition. Let \(R\) and \(S\) be rings. We say a ring homomorphism
\(\varphi\colon R\to S\) is a Isomorphism if it is
bijective (
RingIsomorphismis a synonym forisomorphismdefined inmod_4:def 12) - Implicit Definition. Let \(R\) and \(S\) be rings. We say \(R\) is
Isomorphic to \(S\) if there exists an isomorphism \(R\to S\)
(
quofield:def 23). This is usually denoted \(R\cong S\). - Example. Let \(R\) be a ring, let \(c_{1},\dots,c_{n}\in R\) be elements of \(R\). The
function \(e_{c_{1},\dots,c_{n}}\colon R[x_{1},\dots,x_{n}]\to R\),
defined by sending \(x_{i}\mapsto c_{i}\) then evaluating the
polynomial, is a ring homomorphism. (
polynom2:def 5) - Proposition 5. Let \(R\) and \(S\) be rings, let \(\varphi\colon R\to S\)
be a ring morphism. Then
- The image of \(\varphi\) is a subring of \(S\) (
ring_2:def 6) - The kernel of \(\varphi\) is a subring of \(R\).
- The image of \(\varphi\) is a subring of \(S\) (
- Definition. Let \(R\) be a ring. Let \(I\subset R\) be a subset of
\(R\), and let \(r\in R\) be an element of \(R\).
- (
ideal_1:def 18) \(rI = \{ra\mid a\in I\}\) and - Missing from Mizar \(Ir = \{ar\mid a\in I\}\)
- (
- A subset \(I\) of \(R\) is a Left Ideal of \(R\) (
ideal_1) if\(I\) is a subring of \(R\) and\(I\) isadd-closed: for any \(x,y\in I\), \(x+y\in I\)left-ideal: \(I\) is closed under left multiplication by elements from \(R\), i.e., \(rI\subset I\) for all \(r\in R\)
- A subset \(I\) of \(R\) is a Right Ideal of \(R\) (
ideal_1) if- \(I\) is
add-closed: for any \(x,y\in I\), \(x+y\in I\) right-ideal: \(Ir\subset I\) for all \(r\in R\)
- \(I\) is
- A subset \(I\) of \(R\) is a Two-Sided Ideal (
ideal_1) if it is both a left and right ideal of \(R\)
- Notation: We write \(I\trianglelefteq R\) is \(I\) is a two-sided ideal of \(R\), and \(I\triangleleft R\) if \(I\neq R\) is a proper two-sided ideal.
- Proposition 6. Let \(R\) be a (possibly nonassociative) ring, let
\(I\) be an ideal of \(R\). Then \(R/I\) is a ring consisting of \(a+I\) for
\(a\in R\), addition is defined using \((a+I)+(b+I)=(a+b)+I\),
multiplication is defined by \((a+I)(b+I)=ab+I\). (
ring_1:def 6) - Theorem 7.
- First isomorphism theorem (
ring_2:15): If \(\varphi\colon R\to S\) is a ring morphism, then \(R/\ker(\varphi)\cong\varphi(R)\) - If \(I\) is any ideal of \(R\), then the map \(R\to R/I\) defined by
\(r\mapsto r+I\) is a surjective ring morphism with kernel \(I\)
which Mizar calls the Canonical Morphism of \(I\)
(
ring_2:def 5). Thus every ideal is the kernel of a morphism, and vice-versa.
- First isomorphism theorem (
- Examples. Let \(R\) be a ring.
The subrings \(R\) and \(\{0\}\) are ideals (
ideal_1:10).An ideal \(I\) is proper if \(I\neq R\).
The ideal \(\{0\}\) is called the Trivial Ideal and denoted by \(0\) (
ideal_1:def 4).- The sets \(n\mathbb{Z}\) are ideals of \(\mathbb{Z}\) for any \(n\in\mathbb{Z}\) (although this is Missing from Mizar, its significance for Dummit and Foote is only to construct \(\mathbb{Z}/n\mathbb{Z}\) which is done explicitly in Mizar)
- Let \(R=\mathbb{Z}[x]\) be the ring of polynomials in \(x\) with integer coefficients. Let \(I\) be the set of polynomials whose terms are of degree at least 2 (i.e., they all look like \(a_{n}x^{n}+\dots+a_{2}x^{2}\)). Then \(I\) is an ideal of \(R\). Missing from Mizar
- Theorem 8. Let \(R\) be a ring.
