Ideals and quotients

AbbreviationIdeal of a commutative ring#

An ideal of a commutative ring \(R\) is a nonempty subset of \(R\) that is closed under addition and under multiplication by arbitrary elements of \(R\). Ideals play for rings the part that normal subgroups play for groups: they are exactly the kernels of ring morphisms, and they are what one divides by to form quotient rings. They also carry the arithmetic of divisibility over from elements to sets, which is how factorization is studied in rings where elements fail to factor uniquely.

Definition

Let \(R\) be a commutative ring. An ideal of \(R\) is a subset \(I \subseteq R\) such that

  • \(0 \in I\);
  • if \(a, b \in I\), then \(a + b \in I\);
  • if \(b \in I\) and \(a \in R\), then \(ab \in I\).

Taking \(a = -1\) in the last condition shows that \(I\) is closed under negation, so an ideal is an additive subgroup of \(R\) that absorbs multiplication by elements of \(R\). Equivalently, an ideal is a submodule of \(R\) regarded as a module over itself. Every ring has the zero ideal \(\{0\}\) and the ideal \(R\) itself; an ideal different from \(R\) is called proper.

For a subset \(S \subseteq R\), the ideal generated by \(S\) is the intersection of all ideals containing \(S\); it consists of the finite sums \(a_1 s_1 + \dots + a_n s_n\) with \(a_i \in R\) and \(s_i \in S\), including the empty sum \(0\), and it is the smallest ideal containing \(S\): it lies inside an ideal \(I\) exactly when \(S \subseteq I\) (see generating sets). For instance, \(z\) lies in the ideal generated by \(x\) and \(y\) exactly when \(z = ax + by\) for some \(a, b \in R\). An ideal generated by a single element \(x\) is principal and is written \((x)\) or \(xR\); it consists of the multiples of \(x\), since \(z \in (x)\) exactly when \(x\) divides \(z\).

Examples
Properties

Ideals and units. An ideal is the whole ring as soon as it contains a unit: an ideal containing an invertible element is all of \(R\), and in particular \(I = R\) exactly when \(1 \in I\). For principal ideals, \((x) = R\) exactly when \(x\) is a unit.

Divisibility. Inclusion of principal ideals reverses divisibility: \((x) \subseteq (y)\) exactly when \(y\) divides \(x\). In an integral domain, \((x) = (y)\) exactly when \(x\) and \(y\) are associates, that is, when \(x = uy\) for a unit \(u\). Elements \(x\) and \(y\) are coprime, meaning \(ax + by = 1\) for some \(a, b \in R\), exactly when \((x) + (y) = R\).

Operations. The ideals of \(R\) are closed under several operations.

Kernels and quotients. For a ring morphism \(f : R \to S\), the kernel, the set of \(r \in R\) with \(f(r) = 0\), is an ideal of \(R\). More generally the preimage of an ideal of \(S\) is an ideal of \(R\), though the image of an ideal need not be an ideal: the image of \(\mathbb{Z}\) in the rational numbers is not one. Conversely, every ideal \(I\) is a kernel: in the quotient ring \(R/I\), two elements \(a\) and \(b\) have the same class exactly when \(a - b \in I\), and the kernel of the quotient map \(R \to R/I\) is \(I\).

Maximal ideals. By Zorn's lemma, every proper ideal is contained in a maximal ideal, an ideal that is proper and contained in no larger proper ideal.

Special kinds of ideals are defined through their quotients: for a commutative ring, \(I\) is a prime ideal exactly when \(R/I\) is an integral domain, and a maximal ideal exactly when \(R/I\) is a field. A ring whose ideals are all principal is a principal ideal ring; an integral domain with this property is a principal ideal domain, a class that includes every Euclidean domain. A ring in which every ideal is generated by finitely many elements is a Noetherian ring. The first isomorphism theorem and the Chinese remainder theorem are statements about ideals and their quotients, and fractional ideals extend the notion to certain submodules of the field of fractions of a domain, those that become ideals after multiplication by a single nonzero element of the domain.

