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Joined 2026-09-21 Message
Incoming links: Rational number
Map the integers into a field by . Its kernel is a prime ideal of , since the image has no zero divisors. Thus the kernel is either zero or for a prime . In the zero case, the map extends injectively to the rational numbers by taking quotients of nonzero integers. In the other case, its image is the field . This proves the prime-field assertion.
A finite field cannot contain the infinite rational field, so it has characteristic . It is a finite-dimensional vector space over its prime field , say of dimension , and choosing a basis counts its elements asFor existence, put and take a splitting field of over . Its derivative is , so it has exactly distinct polynomial roots. The polynomial roots are closed under addition and multiplication because the Frobenius identity gives and . They include , are closed under additive inverses, and for a nonzero polynomial root , . They therefore form a field of exactly elements. Indeed this field is the splitting field because it contains all polynomial roots and is generated by them.
In order 4, characteristic is 2. For , the multiplicative group has order 3, so and . Factorization of gives . The quadratic has no polynomial root in and is the minimal polynomial. Write . The complete tables areFor an element in order 8, the tower law makes divide3. It is not1, so its minimal polynomial is an irreducible monic cubic. Its constant term must be1. A cubic has no polynomial root at 1 exactly when , and absence of a polynomial root is equivalent to cubic irreducibility. Consequently the two possibilities areBoth occur: together with and they factor over , whose distinct polynomial roots are exactly the eight elements of the field. Each cubic contributes its three polynomial roots.
Every point belongs to its component, so the components cover ; by part (ii) they are pairwise disjoint, and by part (iii) they are nonempty open intervals. Each contains a rational number. Fix an enumeration of and choose the first rational belonging to each component. Distinct components receive distinct rationals, giving an injection from the component family into a countable set. ThereforeThe empty open set is represented by the empty union.
The components cover and are pairwise disjoint by (b)(ii); by (iii) they are open intervals. Each nonempty interval contains a rational number. Enumerate the rationals and assign to each interval the first rational it contains. Disjointness makes these assigned rationals distinct, giving an injection from the set of components into . Thus , with the empty union for .
Past exam of the rings and modules course of the University of Oxford 2023 3 a iii Solution 2026-10-25
Let be nonzero. If the columns were linearly dependent over the rational numbers, then for some . Applying once more would giveso . Since , this forces , impossible in . Thus the two columns are linearly independent, and the square matrixis an invertible matrix.
Past exam of the rings and modules course of the University of Oxford 2023 3 a i Solution 2026-10-25
Since , its minimal polynomial divides . The polynomial is an irreducible polynomial over the rational numbers: a reducible quadratic would have a rational root, whereas for every .
Because , the constant polynomial one evaluates to the nonzero identity matrix , so . Both and are monic polynomials; hence the only remaining divisor is
Past exam of the rings and modules course of the University of Oxford 2023 3 b iii Solution 2026-10-25
Regard as a matrix over the rational numbers. Part (a)(ii), with , gives its characteristic polynomial . Therefore, writing for the imaginary unit,The matrix is not zero: otherwise would have noninteger diagonal entries. It consequently has rank of a matrix one over the fraction field of the Gaussian integers.
By part (b)(i), is a euclidean domain, hence a principal ideal domain, so apply Smith normal form. There exist such thatIndeed, invertible multiplication preserves rank of a matrix, so exactly one diagonal entry is nonzero. For , multiplication on the right gives . Since is an invertible matrix,The Smith normal form supplies a kernel basis lemma ensures that the null vector is a column of an invertible matrix over ; an arbitrary nonzero solution over would not itself establish that requirement.
False, with “disjoint union” understood as a topological disjoint union. Take the rational numbers with the usual subspace topology from the real line.
Every connected subset of has at most one point. Indeed, if it contained , choose an irrational number . The two relatively open setspartition and separate any subset containing both and . Hence the connected components of are singletons.
If a topological disjoint union of nonempty connected spaces were homeomorphic to , the image of each summand would be connected, hence a singleton. Every summand would therefore be a one-point topological space. Their topological disjoint union is a discrete space, since each summand is open. But is not a discrete space: for each , the distinct rational numbers approach , so is not an isolated point. This is a contradiction. The set decomposition into connected components always exists; giving those pieces the topological disjoint union topology need not recover the original topology.
True. Choose the irrational number , and partition the rational numbers intoNo endpoint is rational, so these disjoint sets cover . Each is nonempty by the density of the rational numbers, and each is clopen in the subspace topology on .
Let two rational numbers be equivalent exactly when they belong to the same . The bijectionis well-defined. For every subset , the preimage of under the canonical projection of a quotient set is , which is open in . Thus every subset of is open: its quotient topology is the discrete topology. The usual subspace topology on the integers is also the discrete topology, because . Consequently is a homeomorphism.
For positive integer pairs, reflexivity is . If , reversing the equality gives , establishing symmetry. For transitivity, suppose and . Then and , whenceSince , cancellation gives . Thus , and is an equivalence relation.