- Second isomorphism theorem for rings: Let \(A\) be a subring of \(R\), let \(I\) be an ideal of \(R\). Then \(A+I\) is a subring of \(R\), \(A\cap I\) is an ideal of \(A\), and \((A+I)/I\cong A/(A\cap I)\)
- Third isomorphism theorem for rings: Let \(I\) and \(J\) be ideals of \(R\) with \(I\subset J\). Then \(J/I\) is an ideal of \(R/I\) and \((R/I)/(J/I)\cong R/J\).
- Fourth isomorphism theorem for rings: Let \(I\) be an ideal of \(R\). The correspondence \(A\leftrightarrow A/I\) is an inclusion-preserving bijection between the set of subrings \(A\) of \(R\) which contain \(I\) and the set of subrings of \(R/I\). Furthermore, \(A\) (a subring containing \(I\)) is an ideal of \(R\) if and only if \(A/I\) is an ideal of \(R/I\).
- Definition. Let \(I\) and \(J\) be ideals of \(R\).
- The Sum of \(I\) and \(J\) is \(I+J=\{a+b\mid a\in I,b\in J\}\)
(
ideal_1:def 19) - The Product of \(I\) and \(J\) is the set \(IJ\) of all finite
sums of terms of the form \(ab\) with \(a\in I\) and \(b\in J\)
(
ideal_1:def 21) - For any \(n\geq1\), define the n-th power of \(I\),
denoted by \(I^{n}\), to be the set consisting of all finite sums
of elements of the form \(a_{1}a_{2}(\dots)a_{n}\) for \(a_{i}\in I\)
for all \(i\). That is to say \(I^{1}=I\) and \(I^{n+1}=II^{n}\)
(
ideal_2:def 2)
- The Sum of \(I\) and \(J\) is \(I+J=\{a+b\mid a\in I,b\in J\}\)
(
§7.4 Properties of Ideals
- Definition. Let \(A\) be any subset of the ring \(R\).
- The Ideal Generated by \(A\) is the smallest ideal \((A)\)
of \(R\) containing \(A\) (
ideal_1:def 14); this also generalizes to "left ideals generated by \(A\)" and "right ideals generated by \(A\)" - Let \(AR=\{\sum^{n}_{i=1}a_{i}r_{i}\mid n\in\mathbb{N},a_{i}\in A,r_{i}\in R\}\) and \(RA=\{\sum^{n}_{i=1}r_{i}a_{i}\mid n\in\mathbb{N},a_{i}\in A,r_{i}\in R\}\)
- A Principal Ideal is the ideal generated by a single
element
(
ideal_1:def 27) - An ideal generated by a finite set is called a
Finitely-Generated Ideal
(
ideal_1:def 25)
- The Ideal Generated by \(A\) is the smallest ideal \((A)\)
of \(R\) containing \(A\) (
- Proposition 9. Let \(I\) be an ideal of \(R\).
- \(I=R\) if and only if \(I\) contains a unit (
ideal_1:20) - Assume \(R\) is commutative. Then \(R\) is a field if and only if its
only ideals are \(0\) and \(R\). (
ideal_1:22)
- \(I=R\) if and only if \(I\) contains a unit (
- Corollary 10. If \(R\) is a field, then any nonzero ring homomorphism from \(R\) into another ring is an injection. Missing from Mizar
- Definition. An ideal \(M\) in an arbitrary ring \(S\) is called a
Maximal Ideal if \(M\neq S\) and the only ideals containing
\(M\) are \(M\) and \(S\). (
ring_1:def 4) - Proposition 11. In a unital ring, every proper ideal is contained
in a maximal ideal. (
field_11:def 8) - Proposition 12. Let \(R\) be a commutative ring. The ideal \(M\) is a
maximal ideal if and only if the quotient ring \(R/M\) is a field.
(
ring_1:21) - Definition. Let \(R\) be a commutative ring. An ideal \(P\) of \(R\) is
called a Prime Ideal if \(P\neq R\) and whenever \(ab\in P\)
then either \(a\in P\) or \(b\in P\)
(
ring_1:def 2) - Proposition 13. Let \(R\) be a commutative ring. Then \(P\) is a prime
ideal in \(R\) if and only if the quotient ring \(R/P\) is an integral domain
(
ring_1:18) - Corollary 14. Let \(R\) be a commutative ring. Then every maximal
ideal of \(R\) is a prime ideal. This is a registered fact in
ring_1.