In Mathlib

Mathlib defines an ideal for any semiring, not only for commutative rings, as a submodule of the ring viewed as a module over itself. In a noncommutative ring this is a left ideal; whether it is also closed under multiplication on the right is a separate property, which holds automatically for every ideal of a commutative ring, so results stated for two-sided ideals apply to commutative rings without extra hypotheses. Because ideals are submodules, the general facts about submodules apply to them directly. The ideals of a ring are ordered by inclusion and form a complete lattice, in which the zero ideal is the least element, the whole ring is the greatest, intersection is the meet and sum is the join; the ideal generated by a set is the span of that set. Images of ideals under a morphism are defined as the ideal generated by the image, which may be larger than the image itself.

Written for Trough, under CC BY-SA 4.0.

Sources
MathlibIdeal
Uses 2 · Used by 19
Statement usesRing · Set
Used byDirichlet class number formula · Dirichlet unit theorem · Finiteness of the class number · Coprime element · Principal ideal domain · Chinese remainder theorem · First isomorphism theorem for commutative rings · Fractional ideal · Maximal ideal · Prime ideal · Quotient ring · Hilbert basis theorem · Local ring · Noetherian ring · Nullstellensatz · Lasker–Noether theorem · Wedderburn–Artin theorem · Ax–Grothendieck theorem · Annihilating polynomials

InstanceQuotient ring#

A quotient ring is the ring obtained from a ring \(R\) by declaring the elements of a two-sided ideal \(I\) (any ideal, when \(R\) is commutative) to be zero. Its elements are the cosets \(a + I\), added and multiplied through representatives. Quotient rings are the ring-theoretic counterpart of quotient groups: they turn congruence modulo an ideal into equality, they describe the image of every ring morphism, and they are the standard way to build new rings, such as the integers modulo \(n\) or fields containing a root of a given polynomial.

Definition

Let \(R\) be a ring and \(I\) a two-sided ideal of \(R\), that is, an additive subgroup with \(ra \in I\) and \(ar \in I\) for all \(a \in I\) and \(r \in R\). As a set, \(R/I\) is the set of cosets \(a + I\) of \(I\) in the additive group of \(R\), and the operations are

\[(a + I) + (b + I) = (a + b) + I, \qquad (a + I)(b + I) = ab + I.\]

These do not depend on the chosen representatives. Addition is handled by the quotient group construction. For multiplication, if \(a - a' \in I\) and \(b - b' \in I\), then

\[ab - a'b' = a(b - b') + (a - a')b',\]

and both terms lie in \(I\), the first because \(I\) absorbs multiplication on the left and the second because it absorbs multiplication on the right. The ring axioms then pass from \(R\) to the cosets, so the quotient of \(R\) by a two-sided ideal is a ring, with zero \(0 + I = I\) and identity \(1 + I\). If \(R\) is commutative, every ideal is two-sided and \(R/I\) is again commutative.

The quotient map \(a \mapsto a + I\) is a ring morphism, and it is surjective. Two elements have the same image exactly when their difference lies in \(I\), so the kernel of the quotient map is \(I\). Together with the fact that kernels of ring morphisms are two-sided ideals, this shows that the two-sided ideals of \(R\) are exactly the kernels of ring morphisms out of \(R\).

Examples
Properties

Universal property. Let \(f\colon R \to S\) be a ring morphism with \(f(a) = 0\) for every \(a \in I\). Then \(f\) induces a ring morphism \(\bar f\colon R/I \to S\) with \(\bar f(a + I) = f(a)\) for all \(a \in R\), and since every element of \(R/I\) is a coset, a ring morphism out of \(R/I\) is determined by its composite with the quotient map. Applied with \(I = \ker f\), the induced map is also injective, since \(\bar f(a + I) = 0\) means \(f(a) = 0\), that is, \(a \in I\) and \(a + I = I\). So for a surjective ring morphism \(f\colon R \to S\), \(R/\ker f\) is isomorphic to \(S\); this is the first isomorphism theorem.

Prime and maximal ideals. Let \(R\) be commutative. Then \(R/I\) is an integral domain exactly when \(I\) is a prime ideal, and \(I\) is a maximal ideal exactly when \(R/I\) is a field (see prime ideal, maximal ideal and field). For instance \(\mathbb{Z}/p\mathbb{Z}\) is a field for every prime \(p\), while \(\mathbb{Z}/6\mathbb{Z}\) is not even an integral domain, since \(2 \cdot 3 = 0\) there. Since every field is an integral domain, every maximal ideal is prime.

