Among integral domains, fields \(\subsetneq\) Euclidean domains \(\subsetneq\) principal ideal domains \(\subsetneq\) unique factorization domains \(\subsetneq\) integral domains. In an integral domain every prime element is irreducible, and in a unique factorization domain every irreducible element is prime. In \(\ZZ[\sqrt{-5}]\) the element \(2\) is irreducible and not prime: it divides \((1+\sqrt{-5})(1-\sqrt{-5})=6\) and neither factor. Gauss’s lemma: a primitive polynomial of positive degree in \(\ZZ[x]\) is irreducible in \(\ZZ[x]\) if and only if it is irreducible in \(\QQ[x]\). Eisenstein’s criterion: if a prime \(p\) divides every coefficient of \(f\in\ZZ[x]\) except the leading one, does not divide the leading coefficient, and \(p^2\) does not divide the constant term, then \(f\) is irreducible in \(\QQ[x]\). If \(p\) does not divide the leading coefficient of \(f\in\ZZ[x]\) and the reduction of \(f\) in \(\FF_p[x]\) is irreducible, then \(f\) is irreducible in \(\QQ[x]\).
Divisibility of elements (Tag D-AVBIP)
Let \(R\) be a commutative ring and let \(q, r\in R\). The element \(r\) is divisible by \(q\), written \(q\divides r\), if there exists \(c \in R\) such that \(r = qc\).
Associate elements (Tag D-R4H6F)
Let \(R\) be a commutative ring. Elements \(a, b\in R\) are associates if there exists a unit \(u\in R^{\times}\) such that \(a = ub\).
Let \(R\) be an integral domain and \(a,b\in R\). Then \(a\) and \(b\) are associates if and only if \(a\divides b\) and \(b\divides a\).
If \(a=ub\) with \(u\) a unit, then \(b\divides a\), and \(b=u^{-1}a\) gives \(a\divides b\). Conversely, suppose \(b=ac\) and \(a=bd\) with \(c,d\in R\), in the sense of divisibility. If \(a=0\), then \(b=0\) and \(a=1\cdot b\). If \(a\neq 0\), then \(a=acd\) gives \(a(1-cd)=0\), so \(cd=1\) because \(R\) is an integral domain; thus \(d\) is a unit and \(a=db\).
Irreducible element (Tag D-TO3IY)
Let \(R\) be a commutative ring. A nonzero element \(r\in R\) that is not a unit is irreducible if for all \(a,b\in R\), \[ r=ab \implies a \in R^{\times} \text{ or } b\in R^{\times}. \]
Prime element (Tag D-AWSKI)
Let \(R\) be a commutative ring. A nonzero element \(p\in R\) that is not a unit is prime if for all \(a,b \in R\), \[ p\divides ab \implies p\divides a \text{ or } p\divides b, \] where \(\divides\) denotes divisibility.
Prime and irreducible agree in a UFD (Tag PR-LX7GH)
Let \(R\) be an integral domain.
Every prime element of \(R\) is irreducible.
If \(R\) is a unique factorization domain, a nonzero nonunit \(r \in R\) is prime if and only if \(r\) is irreducible.
In the integral domain \(\ZZ[\sqrt{-5}]\), the element \(2\) is irreducible but not prime: \(2\) divides \(6 = (1+\sqrt{-5})(1-\sqrt{-5})\) but divides neither factor.
Unique factorization domain (Tag D-INULL)
An integral domain \(R\) is a unique factorization domain if every nonzero \(r\in R\) admits a decomposition \[ r = u \prod_{i=1}^n p_i \] with \(n\geq 0\), \(u\in R^{\times}\) a unit, and \(p_1,\ldots,p_n\) irreducible, and this decomposition is unique up to order and associates: if also \(r = v\prod_{j=1}^m q_j\) with \(v\in R^{\times}\) and each \(q_j\) irreducible, then \(m=n\) and there is a permutation \(\sigma\) of \(\theset{1,\ldots,n}\) such that \(p_i\) and \(q_{\sigma(i)}\) are associates for every \(i\).