§7.5 Rings of Fractions
- Theorem 15. Let \(R\) be a commutative ring, let \(D\) be any nonempty
subset of \(R\) which does not contain \(0\notin D\), and \(D\) does not
contain any zero-divisors, and \(D\) is closed under
multiplication. Then there exists a commutative unital ring \(Q\) such
that \(Q\) contains \(R\) as a subring and every element of \(D\) is a
unit in \(Q\). The ring \(Q\) also has the following properties:
- Every element of \(Q\) is of the form \(rd^{-1}\) for some \(r\in R\) and \(d\in D\).
- (Uniqueness of \(Q\)) The ring \(Q\) is the "smallest" ring containing \(R\) in which all elements of \(D\) are units, in the sense that for any commutative unital ring \(S\) and for any injective ring morphism \(\varphi\colon R\to S\) such that \(\varphi(d)\) is a unit for all \(d\in D\), then there exists an injective ring morphism \(\Phi\colon Q\to S\) such that \(\Phi|_{R}=\varphi\)
- Definition. Let \(R\), \(D\), \(Q\) be as in Theorem 15.
- The ring \(Q\) is called the Ring of Fractions of \(D\)
with respect to \(R\) and is denoted \(Q^{-1}R\)
(
ringfrac:def 18). Mizar uses the notationQ ~ Rinstead of \(Q^{-1}R\). - If \(R\) is an integral domain and \(D=R\setminus\{0\}\), then \(Q\) is
called the Field of Fractions or Quotient Field
of \(R\); this is defined for any nondegenerate unital commutative
ring, where it's called the Total Quotient Ring
(
ringfrac:def 24).
- The ring \(Q\) is called the Ring of Fractions of \(D\)
with respect to \(R\) and is denoted \(Q^{-1}R\)
(
- Corollary 16. Let \(R\) be an integral domain, let \(Q\) be the field
of fractions of \(R\). If a field \(F\) contains a subring \(R'\)
isomorphic to \(R\), then the subfield of \(F\) generated by \(R'\) is
isomorphic to \(Q\). (This is the universal property of the ring of
fractions, and is formalized in
ringfrac:def 23)
Ch. 8 Euclidean Domains, Principal Ideal Domains, Unique Factorization Domains
§8.1 Euclidean Domains
Definition. The integral domain \(R\) is said to be a Euclidean Domain if there exists a function \(N\colon R\to\mathbb{N}_{0}\) with \(N(0)=0\) and for any \(a,b\in R\) with \(b\neq0\) there exists elements \(q,r\in R\) such that \(a=qb+r\) and \(r=0\) or \(N(r)\lt N(b)\) (
int_3:def 8)The function \(N\) is called the Degree Function in Mizar (
int_3:def 9).- Proposition 1. Every ideal in a Euclidean domain is principal;
i.e., if \(I\) is any nonzero ideal in the Euclidean domain \(R\), then
there exists a \(d\in R\) such that \(I=(d)\).
(
ideal_1:94) - Definition. Let \(R\) be a commutative ring, let \(a,b\in R\) with \(b\neq0\).
- We say \(a\) is a multiple of \(b\) if there exists an
element \(x\in R\) with \(a=bx\). In this case, we also say \(b\)
Divides \(a\)
(
gcd_1:def 1) - A Greatest Common Divisor of \(a\) and \(b\) is a nonzero
element \(d\) such that \(d\) divides \(a\), and \(d\) divides \(b\), and
for any other \(d'\in R\) if \(d'\) divides both \(a\) and \(b\) then
\(d'\) divides \(d\).
(This is generalized in
gcd_1:def 12and especiallyring_4:def 10)
- We say \(a\) is a multiple of \(b\) if there exists an
element \(x\in R\) with \(a=bx\). In this case, we also say \(b\)
Divides \(a\)
(
- Proposition 2. If \(a\) and \(b\) are elements of a commutative ring \(R\) such that the ideal generated by \(a\) and \(b\) is a principal ideal \((d)\), then \(d\) is a greatest common divisor of \(a\) and \(b\) Missing from Mizar
- Proposition 3. Let \(R\) be an integral domain. If two elements \(d\)
and \(d'\) of \(R\) generate the same principal ideal \((d)=(d')\), then
\(d'=ud\) for some unit \(u\in R\).
(
ring_2:21) - Theorem 4. Let \(R\) be a Euclidian domain, let \(a\) and \(b\) be
nonzero elements of \(R\). Let \(d=r_{n}\) be the last nonzero remainder
in the Euclidian algorithm. Then
- \(d\) is a greatest common divisor of \(a\) and \(b\) (Missing from Mizar), and
- The principal ideal \((d)\) is the ideal generated by \(a\) and \(b\)
(Missing from Mizar but seems to follow from
ring_2:18)
- Proposition 5. Let \(R\) be an integral domain that is not a field. If \(R\) is a Euclidian domain, then there are universal side divisors in \(R\).