Ideals of a quotient. Let \(R\) be commutative. For a surjective ring morphism \(f\colon R \to S\), the ideals of \(S\) correspond to the ideals of \(R\) containing \(\ker f\), by taking preimages, and the correspondence preserves inclusion. Applied to the quotient map, the ideals of \(R/I\) are the images of the ideals of \(R\) that contain \(I\). For ideals \(I \subseteq J\) of \(R\), the image \(J/I\) of \(J\) in \(R/I\) satisfies \((R/I)/(J/I) \cong R/J\).

Products. The Chinese remainder theorem states that for finitely many ideals \(I_1, \dots, I_k\) of a commutative ring with \(I_i + I_j = R\) whenever \(i \neq j\), the quotient by \(I_1 \cap \dots \cap I_k\) is isomorphic to the product of the quotients \(R/I_j\). For coprime integers \(m\) and \(n\) this gives \(\mathbb{Z}/mn\mathbb{Z} \cong \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}\).

In Mathlib

Mathlib defines an ideal of a possibly noncommutative ring as a left ideal, a subset containing \(0\) and closed under addition and under multiplication on the left, and treats being two-sided as a separate condition. The quotient \(R/I\) is first built as a set by reusing the quotient of a module by a submodule, viewing \(R\) as a module over itself; its ring structure is then supplied whenever the ideal is two-sided, which is automatic for commutative rings. No properness is assumed, so the quotient by the whole ring is allowed and is the zero ring. The field structure on the quotient of a commutative ring by a maximal ideal is provided as a definition to be invoked by hand rather than found automatically, so that in specific cases a computable inverse can be used instead. The integers modulo \(n\) are defined independently of quotients and identified with \(\mathbb{Z}/n\mathbb{Z}\) by an explicit isomorphism, while the ring obtained by adjoining a root of a polynomial is defined directly as a quotient.

Written for Trough, under CC BY-SA 4.0.

Sources
MathlibIdeal.Quotient.ring
Uses 3 · Used by 7
Statement usesRing · Ideal of a commutative ring · Quotient group
Used byExistence of algebraic closure of a field · Rupture field · Chinese remainder theorem · First isomorphism theorem for commutative rings · Prime ideal · Eisenstein's criterion · Weierstrass preparation theorem

ClassPrime ideal#

A prime ideal of a commutative ring is a proper ideal with the property that whenever a product of two elements lies in it, one of the two factors already does. The definition transfers a characteristic property of a prime number (Euclid's lemma: it divides a product only if it divides one of the factors) from elements to ideals. Prime ideals are exactly the ideals whose quotient ring is an integral domain, a nonzero commutative ring in which a product of nonzero elements is nonzero, and they are the points of the prime spectrum, the space on which algebraic geometry is built.

Definition

Let \(R\) be a commutative ring. An ideal \(P\) of \(R\) is prime if

  • \(P \neq R\), and
  • for all \(a, b \in R\), if \(ab \in P\) then \(a \in P\) or \(b \in P\).

Since an ideal equals \(R\) exactly when it contains \(1\), the first condition says that \(1 \notin P\). Read contrapositively, the second condition says that the product of two elements outside \(P\) is again outside \(P\). Together with \(1 \notin P\), this means that the complement \(R \setminus P\) is a multiplicatively closed set.

The leading characterization is through quotients: for an ideal \(P\) of \(R\), the quotient \(R/P\) is an integral domain exactly when \(P\) is prime. The condition \(P \neq R\) corresponds to the requirement that an integral domain have \(0 \neq 1\), and the product condition to the absence of zero divisors in \(R/P\).

Examples
Properties

Maximal ideals are prime. In a commutative ring, every maximal ideal is prime, as one also sees from the quotients: the quotient by a maximal ideal is a field, and every field is an integral domain. The converse fails: \(\{0\}\) is a prime ideal of \(\mathbb{Z}\) that is not maximal, and so is \((x)\) in \(\mathbb{Z}[x]\), which lies strictly inside the proper ideal \((2, x)\). Since every proper ideal is contained in a maximal ideal, every proper ideal is contained in a prime ideal, and in particular every nonzero commutative ring has at least one prime ideal.