Unique factorization domain (Tag FD-25SUQ)
An integral domain \(R\) is a unique factorization domain if every nonzero \(x\in R\) can be written as \[ x=u\prod_{i=1}^n p_i \] with \(n\geq 0\), \(u\in R^\times\) a unit, and \(p_1,\ldots,p_n\) irreducible, and this decomposition is unique up to units and order: if also \(x=v\prod_{j=1}^m q_j\) with \(v\in R^\times\) and each \(q_j\) irreducible, then \(m=n\) and there is a permutation \(\sigma\) of \(\theset{1,\ldots,n}\) such that \(p_i\) and \(q_{\sigma(i)}\) are associates for every \(i\).
Euclidean domain (Tag D-NKRGN)
An integral domain \(R\) is Euclidean if there exists a function \(d\colon R\to \ZZ_{\geq 0}\) such that for all \(x,y\in R\) with \(y\neq 0\) there exist \(q,r\in R\) with \(x = qy + r\) and either \(r=0\) or \(d(r) < d(y)\). Such a function \(d\) is a Euclidean function on \(R\).
Euclidean domain (Tag FD-GXXBV)
An integral domain \(R\) is a Euclidean domain if there exists a function \(f\colon R\setminus\theset{0} \to \ZZ^{\geq 0}\), a Euclidean function, such that for all \(a\in R\) and all nonzero \(b\in R\) there exist \(q, r\in R\) with \(a = bq + r\) and either \(r=0\) or \(f(r) < f(b)\).
Principal ideal domain (Tag D-HTIL5)
An integral domain \(R\) is a principal ideal domain if every ideal of \(R\) is principal.
Inclusions among fields, Euclidean domains, PIDs, UFDs, and integral domains (Tag PR-LLBRB)
Every field is a Euclidean domain, every Euclidean domain is a principal ideal domain, every principal ideal domain is a unique factorization domain, every unique factorization domain is an integral domain, and every integral domain is a commutative ring: \[ \text{Fields} \subset \text{Euclidean domains} \subset \text{PIDs} \subset \text{UFDs} \subset \text{Integral domains} \subset \text{Commutative rings}. \]
Each inclusion is proper: \(\ZZ\) is a Euclidean domain that is not a field; \(\ZZ\big[\tfrac{1+\sqrt{-19}}{2}\big]\) is a PID that is not Euclidean; \(\ZZ[x]\) is a UFD that is not a PID; \(\ZZ[\sqrt{-5}]\) is an integral domain that is not a UFD; and \(\ZZ/4\ZZ\) is a commutative ring that is not an integral domain.
Implications among fields, PIDs, UFDs, and Noetherian rings (Tag PR-GEHJF)
Let \(R\) be a commutative ring with \(1\).
If \(R\) is a commutative division ring, then \(R\) is a field.
If \(R\) is a finite integral domain, then \(R\) is a field.
\(R\) is a field if and only if \(R[x]\) is a principal ideal domain.
\(R\) is a field if and only if \(R\) is a simple ring with \(1 \neq 0\).
\(R\) is a unique factorization domain if and only if \(R[x]\) is a unique factorization domain.
If \(R\) is a principal ideal domain, then \(R[x]\) is a unique factorization domain.
If \(R\) is a principal ideal domain, then \(R\) is Noetherian.
An ideal of an integral domain is free if and only if it is principal (Tag PR-ASW5L)
Let \(R\) be an integral domain and \(I \subseteq R\) an ideal. Then \(I\) is a free \(R\)-module if and only if \(I\) is a principal ideal.
Over a ring with zero divisors, a principal ideal need not be free. In \(\ZZ/6\ZZ\), the principal ideal \(\gens{2}\) is not a free \(\ZZ/6\ZZ\)-module, since \(3 \cdot 2 = 0\) with \(3 \neq 0\), whereas a nonzero element of a basis of a free module has zero annihilator.
