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The 2-adic absolute value is the p-adic absolute value for the prime number $p=2$. It is defined on the rational numbers by $|x{|}_2=2^{-v_2(x)}$ for $x\ne 0$ and $|0{|}_2=0$, and extends to their P-adic number completion ${\mathbb{Q}}_2$. For a nonzero integer, count factors of two until the remaining factor is odd; for a rational number, subtract the denominator's count from the numerator's count.

The associated metric reverses a familiar comparison:Although $13$ is eight units from $5$ on the ordinary number line, it agrees with $5$ modulo $8$, whereas $6$ already disagrees modulo $2$. The 2-adic absolute value measures that agreement. The value $|3/8{|}_2=8$ also shows why this Non-Archimedean absolute value must be distinguished from the ordinary absolute value $|3/8|=3/8$.

In the left panel, differences farther from the diagonal grow in ordinary absolute value. In the right panel, nonzero even differences are closer than odd differences, and each further factor of two halves the 2-adic absolute value. The highlighted pairs have opposite distance rankings. The repeated pattern is the nested congruence class structure of an ultrametric space, visible without adding any new points.

The 2-adic valuation is the P-adic valuation at the prime number $p=2$.

Let $p<q$ be distinct prime numbers. Every group of group order $p^2{q}^2$ is Abelian if and only if $p$ does not divide $q^2-1$.

For sufficiency, the Sylow theorems make the number of Sylow p-subgroups one of $1,q,q^2$, congruent to one modulo $p$. The divisibility hypothesis excludes both nontrivial choices, so that subgroup is normal. The number of Sylow subgroups for $q$ is one of $1,p,p^2$, congruent to one modulo $q$. The choice $p$ is impossible because $p<q$. If $q$ divided $p^2-1=(p-1)(p+1)$, primality and $q>p$ would force $q=p+1$, hence $(p,q)=(2,3)$; this pair violates the hypothesis. Thus the other Sylow subgroup is normal too. Their intersection is trivial, and their group commutators lie in that intersection, so the group is their direct product of groups. A group of prime-square order is abelian: its nontrivial center of a group is either the whole group or has cyclic quotient, and a cyclic central quotient also forces commutativity. Both factors are therefore abelian.

For necessity, suppose $p\mid q^2-1$. The multiplicative group of a finite field is cyclic, so ${\mathbb{F}}_{q^2}^{\times }$ contains $\alpha$ of order $p$. Let the cyclic group $C_{p^2}$ act on the additive group of ${\mathbb{F}}_{q^2}$ by mapping its generator to multiplication by $\alpha$. This is a well-defined nontrivial group action, since ${\alpha }^{p^2}=1$. The resulting semidirect product has order $p^2{q}^2$ and is nonabelian, because the generator does not commute with a nonzero field element. Thus a nonabelian group of this order exists exactly when $p\mid q^2-1$.

Fix a prime number $p$. Take the full subcategory of finitely generated abelian groups with no element of order $p^2$. It is closed under finite direct sums and subgroups. If $t_p(G)$ is the $p$-primary torsion subgroup, the map $G\to R(G)=G/(p{t}_p(G))$ is a reflection into this subcategory: every map to an allowed group kills $p{t}_p(G)$. Therefore ordinary kernels in a category and reflected cokernels in a category supply all finite categorical limits and colimits. No nonzero finitely generated group has zero reflection, so the epimorphisms are precisely surjections and are all conormal epimorphisms. Multiplication by $p^2$ on $\mathbb{Z}$ is a monomorphism but not a normal monomorphism: its cokernel here is $\mathbb{Z}/p\mathbb{Z}$, whose kernel is $p\mathbb{Z}$, not $p^2\mathbb{Z}$. Multiplication by $p$ is normal; composing it with itself gives this nonnormal monomorphism.

For distinct or equal prime numbers $\ell ,\lambda$, the polynomial $X^{\ell }-\lambda$ is Eisenstein at $\lambda$. Its splitting field over $\mathbb{Q}({\zeta }_{\ell })$ has Galois group a subgroup of $C_{\ell }$, which is nontrivial because a degree-$\ell$ radical cannot belong to a degree-$(\ell -1)$ cyclotomic field. Thus the full Galois group has order $\ell (\ell -1)$. On roots indexed by $j\in {\mathbb{F}}_{\ell }$, all automorphisms act as $j\mapsto uj+b$. Equality of orders identifies the group with the full affine group over a finite field.

// Target: algebra.bigb

Let $G$ have group order equal to a prime number $p$. Choose any $g\ne e$. By Lagrange's theorem, the order of a group element $g$ divides $p$, and it is greater than one. It is therefore $p$, so the cyclic subgroup generated by $g$ is all of $G$. This proves that every nonidentity element generates $G$, and $G$ is isomorphic to the additive group of integers modulo $p$.