Products and powers. By induction on the number of factors, a finite product of elements lies in a prime ideal \(P\) exactly when one of the factors does. In particular if \(r^n \in P\) for some \(n\), then \(r \in P\). The radical of an ideal \(I\) is the set of elements \(r\) with \(r^n \in I\) for some \(n \geq 1\), and \(I\) is a radical ideal when it equals its radical; so a prime ideal is a radical ideal: it equals its own radical. The same property holds one level up, for ideals: for ideals \(I\) and \(J\), the product \(IJ\) is contained in \(P\) exactly when \(I \subseteq P\) or \(J \subseteq P\), and likewise \(I \cap J \subseteq P\) exactly when \(I \subseteq P\) or \(J \subseteq P\).

Radicals. Prime ideals determine radicals: for every ideal \(I\) of \(R\), the radical of \(I\) is the intersection of all prime ideals containing \(I\).

Ring morphisms. If \(f \colon R \to S\) is a ring morphism of commutative rings and \(Q\) is a prime ideal of \(S\), then the preimage \(f^{-1}(Q)\) is a prime ideal of \(R\). In particular, taking \(Q = \{0\}\) in an integral domain, the kernel of a ring morphism into an integral domain is a prime ideal. Images of prime ideals, by contrast, need not be prime.

Existence by maximality. Let \(S\) be a multiplicatively closed subset of \(R\) and \(I\) an ideal disjoint from \(S\) such that every ideal strictly containing \(I\) meets \(S\). Then \(I\) is prime. Combined with Zorn's lemma, this shows that every ideal disjoint from \(S\) is contained in a prime ideal disjoint from \(S\); such an ideal exists only when \(0 \notin S\). Similarly, the intersection of a nonempty chain of prime ideals is prime, so, by Zorn's lemma applied to reverse inclusion, every prime ideal contains a minimal prime ideal, one that contains no strictly smaller prime ideal.

  • Localization. Because \(R \setminus P\) is multiplicatively closed, one can localize \(R\) at it, inverting every element outside \(P\). The resulting ring \(R_P\) is a local ring (see local ring), whose unique maximal ideal comes from \(P\).
  • Prime spectrum. The set of all prime ideals of \(R\), equipped with the Zariski topology, is the prime spectrum of \(R\).
  • Prime elements. A nonzero element \(p\) is prime exactly when the principal ideal \((p)\) is prime, so prime ideals generalize prime elements; in rings without unique factorization, such as many rings of integers of number fields (for instance \(\mathbb{Z}[\sqrt{-5}]\)), factorization of ideals into prime ideals takes the place of factorization of elements.
In Mathlib

Mathlib states primality as a property of an ideal that may be supplied automatically, so that facts such as the primality of maximal ideals, of the zero ideal of an integral domain and of preimages are found without being invoked by name. The definition is made for an arbitrary semiring rather than only for commutative rings. For a noncommutative ring an ideal in Mathlib means a left ideal, and the elementwise product condition is then stronger than the usual noncommutative notion of a prime ideal, which is phrased with products of ideals; most results about prime ideals, such as the description of the radical, assume commutativity. Prime ideals of a lattice in the sense of order theory are a separate notion with their own definition.

Written for Trough, under CC BY-SA 4.0.

Sources
MathlibIdeal.IsPrime
Uses 3 · Used by 6
Statement usesRing · Ideal of a commutative ring · Quotient ring
Used byFiniteness of the class number · Eisenstein's criterion · Nullstellensatz · Prime spectrum · Structure theorem · Ostrowski's theorem

ClassMaximal ideal#

A maximal ideal of a commutative ring is a proper ideal that is not contained in any larger proper ideal. Maximal ideals are exactly the ideals whose quotient ring is a field, which makes them the standard way to produce fields from rings: the integers modulo a prime, finite fields, and fields obtained by adjoining a root of a polynomial all arise this way. They correspond to the closed points of the prime spectrum, which makes them central in commutative algebra and algebraic geometry, and every nonzero commutative ring has at least one.

Definition

Let \(R\) be a commutative ring. An ideal \(M\) of \(R\) is maximal if \(M \neq R\) and for every ideal \(J\) with \(M \subseteq J \subseteq R\), either \(J = M\) or \(J = R\). In the language of order theory, the maximal ideals are the coatoms of the lattice of ideals, the elements just below the top element \(R\).

Since an ideal equals \(R\) exactly when it contains \(1\), the definition can be restated elementwise: \(M\) is maximal if and only if \(1 \notin M\) and every ideal that contains \(M\) together with some element outside \(M\) also contains \(1\).