Polynomial ring $K[x]$ (Tag FD-CI4NB)
Let \(K\) be a commutative ring. The polynomial ring in the variable \(x\) over \(K\) is \[ K[x] \coloneqq \theset{\sum_{i=0}^n a_ix^i \st n\in \ZZ^{\geq 0},\ a_0,\ldots,a_n\in K}, \] where two polynomials are equal when their coefficients of \(x^i\) agree for every \(i\geq0\), with missing coefficients taken to be \(0\). Addition is coefficientwise, and multiplication is \(\bigl(\sum_i a_ix^i\bigr)\bigl(\sum_j b_jx^j\bigr)=\sum_k\bigl(\sum_{i+j=k}a_ib_j\bigr)x^k\).
Field of rational functions $K(x)$ (Tag FD-24RNF)
Let \(K\) be a field. The field of rational functions in the variable \(x\) over \(K\) is the field of fractions of the polynomial ring \(K[x]\): \[ K(x)\coloneqq\theset{\frac{P(x)}{Q(x)} \st P,Q \in K[x],\ Q\neq 0}. \]
\(\CC(x)\) is the field of meromorphic functions on the Riemann sphere \(\hat\CC=\CC\cup\theset{\infty}\). Every rational function is meromorphic on \(\hat\CC\). Conversely, a meromorphic function \(h\) on the compact surface \(\hat\CC\) has finitely many poles; subtracting from \(h\) its principal parts at the poles in \(\CC\), which are rational functions, and the polynomial principal part at \(\infty\) leaves a holomorphic function on \(\hat\CC\), which is constant by Liouville’s theorem, so \(h\in\CC(x)\).
Reducible and irreducible polynomials (Tag D-BVMTZ)
Let \(k\) be a field and let \(f\in k[x]\) be nonconstant. The polynomial \(f\) is reducible if there exist \(g, h\in k[x]\) with \(\deg g, \deg h \geq 1\) and \(f = gh\). The polynomial \(f\) is irreducible if it is not reducible.
Irreducible polynomial over a field (Tag FD-GHY34)
Let \(K\) be a field. A polynomial \(p\in K[x]\) is irreducible if \(\deg p \geq 1\) and whenever \(p= qr\) with \(q,r\in K[x]\), \(q\) or \(r\) is constant.
Every polynomial of degree \(1\) is irreducible: if \(p=qr\), then \(\deg q+\deg r=1\), so one of \(q,r\) has degree \(0\).
Eisenstein's criterion (Tag FT-2P5VV)
Let \(f(x)=\sum_{i=0}^na_ix^i\in\ZZ[x]\) with \(n\ge1\), and let \(p\) be a prime such that
\(p\nmid a_n\),
\(p\divides a_i\) for \(0\le i\le n-1\), and
\(p^2\nmid a_0\).
Then \(f\) is irreducible in \(\QQ[x]\).
Suppose \(f=gh\) with \(g,h\in\QQ[x]\) nonconstant. By Gauss' lemma we may take \(g,h\in\ZZ[x]\) with \(\deg g,\deg h\ge1\). Reducing modulo \(p\) gives \(\overline g\,\overline h=\overline{a_n}x^n\) in \(\FF_p[x]\) with \(\overline{a_n}\ne0\). The leading coefficients of \(g\) and \(h\) multiply to \(a_n\), so neither is divisible by \(p\), and \(\deg\overline g=\deg g\ge1\), \(\deg\overline h=\deg h\ge1\). Since \(\FF_p[x]\) is a unique factorization domain, \(\overline g\) and \(\overline h\) are nonzero multiples of positive powers of \(x\), so \(p\) divides the constant terms of \(g\) and \(h\). Then \(p^2\) divides their product \(a_0\), a contradiction.