Membership in the equivalence class of means , or . Since and are coprime, and for a positive integer . Conversely every such pair satisfies the equality. ThereforeThese pairs are precisely the positive numerator-denominator representations of the rational number .
Convert the tail first and apply the reciprocal identity. Two output parameters hold the rational number's numerator and denominator.The singleton case is immediate. If the recursive call returns the tail as , then because every tail head is positive. The full value isStructural induction on the finite nonempty linked list proves correctness, and the denominator is always positive even if the first head is negative. If the tail pair is coprime, then ; the base pair is coprime, so the result is in lowest terms. Return
PROCEDURE cf2rat(f: fraction; VAR n, d: INTEGER);
VAR u, v: INTEGER;
BEGIN
IF f.tail = NIL THEN
n := f.head; d := 1
ELSE
cf2rat(f.tail, u, v);
n := f.head * u + v;
d := u
END
END cf2rat;(head*u+v,u) after converting the tail, using linear node-processing time and linear recursive stack space.function GCD(m,n): integer
a := m
b := n
while b != 0 do
r := a mod b
a := b
b := r
return a
procedure Reduce(var m,n)
require m >= 0 and n > 0
d := GCD(m,n)
m := m div d
n := n div dThe loop invariant is with . The nonnegative loop variant is : when positive, the new remainder is strictly smaller. Thus the final is the positive greatest common divisor. Dividing both numerator and denominator by it preserves the rational number and makes them coprime. The stated proper fraction case is included, but the same code works for every nonnegative numerator, a fact useful in part (d). In particular, zero reduces to .
Use the supplied pointer-to-record representation and allocate a fresh record:
PROCEDURE mkRat(n,d: INTEGER): rational;
VAR z: rational;
BEGIN
ASSERT(d # 0);
NEW(z); z.num := n; z.den := d;
reduce(z);
RETURN z
END mkRat;The rational number constructor returns a value with positive denominator in lowest terms, using part (b). Zero is represented as . Arithmetic below uses exact integers, or inputs whose intermediate values fit the chosen implementation; fixed-width overflow would otherwise invalidate exact arithmetic.
Use the remainder form of the Euclidean algorithm to compute the greatest common divisor, then divide both components of the fraction by it:The loop invariant is , , and . Initially it holds by symmetry of the greatest common divisor. For , the integer division identity , with , shows that and have exactly the same common integer divisors. Thus the update preserves the loop invariant. The loop variant is the nonnegative integer , which strictly decreases on each iteration. At exit , so . Dividing by preserves the rational number and makes the returned components coprime. In particular, zero is returned as . No multiplication is needed, and all remainders remain within the original nonnegative
def Reduce(num: Int, den: Int): (Int, Int) = {
require(0 <= num && num < den)
var a = den
var b = num
while (b != 0) {
val r = a % b
a = b
b = r
}
(num / a, den / a)
}Int range.The inclusive interval is nonempty exactly when . Reject with a failed precondition, rather than assigning a nonexistent normalization to an empty interval. Compute its width in The loop invariant says that and
Long before taking a reciprocal:class Uniform(l: Int, u: Int) extends FiniteDist[Int] {
require(l <= u, "Empty uniform interval")
private val width = u.toLong - l.toLong + 1L
private val mass = 1.0 / width.toDouble
def getProb(x: Int): Double =
if (l <= x && x <= u) mass else 0.0
def getDom: List[Int] = {
var x = l.toLong
var rev = List.empty[Int]
while (x <= u.toLong) {
rev = x.toInt :: rev
x += 1L
}
rev.reverse
}
}rev contains exactly the integers in reverse order. The loop variant is while the guard is true. Long handles both the full interval width and the increment past Int.MaxValue. At exit, reversing yields each support value once. All support values have mass , so their exact sum is one; Double approximates that rational number. Very large valid domains require correspondingly large lists and sufficient memory; that resource issue does not make the mathematical endpoint values invalid.Use the intended first-order structure of real arithmetic, with
Int selecting the integers; is the parameter in . A suitable first-order formula isThe first condition requires a real square root of . The second excludes every representation as a rational number with a nonzero integer denominator. This uses only the listed function and constant symbols, the Int predicate and logical equality; in particular, is written and there is no extra division symbol. A square root exists exactly when , and either sign of a root is a rational number exactly when the other sign is a rational number, so the absence of a positivity condition on does not affect the claim about the principal square root. At the formula is false; for there is no real witness. The meanings of and as real arithmetic, not just the meaning of Int, are part of this interpretation.Define the indicator function of the rational numbers on the real line by for rational and otherwise. Every rational shift preserves rationality and irrationality, so every nonzero rational number is a period of a function. Conversely, an irrational shift sends from value one to value zero, so it is not a period. The period group of a function is exactly , and the positive periods approach zero without a smallest one. The function is nonconstant and discontinuous everywhere; this is why continuity matters in continuous periodic functions have a least positive period.