The main characterization is through quotients: for a commutative ring \(R\), an ideal \(M\) is maximal if and only if \(R/M\) is a field. The reason is that the ideals of \(R/M\) correspond to the ideals of \(R\) containing \(M\), and that a nonzero commutative ring is a field exactly when its zero ideal is maximal, that is, when it has no ideals other than \(0\) and itself.

Examples
Properties

A commutative ring with exactly one maximal ideal is a local ring: a commutative ring with a unique maximal ideal is local, and in a local ring every maximal ideal equals the maximal ideal of the ring, the ideal of non-units. Examples are fields, the ring of rational numbers whose denominators are prime to a fixed prime \(p\), and rings of formal power series over a field; localization at a prime ideal produces local rings in general. In the prime spectrum of a commutative ring, the set of its prime ideals with the Zariski topology, the maximal ideals are exactly the closed points. The intersection of all maximal ideals is called the Jacobson radical of the ring.

In Mathlib

Mathlib defines maximality for ideals of an arbitrary semiring, as the property of being a coatom in the complete lattice of ideals. An ideal there is a submodule of the ring over itself, so for a noncommutative ring the notion is that of a maximal left ideal; for commutative rings it is the usual one. Several basic facts, including Krull's theorem and the comaximality of distinct maximal ideals, are proved at this generality, while the field structure on the quotient and the primality of maximal ideals assume commutativity. The field structure on \(R/M\) is supplied as a construction to be invoked explicitly rather than found automatically. The maximal ideal of a local ring is a separately named ideal, defined as the set of non-units, and is shown to be maximal.

Written for Trough, under CC BY-SA 4.0.

Sources
MathlibIdeal.IsMaximal
Uses 2 · Used by 4
Statement usesRing · Ideal of a commutative ring
Used byExistence of algebraic closure of a field · Local ring · Structure theorem · Ostrowski's theorem

DefinitionChinese remainder theorem#

The Chinese remainder theorem says that a finite system of congruences with pairwise coprime moduli always has a solution, and that the solution is unique modulo the product of the moduli. In the language of rings it says that the quotient of a commutative ring by the intersection of finitely many pairwise coprime ideals splits as a product of the separate quotients. It is the standard tool for reducing a question modulo a composite number to questions modulo prime powers, and it reappears in algebraic number theory, in polynomial interpolation and in computer arithmetic.

Statement

Let \(R\) be a commutative ring. Two ideals \(I\) and \(J\) of \(R\) are coprime when \(I + J = R\), that is, when \(i + j = 1\) for some \(i \in I\) and \(j \in J\). Let \(I_1, \dots, I_k\) be finitely many ideals that are pairwise coprime, meaning \(I_i\) and \(I_j\) are coprime whenever \(i \neq j\). Then the map sending the class of \(x\) modulo \(I_1 \cap \dots \cap I_k\) to the tuple of its classes modulo each \(I_i\) is an isomorphism of rings

\[R/(I_1 \cap \dots \cap I_k) \;\cong\; R/I_1 \times \dots \times R/I_k.\]

Equivalently, for any elements \(x_1, \dots, x_k\) of \(R\) there is an \(r \in R\) with \(r - x_i \in I_i\) for every \(i\), and \(r\) is unique modulo the intersection. For pairwise coprime ideals the intersection equals the product: \(I_1 \cdots I_k = I_1 \cap \dots \cap I_k\), so the left-hand side may also be written \(R/(I_1 \cdots I_k)\); for two ideals, \(R/IJ \cong R/I \times R/J\) when \(I\) and \(J\) are coprime.

The integers

In the ring of integers the ideals \(m\mathbb{Z}\) and \(n\mathbb{Z}\) are coprime exactly when \(\gcd(m, n) = 1\). For coprime natural numbers \(m\) and \(n\) the theorem gives \(\mathbb{Z}/mn\mathbb{Z} \cong \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}\), and more generally, for pairwise coprime natural numbers \(n_1, \dots, n_k\), \(\mathbb{Z}/(n_1 \cdots n_k)\mathbb{Z} \cong \prod_i \mathbb{Z}/n_i\mathbb{Z}\). In terms of congruences: if \(\gcd(m, n) = 1\), then for any natural numbers \(a\) and \(b\) there is a natural number \(k\) with \(k \equiv a \pmod m\) and \(k \equiv b \pmod n\), and when \(m, n \ge 1\) it can be chosen with \(k < mn\). Uniqueness says that, for natural numbers \(x\) and \(y\), \(x \equiv y\) modulo both \(m\) and \(n\) if and only if \(x \equiv y \pmod{mn}\).