Eisenstein's criterion (Tag T-CF6S3)
Let \[ f(x) = \sum_{i=0}^n a_i x^i = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 \in \ZZ[x] \] with \(n\geq1\), and suppose there is a prime \(p\) such that
\(p\) divides \(a_i\) for \(0\leq i\leq n-1\),
\(p\) does not divide \(a_n\), and
\(p^2\) does not divide \(a_0\).
Then \(f\) is irreducible in \(\QQ[x]\). If moreover the coefficients of \(f\) have no common prime factor, then \(f\) is irreducible in \(\ZZ[x]\).
Suppose \(f\) is reducible in \(\QQ[x]\). By Gauss’s lemma \(f=gh\) with \(g,h\in\ZZ[x]\) of degrees \(r,s\geq1\). Reducing modulo \(p\) gives \(\bar a_nx^n=\bar g\bar h\) in \(\FF_p[x]\) with \(\bar a_n\neq0\). The leading coefficients of \(g\) and \(h\) multiply to \(a_n\), so they are not divisible by \(p\) and \(\deg\bar g=r\), \(\deg\bar h=s\). Since \(\FF_p[x]\) is a unique factorization domain, \(\bar g=bx^r\) and \(\bar h=cx^s\) with \(b,c\in\FF_p^\times\), so \(p\) divides both constant terms \(g(0)\) and \(h(0)\), and \(p^2\) divides \(g(0)h(0)=a_0\), a contradiction. If the coefficients of \(f\) have no common prime factor, a factorization of \(f\) in \(\ZZ[x]\) into nonunits has both factors of positive degree, which the first statement excludes.
Primitive polynomials (Tag D-4VC6X)
Let \(R\) be a unique factorization domain. A polynomial \(p\in R[x]\) is primitive if a greatest common divisor of its coefficients is a unit.
Gauss's lemma (Tag T-JEZZY)
Let \(R\) be a unique factorization domain with field of fractions \(F\), and let \(p\in R[x]\) be primitive. Then \(p\) is irreducible in \(R[x]\) if and only if \(p\) is irreducible in \(F[x]\).
More precisely, if \(p=AB\) with \(A,B\in F[x]\) nonconstant, then there exist \(r,s\in F^\times\) with \(rs=1\) such that \(rA,sB\in R[x]\), so \(p=(rA)(sB)\) is a factorization of \(p\) in \(R[x]\) into nonconstant polynomials.
Every monic polynomial in \(R[x]\) is primitive, since its leading coefficient \(1\) is a unit; so a monic \(p\in\ZZ[x]\) is irreducible in \(\ZZ[x]\) if and only if it is irreducible in \(\QQ[x]\).
Gauss' lemma (Tag FT-OXN3Y)
Let \(R\) be a unique factorization domain with field of fractions \(F\), and let \(p\in R[x]\). If \(p=gh\) with \(g,h\in F[x]\) and \(\deg g,\deg h\ge1\), then there are \(g',h'\in R[x]\) with \(\deg g'=\deg g\), \(\deg h'=\deg h\), and \(p=g'h'\). In particular, if \(p\) is reducible in \(F[x]\), then \(p\) is reducible in \(R[x]\).
If \(R\) is a unique factorization domain, then so is \(R[x]\); the proof uses Gauss’ lemma to reduce factorization in \(R[x]\) to factorization in \(R\) and in the principal ideal domain \(F[x]\). By induction, \(R[x_1,\ldots,x_n]\) is a unique factorization domain, and so is the polynomial ring over \(R\) in any set of variables, since each polynomial involves finitely many variables.
Discriminant of a polynomial (Tag D-W3DSO)
Let \(k\) be a field and \(f\in k[x]\) a monic polynomial of degree \(n\geq 1\), and write \(f=\prod_{i=1}^n (x-r_i)\) with roots \(r_1,\ldots,r_n\), listed with multiplicity, in a splitting field of \(f\) over \(k\). The discriminant of \(f\) is \[ \Delta_f \coloneqq \prod_{1\leq i<j\leq n} (r_i - r_j)^2 . \]
\(\Delta_f = 0\) if and only if \(f\) has a repeated root.