Algebraic number theory studies integer factorization, divisibility, prime ideals, and units after extending the rational numbers to a number field. Its basic objects are a number field $K$ and its ring of integers ${\mathcal{O}}_K$: the counterparts of $\mathbb{Q}$ and $\mathbb{Z}$. The field norm turns multiplication in $K$ into multiplication of rational numbers, and takes algebraic integers to integers. These connections let linear algebra and Euclidean geometry answer arithmetic questions.^[ref]

A first example is $K=\mathbb{Q}(\sqrt{2})$ with ring of integers $\mathbb{Z}[\sqrt{2}]$. The integer $7$ acquires the factorizationEach factor has field norm $7$. Neither is a unit, and each is an irreducible element: any factorization into two nonunits would factor the prime number $7$ into two integers of absolute value greater than one. Arithmetic in a larger ring thus reveals structure invisible in ordinary integer factorization.

There is also a complication: unique factorization of elements can fail, as in the failure of unique factorization in the ring of integers generated by square root of minus five. The general replacement is prime ideal factorization. The ideal class group measures the obstruction to making every ideal a principal ideal.

The historical motivation is visible in Richard Dedekind's original 1877 exposition: he begins with ordinary integer factorization, passes to Gaussian integers, and then uses a quadratic example to explain why ideals are needed. His later section establishes unique factorization of ideals in a number field. This is a direct record of the development of the theory, rather than a claim that all its ideas originated in one year.^[ref][ref]

Analytic number theory studies integers, prime numbers, and arithmetic functions using tools from real analysis and complex analysis, especially Dirichlet series and their singularities.

For a nonnegative integer $n$ and integer base $p\ge 2$, write its base-p expansion asThe base-$p$ digit sum is $d_p(n)={\sum }_r{\alpha }_r$, with $d_p(0)=0$. When $p$ is a prime number, the Legendre formula becomesTo see the connection, divide the base-p expansion by $p^j$ and discard the fractional part. Swapping the resulting finite sums and using a finite geometric series givesIn particular, $50=1\cdot 27+2\cdot 9+1\cdot 3+2=(1212{)}_3$, so $d_3(50)=6$ and $v_3(50!)=(50-6)/2=22$. The leading numeral digit contributes thirteen, the next contributes eight, the next one, and the units contribute zero. Thus the base-p digit sum compresses the same contribution ledger into information already present in positional notation.

For every integer $n>1$, there is a prime number strictly between $n$ and $2n$. This supplies cyclic prime moduli comparable to an integer interval's length in additive combinatorics.

A nonzero residue class $a$ modulo an odd prime number $p$ is a biquadratic residue if $x^4\equiv a(\mathrm{mod}p)$ for some integer $x$.

For an integer $n>1$, the Carmichael function $\lambda (n)$ is the exponent of a finite group for the unit group $U(n)=(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$: the least positive integer $m$ for whichHere the greatest common divisor condition selects the units. Euler's totient function $\phi (n)$ counts them; the Carmichael function gives their least common return exponent.^[ref]

Modulo eight, the units are $1,3,5,7$, and their squares are all congruent to one. Their multiplicative orders are $1,2,2,2$, so $\lambda (8)=2$ although $\phi (8)=4$. Modulo fifteen, the units $1,2,4,7,8,11,13,14$ have multiplicative orders $1,4,2,4,4,2,4,2$. For example, the powers of two return through $1,2,4,8,1$. Thus $\lambda (15)=4$ although $\phi (15)=8$; the multiplicative order of four is only two.

The three quantities are related byA common return exponent must be a multiple of every multiplicative order, hence of their least common multiple. Conversely, that least common multiple makes every unit return to one. Lagrange's theorem makes each multiplicative order divide the group order $\phi (n)$, proving the final divisibility. The Carmichael function applies to all units at once, while the multiplicative order concerns a chosen unit.

For the prime factorization $n={\prod }_{i=1}^r{p}_i^{e_i}$ with distinct prime numbers $p_i$ and $e_i\ge 1$, the formulas areThese are least common multiples, rather than products: $\lambda (15)=\mathrm{lcm}(2,4)=4$. The convention $\lambda (1)=1$ handles the trivial case separately.^[ref]

The Chinese remainder theorem for unit groups explains the least common multiple: a power returns to one in every prime power component exactly when it is a multiple of each component's exponent of a finite group. For an odd prime power, the unit group is a cyclic group of group order $p^{e-1}(p-1)$. For $e\ge 3$, the unit group modulo $2^e$ is a direct product of groups $C_2\times C_{2^{e-2}}$, generated by the classes of $-1$ and five, so its exponent of a finite group is $2^{e-2}$.^[ref]

This gives a uniform reduction for modular exponentiation: for a unit $a$ and a nonnegative integer $k$, reduce $k$ modulo $\lambda (n)$. For example, $2^{1001}\equiv 2(\mathrm{mod}15)$ because $1001\equiv 1(\mathrm{mod}4)$. The unit hypothesis matters: $3^2\equiv 1(\mathrm{mod}8)$, but $2^2\not\equiv 1(\mathrm{mod}8)$. Computing the formula assumes the prime factorization is available. The same common-exponent viewpoint explains Carmichael numbers: a composite number $n$ passes the Fermat primality test for every unit base exactly when $\lambda (n)\mid n-1$. For example, $\lambda (561)=\mathrm{lcm}(2,10,16)=80\mid 560$.