Examples
  • The system \(x \equiv 2 \pmod 3\), \(x \equiv 3 \pmod 5\), \(x \equiv 2 \pmod 7\) has the solution \(x = 23\), and its solutions are exactly the integers congruent to \(23\) modulo \(105\).
  • Coprimality cannot be dropped. For the non-coprime ideals \(I_1 = I_2 = 2\mathbb{Z}\), the ring \(\mathbb{Z}/(2 \cdot 2)\mathbb{Z} = \mathbb{Z}/4\mathbb{Z}\) is not isomorphic to \(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}\): in the first, \(1 + 1 \neq 0\), while in the second every element \(y\) satisfies \(y + y = 0\). The intersection form fails as well, since \(2\mathbb{Z} \cap 2\mathbb{Z} = 2\mathbb{Z}\) and \(\mathbb{Z}/2\mathbb{Z}\) has two elements while the product has four.
  • If \(a_1, \dots, a_k\) are distinct elements of a field \(K\), the ideals generated by \(X - a_i\) in the polynomial ring \(K[X]\) are pairwise coprime, and \(K[X]/(X - a_i) \cong K\) by evaluation at \(a_i\). The theorem then says that any values \(b_1, \dots, b_k\) are taken at \(a_1, \dots, a_k\) by some polynomial, unique modulo \((X - a_1) \cdots (X - a_k)\). Choosing the representative of degree less than \(k\) gives the polynomial of Lagrange interpolation.
Proof

The map \(R \to \prod_i R/I_i\), \(x \mapsto (x + I_i)_i\), is a ring homomorphism whose kernel is \(I_1 \cap \dots \cap I_k\), so it induces an injective homomorphism from \(R/(I_1 \cap \dots \cap I_k)\); by the first isomorphism theorem it remains to show surjectivity. Fix \(i\). For each \(j \neq i\) write \(1 = u_j + v_j\) with \(u_j \in I_i\) and \(v_j \in I_j\). Multiplying these equations over all \(j \neq i\) gives \(1 = u + e_i\), where \(u \in I_i\) and \(e_i = \prod_{j \neq i} v_j\) lies in every \(I_j\) with \(j \neq i\). So \(e_i \equiv 1 \pmod{I_i}\) and \(e_i \equiv 0 \pmod{I_j}\) for \(j \neq i\). Given targets \(x_1, \dots, x_k\), the element \(r = x_1 e_1 + \dots + x_k e_k\) satisfies \(r \equiv x_i \pmod{I_i}\) for each \(i\).

Consequences
Generalizations

In a Dedekind domain, such as the ring of integers of a number field, every nonzero ideal \(I\) factors as a product of powers \(P_1^{e_1} \cdots P_s^{e_s}\) of distinct prime ideals, and the theorem gives \(R/I \cong \prod_i R/P_i^{e_i}\), which generalizes the prime power decomposition of \(\mathbb{Z}/n\mathbb{Z}\).

In Mathlib

Mathlib states the theorem for a commutative ring and a family of ideals indexed by an arbitrary finite type, with pairwise coprimality taken in the sense of coprime elements of the semiring of ideals, which is equivalent to \(I + J = R\). The isomorphism is the canonical map, not merely an abstract isomorphism, and its value on the class of \(x\) is the family of classes of \(x\). Half of the statement holds with no hypotheses: for any family of two-sided ideals in any ring, finite or infinite, the canonical map from the quotient by the intersection to the product of the quotients is injective; coprimality and finiteness are needed only for surjectivity. The integer versions are stated for natural-number moduli, and Mathlib's ring of integers modulo \(0\) is the integers themselves, so the two-modulus version is a true statement for every coprime pair, including the degenerate pair \(0\) and \(1\).

Written for Trough, under CC BY-SA 4.0.