A product in a field vanishes if and only if one of its factors does, and \((r_i-r_j)^2=0\) if and only if \(r_i=r_j\) for some \(i<j\).
Let \(f\in \RR[x]\) be a monic cubic with \(\Delta_f\neq 0\). Then \(\Delta_f>0\) if and only if \(f\) has three distinct real roots, and \(\Delta_f<0\) if and only if \(f\) has one real root and a pair of complex conjugate non-real roots.
Since \(\Delta_f\neq 0\), the roots \(r_1,r_2,r_3\in\CC\) are distinct, and since \(f\) has real coefficients, its non-real roots occur in conjugate pairs. So either all three roots are real, or \(r_1\in\RR\) and \(r_{2,3}=a\pm bi\) with \(a,b\in\RR\) and \(b\neq 0\). In the first case each factor \(r_i-r_j\) is a nonzero real number, so \(\Delta_f>0\). In the second case \((r_1-r_2)(r_1-r_3)=\abs{r_1-r_2}^2>0\) and \((r_2-r_3)^2=(2bi)^2=-4b^2<0\), so \[ \Delta_f=\bigl((r_1-r_2)(r_1-r_3)\bigr)^2(r_2-r_3)^2<0 . \] The two cases are exhaustive and give opposite signs, which proves both equivalences.
Elementary symmetric functions (Tag D-FK47C)
Let \(R\) be a commutative ring and let \(x_1, \ldots, x_n\) be indeterminates over \(R\). The elementary symmetric functions in \(x_1, \ldots, x_n\) are \(e_0 \coloneqq 1\) and \[ e_k \coloneqq \sum_{1\leq i_1 < i_2 < \cdots < i_k\leq n} x_{i_1} x_{i_2}\cdots x_{i_k} \in R[x_1, \ldots, x_n], \qquad 1 \leq k \leq n. \]
In \(R[x_1,\ldots,x_n][t]\), \[ \prod_{i=1}^n (t - x_i) = \sum_{k=0}^n (-1)^k e_k t^{n-k}. \] Consequently, if a monic polynomial of degree \(n\) over a commutative ring factors as \(\prod_{i=1}^n(t-r_i)\), its coefficient of \(t^{n-k}\) is \((-1)^k e_k(r_1,\ldots,r_n)\).
Let \(R\) be a commutative ring. Every symmetric polynomial in \(R[x_1, \ldots, x_n]\) is equal to \(g(e_1, \ldots, e_n)\) for a unique polynomial \(g\in R[y_1,\ldots,y_n]\).
See [@DF04].
Descartes' rule of signs (Tag FF-HK72Z)
Let \(p\in\RR[x]\) be a nonzero polynomial, and let \(V(p)\) be the number of sign changes in the sequence of its nonzero coefficients, listed in order of degree. The number \(r^+\) of positive real roots of \(p\), counted with multiplicity, satisfies \[ r^+\le V(p)\qquad\text{and}\qquad V(p)-r^+\in2\ZZ. \] The negative real roots of \(p\) are the negatives of the positive real roots of \(p(-x)\), so their number \(r^-\), counted with multiplicity, satisfies \(r^-\le V(p(-x))\) and \(V(p(-x))-r^-\in2\ZZ\).
Consequently, if \(p(0)\ne0\) and \(\deg p=d\), then \(p\) has at least \(d-V(p)-V(p(-x))\) nonreal complex roots, counted with multiplicity.