§8.2 Principal Ideal Domains
- Definition. A Principal Ideal Domain (PID) is an
integral domain in which every ideal is principal (
ideal_1:def 28andring_2). - Proposition 6. Let \(R\) be a principal ideal domain, let \(a\) and
\(b\) be nonzero elements of \(R\). Let \(d\) be a generator for the ideal
generated by \(a\) and \(b\). Then
- \(d\) is a greatest common divisor of \(a\) and \(b\) Missing from Mizar
- \(d\) can be written as an $R$-linear combination of \(a\) and \(b\), i.e., there exists \(x\) and \(y\) in \(R\) with \(d=ax+by\) Missing from Mizar
- \(d\) is unique up to multiplication by a unit of \(R\) Missing from Mizar
- Proposition 7. Every nonzero prime ideal in a principal ideal
domain is a maximal ideal.
(
ring_2:26) - Corollary 8. If \(R\) is any commutative ring such that the polynomial ring \(R[x]\) is a principal ideal domain (or a Euclidian domain), then \(R\) is necessarily a field. Missing from Mizar
- Definition. Define \(N\) to be a Dedekind-Hasse norm if \(N\) is a positive norm and for every nonzero \(a,b\in R\) either \(a\) is an element of the ideal \((b)\) or there is a nonzero element in the ideal \((a,b)\) of norm strictly less than the norm of \((b)\) (i.e., either \(b\) divides \(a\) in \(R\), or there exists \(s,t\in R\) with \(0\lt N(sa-tb)\lt N(b)\)) Missing from Mizar
- Proposition 9. The integral domain \(R\) is a PID if and only if \(R\) has a Dedekind-Hasse norm. Missing from Mizar
§8.3 Unique Factorization Domains
- Definition. Let \(R\) be an integral domain
- Suppose \(r\in R\) is nonzero and is not a unit. Then \(r\) is called
Irreducible if whenever \(r=ab\) with \(a,b\in R\), then
either \(a\) is a unit in \(R\) or \(b\) is a unit in \(R\) or both;
otherwise \(r\) is called Reducible
(
ring_2:def 9) - The nonzero element \(p\in R\) is called Prime in \(R\) if
the ideal \((p)\) generated by \(p\) is a prime ideal. In other words,
a nonzero element \(p\) is prime if it is not a unit and whenever
\(p\mid ab\) either \(p\mid a\) or \(p\mid b\)
(
ring_2:def 8) - Two elements \(a,b\in R\) differing by a unit are said to be
Associated if \(a=ub\) for some unit \(u\in R\)
(
gcd_1:def 3)
- Suppose \(r\in R\) is nonzero and is not a unit. Then \(r\) is called
Irreducible if whenever \(r=ab\) with \(a,b\in R\), then
either \(a\) is a unit in \(R\) or \(b\) is a unit in \(R\) or both;
otherwise \(r\) is called Reducible
(
- Proposition 10. In an integral domain a prime element is always
irreducible. (Registered in
ring_2) - Proposition 11. In a PID, a nonzero element is prime iff it is
irreducible (
ring_2:26) - Definition. A Unique Factorization Domain (UFD,
ring_2:def 15) is an integral domain \(R\) in which every nonzero element \(r\in R\) which is not a unit has the following two properties- \(r\) can be written as a finite product of irreducibles \(p_{i}\) of \(R\), i.e., \(r=p_{1}p_{2}\dots p_{n}\); and
- the decomposition in (1) is unique up to associates
- Proposition 12. In a UFD, a nonzero element is prime iff it is irreducible.