Sources
MathlibIdeal.quotientInfRingEquivPiQuotient
Uses 5 · Used by 3
Statement usesIdeal of a commutative ring · Quotient ring · Coprime element · Product of rings
Proof usesFirst isomorphism theorem for commutative rings
Used byEuler's totient function (\(\varphi\)) · Structure of finitely generated abelian groups · Matiyasevič's theorem

DefinitionFractional ideal#

A fractional ideal of an integral domain \(R\) is an \(R\)-submodule of the field of fractions \(K\) of \(R\) whose denominators can be cleared by a single nonzero element of \(R\). Fractional ideals are to ideals what fractions are to integers: they allow ideals to be divided as well as multiplied. In the rings of algebraic number theory, the Dedekind domains, the nonzero fractional ideals form a group under multiplication, and the failure of unique factorization of elements is measured by the quotient of that group by its principal elements, the ideal class group.

Definition

Let \(R\) be an integral domain with field of fractions \(K\). An \(R\)-submodule \(I\) of \(K\) is a fractional ideal if there is a nonzero \(d \in R\) with

\[dI = \{\, dx : x \in I \,\} \subseteq R.\]

Since \(R \subseteq K\), every ideal of \(R\) is a fractional ideal (take \(d = 1\)), and distinct ideals remain distinct as fractional ideals. Conversely, a fractional ideal contained in \(R\) is an ideal of \(R\). Clearing denominators gives a description in terms of ideals alone: every fractional ideal has the form \(a^{-1}J\) for a nonzero \(a \in R\) and an ideal \(J\) of \(R\). Some authors require a fractional ideal to be nonzero; here, as in Mathlib, the zero submodule counts.

For \(x \in K\), the submodule \(Rx\) is a fractional ideal, the principal fractional ideal generated by \(x\).

Examples
Operations

Sums, products and intersections of submodules preserve the defining condition: the sum of two fractional ideals is a fractional ideal, so is their product, the submodule generated by all products \(xy\) with \(x \in I\) and \(y \in J\), and so is the intersection of a fractional ideal with any submodule. Under sum and product the fractional ideals form a commutative semiring whose unit element is \(R\).

Fractional ideals can also be divided. For fractional ideals \(I\) and \(J\) with \(J \neq 0\), the set \((I : J) = \{\, x \in K : xJ \subseteq I \,\}\) is again a fractional ideal. Taking \(R\) in place of \(I\) defines the inverse: for \(J \neq 0\), the inverse \(J^{-1} = (R : J)\) consists of the \(x \in K\) with \(xJ \subseteq R\).

Invertible fractional ideals

A fractional ideal \(I\) is invertible if \(IJ = R\) for some fractional ideal \(J\). Then \(J\) is necessarily \(I^{-1}\), so \(I\) is invertible exactly when \(II^{-1} = R\).

The rings in which the last phenomenon cannot occur are exactly the Dedekind domains, the Noetherian integrally closed domains in which every nonzero prime ideal is maximal: an integral domain is a Dedekind domain if and only if every nonzero fractional ideal \(I\) satisfies \(II^{-1} = R\). The element \((1 + \sqrt{-3})/2\) above is a root of \(x^2 - x + 1\), so \(\mathbb{Z}[\sqrt{-3}]\) is not integrally closed, consistent with this criterion. For a Dedekind domain the nonzero fractional ideals therefore form an abelian group under multiplication; Mathlib records this by making the fractional ideals of a Dedekind domain a semifield.

The ideal class group

The principal fractional ideals \(Rx\) with \(x \neq 0\) form a subgroup of the group of invertible fractional ideals, and the quotient is the ideal class group of \(R\). It is trivial when \(R\) is a principal ideal domain, and for a Dedekind domain whose class group is finite, the class group has exactly one element if and only if \(R\) is a principal ideal domain. For the ring of integers of a number field the class group is always finite.

In Mathlib

Mathlib defines fractional ideals in greater generality. It starts from a commutative ring \(R\), a multiplicatively closed subset \(S\) of \(R\) and an \(R\)-algebra \(P\), typically a localization of \(R\) at \(S\), and calls an \(R\)-submodule \(I\) of \(P\) fractional if some \(a \in S\) multiplies every element of \(I\) into the image of \(R\) in \(P\). A fractional ideal is such a submodule bundled with the proof that it is fractional. The classical notion is the case where \(S\) consists of the nonzero elements of a domain and \(P\) is a field of fractions; the results on inverses, Dedekind domains and class groups assume this case. The choice of localization does not matter: two localizations of \(R\) at \(S\) have isomorphic semirings of fractional ideals. If \(S\) contains \(0\), taking \(a = 0\) satisfies the condition, so every submodule is fractional. The inverse is defined for every fractional ideal, with the convention that the inverse of \(0\) is \(0\), and the class group is defined using one fixed field of fractions of \(R\).