Let \(f(x)=x^3+x^2-x-1\). Its coefficient signs are \(+,+,-,-\), so \(V(f)=1\) and \(f\) has exactly one positive root. The polynomial \(f(-x)=-x^3+x^2+x-1\) has signs \(-,+,+,-\), so \(V(f(-x))=2\) and \(f\) has \(2\) or \(0\) negative roots; in fact \(f=(x-1)(x+1)^2\) has the negative root \(-1\) with multiplicity \(2\).
Let \(g(x)=x^3-1\). Then \(V(g)=1\), so \(g\) has exactly one positive root, and \(g(-x)=-x^3-1\) has \(V(g(-x))=0\), so \(g\) has no negative root. Hence \(g\) has at least \(3-1-0=2\) nonreal roots, so exactly \(2\).
Dedekind domain (Tag D-WUGPG)
A Dedekind domain is an integral domain \(R\) in which every nonzero ideal \(I\) factors as a product \(I=\mfp_1\cdots\mfp_n\) of nonzero prime ideals, with \(n\geq 0\) and the empty product equal to \(R\), and this factorization is unique up to the order of the factors.
Factorization of $x^n - y^n$ (Tag FF-UC7SQ)
For \(n\ge1\), in \(\ZZ[x,y]\), \[ x^n-y^n=(x-y)\qty{x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1}}=(x-y)\sum_{k=0}^{n-1}x^{n-1-k}y^k. \]
This is FF-ED3CD over the commutative ring \(R=\ZZ[y]\) with \(a=y\).
Factorization of $x^n - a^n$ (Tag FF-ED3CD)
Let \(R\) be a commutative ring, let \(a\in R\), and let \(n\ge1\). In \(R[x]\), \[ x^n-a^n=(x-a)\sum_{k=0}^{n-1}a^kx^{n-1-k}. \]
Expanding, \[ (x-a)\sum_{k=0}^{n-1}a^kx^{n-1-k}=\sum_{k=0}^{n-1}a^kx^{n-k}-\sum_{k=1}^{n}a^kx^{n-k}, \] and the two sums cancel except for the terms \(x^n\) and \(-a^n\).
Factorization of $x^n + y^n$ for odd $n$ (Tag FF-2AKVH)
Let \(n\ge1\). If \(n\) is odd, then in \(\ZZ[x,y]\) \[ x^n+y^n=(x+y)\sum_{k=0}^{n-1}(-1)^kx^{n-1-k}y^k . \] If \(n\) is even, then \(x+y\) does not divide \(x^n+y^n\) in \(\ZZ[x,y]\).
For odd \(n\), \((-y)^n=-y^n\), so substituting \(a=-y\) in \(x^n-a^n=(x-a)\sum_{k=0}^{n-1}a^kx^{n-1-k}\) (FF-ED3CD) gives the factorization.
Since \(x+y\) is monic in \(x\) over \(\ZZ[y]\), division by \(x+y\) in \((\ZZ[y])[x]\) shows that \(x+y\) divides \(f\in\ZZ[x,y]\) if and only if \(f(-y,y)=0\). For even \(n\), \((-y)^n+y^n=2y^n\ne0\).
Factorization of $x^n + a^n$ for odd $n$ (Tag FF-HAMDC)
Let \(R\) be a commutative ring, let \(a\in R\), and let \(n\ge1\). If \(n\) is odd, then in \(R[x]\) \[ x^n+a^n=(x+a)\sum_{k=0}^{n-1}(-a)^kx^{n-1-k}. \] If \(n\) is even and \(2a^n\ne0\) in \(R\), for example if \(R\) is a field of characteristic other than \(2\) and \(a\ne0\), then \(x+a\) does not divide \(x^n+a^n\) in \(R[x]\).
For odd \(n\), \((-a)^n=-a^n\), so the factorization is FF-ED3CD applied to \(-a\). Since \(x+a\) is monic, division by \(x+a\) in \(R[x]\) leaves the constant remainder \(f(-a)\) for every \(f\in R[x]\). For even \(n\) the remainder of \(x^n+a^n\) is \((-a)^n+a^n=2a^n\ne0\).