Polynomial division Created 2026-09-24 Updated 2026-10-18
Polynomial division separates a polynomial into a multiple of a chosen divisor and a smaller polynomial remainder. Given over a field , with , there are unique polynomials satisfyingHere is the polynomial quotient. The bound concerns polynomial degree, rather than the sizes of the coefficients.[ref]
Fields and Galois Theory by James S. Milne. Fields and Galois Theory, version 5.10, Chapter 1, p. 10, §1.6: division algorithm, zero remainder alternative and uniqueness
For an original example over the rational numbers, takePolynomial long division first cancels by subtracting :Subtracting one more copy of cancels :ThusThe final polynomial remainder has polynomial degree one, below the divisor's polynomial degree two. Multiplying out the last identity checks every coefficient, including the constant term .
The polynomial degree bound makes this decomposition useful: it replaces a potentially long polynomial with a short polynomial remainder whenever multiples of can be discarded. At any root of a polynomial of , those multiples vanish under polynomial evaluation. Repeated polynomial division also drives the Euclidean algorithm for finding a greatest common divisor of polynomials.
The coefficient setting matters. Over a commutative ring, the same existence and uniqueness hold when the divisor's leading coefficient is a unit in a ring. A monic polynomial therefore works over the integers, even though they do not form a field; monic polynomial division over a ring explains why zero divisors cause no difficulty in that case.[ref]
Chapter 3 Ring theory. Abstract Algebra, Chapter 3 Ring theory, §3.9, Theorem 3.9.1: division with a unit leading coefficient
A nonzero divisor alone is insufficient over the integers. Dividing by with an integer polynomial quotient and a constant polynomial remainder would require for an integer , which is impossible. Over the rational numbers the division instead givesThe obstruction is the nonunit leading coefficient . This example separates existence from uniqueness: over an integral domain, the polynomial degree argument still makes any admissible pair unique, if it exists.
Prime field Created 2026-10-03 Updated 2026-10-14
The prime field of a field is its smallest subfield: the arithmetic forced upon us by the identity element . Every subfield must contain , its repeated sums, their additive inverses, and the multiplicative inverses of nonzero results. In characteristic zero these operations produce a copy of the rational numbers . In positive characteristic of a field , they produce a copy of , with addition and multiplication given by modular arithmetic.
For example, inside the field with four elements, repeated addition of gives . Thus its prime field is , and the other two elements cannot be reached by repeatedly adding . This is why a finite field need not consist entirely of ordinary integer residues: a larger finite field extends the arithmetic of its prime field with additional elements.
The positive characteristic of a field must be a prime number. If its least vanishing multiple of were with , then would exhibit zero divisors. The absence of zero divisors rules this out. The prime field supplies the scalars when viewing as a vector space; its basis then explains the possible sizes of a finite field.
Quotient topological space Created 2026-10-06 Updated 2026-10-25
A quotient topological space is a quotient set equipped with the quotient topology. For a concrete gluing, give the closed interval its usual subspace topology, identify , and leave every interior point in its own singleton equivalence class. The resulting quotient space is a circle: the two endpoints become one point, while interior points remain distinct.
The exact continuous map to the unit circle in the plane isHere sine and cosine give , and holds exactly when or . Thus induces a bijection . [ref] The universal property of the quotient topology makes continuous. The quotient of a compact space is compact, and the unit circle is Hausdorff, so the compact-to-Hausdorff continuous bijection theorem makes a homeomorphism. This verifies the topology, not merely the point identifications.
To inspect an open neighbourhood of the seam, choose and let . Its full preimage isBoth pieces are open in the subspace topology on , so is open in the circle. In contrast, is open in , butis not open in : it contains without any neighbourhood of . Hence is not open in the circle. A quotient map can therefore fail to be an open map.
An open neighbourhood crosses both sides of the glued seam; a one-sided arc fails the full-preimage test
. Details. 
For a continuous function , the only gluing condition is . For example, defines a continuous function on the circle, whereas does not. This lets us use an interval as a coordinate for a periodic function while making the seam invisible. More general quotient spaces encode glued geometric pieces in the same way: describe continuous maps before gluing and check that their values agree on every equivalence class.
The Hausdorff property need not survive a quotient topology. In the non-Hausdorff quotient of the real line by rational translation, means . A nonempty open preimage must contain all translates by rational numbers of any open interval inside it. The density of the rational numbers then forces : for any , some rational number has in that interval. Thus this quotient space has the indiscrete topology. The irrationality of the square root of two gives distinct equivalence classes for and , so this quotient space is not Hausdorff. Unlike the endpoint-gluing circle, this gluing identifies a dense subset with each point.
Rado solution inside an m-p-c set 2026-10-05
Suppose the columns of a matrix over the rational numbers satisfy the columns condition with blocks . Choose rational coefficients such thatChoose a positive integer clearing all denominators and . For generators of any full positive M-p-c set, the coordinateslie in that set and satisfy . Thus the monochromatic m-p-c set theorem supplies the sufficiency direction of Rado's theorem.