Written for Trough, under CC BY-SA 4.0.

Sources
MathlibFractionalIdeal
Uses 3 · Used by 2
Statement usesIdeal of a commutative ring · Module · Field of fractions of an integral domain
Used byDirichlet class number formula · Finiteness of the class number

DefinitionFirst isomorphism theorem for commutative rings#

The first isomorphism theorem for commutative rings says that a surjective ring morphism \(f \colon R \to S\) identifies the quotient ring of \(R\) by the kernel of \(f\) with \(S\). It turns the question of which rings are images of \(R\) into the question of which ideals \(R\) has, and it is the usual way to recognize a quotient ring: to show that \(R/I\) is isomorphic to a known ring \(S\), one finds a surjective morphism from \(R\) onto \(S\) whose kernel is \(I\). It is the ring-theoretic counterpart of the first isomorphism theorem for groups.

Statement

Let \(f \colon R \to S\) be a morphism of commutative rings. Its kernel \(\ker f = \{ a \in R : f(a) = 0 \}\) is an ideal of \(R\), so the quotient ring \(R/\ker f\) is defined. If \(f\) is surjective, then \(R/\ker f\) is isomorphic to \(S\) as a ring, by the isomorphism that sends the coset \(a + \ker f\) to \(f(a)\):

\[R/\ker f \;\cong\; S, \qquad a + \ker f \mapsto f(a).\]

Without surjectivity, the image \(f(R)\) is a subring of \(S\), and \(R/\ker f\) is isomorphic to \(f(R)\).

Proof

Write \(K = \ker f\) and define \(\bar f(a + K) = f(a)\). Two cosets \(a + K\) and \(b + K\) are equal exactly when \(a - b \in K\), that is, when \(f(a) - f(b) = 0\). Read in one direction, this says \(\bar f\) is well defined; read in the other, it says \(\bar f\) is injective. Since sums and products of cosets are computed on representatives and \(f\) preserves sums, products and the identity, \(\bar f\) is a ring morphism. Well-definedness and the morphism property are both instances of the universal property of the quotient: a ring morphism that vanishes on an ideal \(I\) induces a ring morphism on \(R/I\). The morphism induced on \(R/\ker f\) is injective, and it is surjective because \(f\) is. A bijective ring morphism has a ring morphism as its inverse, so \(\bar f\) is an isomorphism.

Examples
Consequences
Generalizations

The same argument gives analogous theorems for groups, modules and algebras, whose quotients are likewise taken by kernels. For a surjective group morphism \(\varphi \colon G \to H\), \(G/\ker\varphi \cong H\). For a surjective linear map \(f \colon M \to N\) of modules, the quotient of \(M\) by the kernel is linearly isomorphic to \(N\). For a surjective morphism of algebras over a commutative ring, the quotient by the kernel is isomorphic to the codomain as an algebra.

In Mathlib

Mathlib states the theorem in more generality than for commutative rings: the domain may be any ring, commutative or not, and the codomain need only be a semiring, a structure in which additive inverses need not exist. When the source ring is noncommutative, the kernel is a two-sided ideal, which is what makes the quotient a ring. The theorem is a construction rather than an existence statement: it builds a specific ring isomorphism from the quotient by the kernel to the codomain. Its inverse is obtained by choosing a preimage of each element, so the definition is noncomputable; a computable variant exists for a morphism equipped with a right inverse, and Mathlib uses it for the integers modulo \(n\) and for evaluation of polynomials. The version onto the image is derived from the surjective one by restricting the codomain of the morphism to its image, and a further version allows a semiring as codomain with the image as a subsemiring.

Written for Trough, under CC BY-SA 4.0.

Sources
MathlibRingHom.quotientKerEquivOfSurjective
Uses 4 · Used by 1
Statement usesRing · Ring morphism · Ideal of a commutative ring · Quotient ring
Used byChinese remainder theorem

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