- Theorem 14. Every principal ideal domain is a unique factorization
domain. (This is a registered fact in
ring_2)
Ch. 9 Polynomial Rings
§9.1 Definitions and Basic Properties
- Implicit definition. The Degree of a polynomial
\(p(x)=a_{n}x^{n}+\dots+a_{1}x+a_{0}\) where \(a_{n}\neq0\) is
\(\deg(p)=n\)
(
hurwitz:def 2) - Proposition 1. Let \(R\) be an integral domain
- \(\deg(p(x)q(x))=\deg(p(x))+\deg(q(x))\)
(
hurwitz:23) - the units of \(R[x]\) is just the units of \(R\) Missing from Mizar
- \(R[x]\) is an integral domain Missing from Mizar
- \(\deg(p(x)q(x))=\deg(p(x))+\deg(q(x))\)
(
- Proposition 2. Let \(I\) be an ideal of the ring \(R\), let \((I)=I[x]\) denote the ideal of \(R[x]\) generated by \(I\) (the set of polynomials with coefficients in \(I\)). Then \(R[x]/(I)\cong(R/I)[x]\). In particular, if \(I\) is a prime ideal of \(R\) then \((I)\) is a prime ideal of \(R[x]\). Missing from Mizar
- Definition. The Polynomial Ring in the variables \(x_{1}\) denoted \(R[x_{1},\dots,x_{n}]\)
is defined inductively by \(R[x_{1},\dots,x_{n-1},x_{n}]=R[x_{1},\dots,x_{n-1}][x_{n}]\)
- Although this is easier for informal mathematics, Mizar formalizes a notion of "bags" to describe monomials as a function from the set of unknowns to the exponent for each unknown. A polynomial is then a formal $R$-linear combination of bags.
§9.2 Polynomial Rings over Fields (Part 1)
- Theorem 3. Let \(F\) be a field. The polynmial ring \(F[x]\) is a Euclidian domain. Specifically, if \(a(x)\) and \(b(x)\) are two polynomials in \(F[x]\) with \(b(x)\) nonzero, then there are unique \(q(x)\) and \(r(x)\) in \(F[x]\) such that \(a(x)=q(x)b(x)+r(x)\) with \(r(x)=0\) or \(\deg(r(x))\lt\deg(b(x))\). Missing from Mizar
- Corollary 4. If \(F\) is a field, then \(F[x]\) is a PID and a UFD. Missing from Mizar
§9.3 Polynomial Rings that are Unique Factorization Domains
- Proposition 5 (Gauss's Lemma). Let \(R\) be a UFD with field of fractions \(F\) and let \(p(x)\in R[x]\). If \(p(x)\) is reducible in \(F[x]\) then \(p(x)\) is reducible in \(R[x]\). More precisely, if \(p(x)=A(x)B(x)\) for some nonconstant polynomials \(A(x),B(x)\in F[x]\), then there are nonzero elements \(r,s\in F\) such that \(rA(x)=a(x)\) and \(sB(x)=b(x)\) both lie in \(R[x]\) and \(p(x)=a(x)b(x)\) is a factorization in \(R[x]\). Missing from Mizar
- Corollary 6. Let \(R\) be a UFD, let \(F\) be its field of fractions, and let \(p(x)\in R[x]\). Suppose the greatest common divisor of the coefficients of \(p(x)\) is 1. Then \(p(x)\) is irreducible in \(R[x]\) if and only if it is irreducible in \(F[x]\). In particular, if \(p(x)\) is a monic polynomial that is irreducible in \(R[x]\), then \(p(x)\) is irreducible in \(F[x]\). Missing from Mizar
- Theorem 7. \(R\) is a UFD if and only if \(R[x]\) is a UFD Missing from Mizar
- Corollary 8. If \(R\) is a UFD, then a polynomial ring in an arbitrary number of variables with coefficients in \(R\) is also a UFD. Missing from Mizar
§9.4 Irreducibility Criteria
- Proposition 9. Let \(F\) be a field, let \(p(x)\in F[x]\). Then \(p(x)\) has a factor of degree one if and only if \(p(x)\) has a root in \(F\), i.e., there is an \(\alpha\in F\) with \(p(\alpha)=0\) Missing from Mizar
- Proposition 10. A polynomial of degree 2 or 3 over a field \(F\) is
reducible if and only if it has a root in \(F\)
(
field_10:5andfield_10:6) - Proposition 12. Let \(I\) be a proper ideal in the integral domain \(R\) and let \(p(x)\) be a nonconstant monic polynomial in \(R[x]\). If the image of \(p(x)\) in \((R/I)[x]\) cannot be factored in \((R/I)[x]\) into two polynomials of smaller degree, then \(p(x)\) is irreducible in \(R[x]\). Missing from Mizar
- Proposition 13 (Eisenstein's criterion). Let \(P\) be a prime ideal of the integral domain \(R\) and let \(f(x)=x^{n}+a_{n-1}x^{n-1}+\dots+a_{1}x+a_{0}\) be a polynomial in \(R[x]\) (where \(n\geq1\)). Suppose \(a_{n-1}\), …, \(a_{1}\), \(a_{0}\) are elements of \(P\) and suppose \(a_{0}\notin P^{2}\). Then \(f(x)\) is irreducible in \(R[x]\). Missing from Mizar