A Carmichael number is a composite number $N>1$ for whichEvery unit modulo $N$ satisfies the same modular congruence supplied by Fermat's little theorem for a prime number. The name refers to Robert Daniel Carmichael. The coprime qualification is essential: a nonunit can expose a prime factor through a greatest common divisor calculation. A Carmichael number is therefore a Fermat pseudoprime to every coprime base, rather than to every nonzero residue class.^[ref]

For example,The factors are distinct prime numbers. Their predecessor numbers $2,10,16$ all divide $560$, so Korselt criterion applies. More explicitly, for any integer $a$ coprime to $561$, Fermat's little theorem givesThe Chinese remainder theorem combines these into $a^{560}\equiv 1(\mathrm{mod}561)$. This proves all coprime bases pass from three short divisibility checks, rather than from a long table of sampled bases.

Equivalently, a composite $N$ is a Carmichael number exactly when $a^N\equiv a(\mathrm{mod}N)$ for every integer $a$, including nonunits. The equivalence follows from Korselt criterion: modulo each prime factor, either $a$ is zero or its nonzero power returns to one, and squarefreeness combines those congruences. This does not say a nonunit satisfies $a^{N-1}\equiv 1$. For instance $a=3$ gives $3^{560}\equiv 375(\mathrm{mod}561)$, while $3^{561}\equiv 3(\mathrm{mod}561)$; the greatest common divisor already supplies the factor three.^[ref]

The Carmichael function supplies another compact description: a composite $N$ is Carmichael exactly when $\lambda (N)\mid N-1$. For $561$, $\lambda (561)=\mathrm{lcm}(2,10,16)=80$ and $560=7\cdot 80$. In contrast, Euler's totient function gives $\phi (561)=320$, which does not divide $560$. The least common return exponent, rather than the number of units, controls this test.

The obstruction is useful to know when interpreting a Fermat primality test: even checking every coprime base cannot certify primality. The passing-base fraction for the Fermat test makes this failure quantitative. Stronger primality tests require additional structure, such as the squaring conditions in a Miller-Rabin primality test.

The historical record predates the name. Václav Šimerka's 1885 paper lists $561,1105,1729,2465,2821,6601,8911$ on printed page 224, together with factorizations. This is an early published list of examples, independently of their later name.^[ref]

Let $p$ be a prime number, let $n={\sum }_r{\alpha }_r{p}^r$ with $0\le {\alpha }_r<p$, and let $c_r$ be the sum of partition sizes at level $r$ of the core tower of a partition $\lambda ⊢n$. ThenIndeed $n={\sum }_r{c}_r{p}^r$. Carrying $p$ units from position $r$ to position $r+1$ decreases ${\sum }_r{c}_r$ by $p-1$. Normalization to the base-$p$ digits therefore gives ${\sum }_r{c}_r\ge {\sum }_r{\alpha }_r$, with equality exactly when no carry is needed. The P-adic valuation of a symmetric-group character degree from the core tower is the difference divided by $p-1$.

The positive characteristic of a ring which is an integral domain is a prime number. The characteristic cannot be one because $1\ne 0$. If its least positive value were $n=ab$ with $1<a,b<n$, minimality would make $a1$ and $b1$ nonzero, while $(a1)(b1)=n1=0$ would give zero divisors.

The analogous result holds for a nonzero commutative nonunital ring without zero divisors, using the characteristic of a rng. If $n=ab$ were composite and minimal, choose $r$ with $ar\ne 0$. For every $s$, $(ar)(bs)=n(rs)=0$, so cancellation forces $bs=0$ for every $s$, contradicting minimality of $n$.

The two Chebyshev functions measure the total natural logarithm weight of prime numbers or of prime powers greater than one below a real cutoff $x\ge 0$:Here $p$ ranges over prime numbers, $n$ over positive integers, and $k$ over positive integers. The Chebyshev theta function $\vartheta$, also written $\theta$, weights each prime number once. The Second Chebyshev function $\psi$ weights every positive power of each prime number separately. Both vanish for $0\le x<2$; the inclusive cutoff makes them right-continuous functions and nondecreasing functions, constant between their jumps. ^[ref]

At $x=10$, the prime numbers are $2,3,5,7$, while the extra prime powers greater than one are $4,8,9$. Exactly,At each prime number $p$, both Chebyshev functions jump by $\log p$. At a prime power $p^k$ with $k\ge 2$, only the Second Chebyshev function jumps, by $\log p$. At $6$ and $10$ neither jumps. Notice that the gap is already positive at $5$ and $7$: a shared jump does not remove the earlier contribution from $4$.

Grouping the prime powers by their exponent gives, for $x\ge 2$,The floor function cutoff is exact: $p\ge 2$ forces $2^k\le x$ whenever $p^k\le x$. For each exponent, $p^k\le x$ is equivalent to $p\le x^{1/k}$. An upper cutoff smaller than the lower one means an empty sum. ^[ref]

The Chebyshev theta function is one of the two Chebyshev functions and is the natural logarithm of the primorial:where both the sum and the product range over prime numbers.

A prime number for which every cyclic rotation of its numerical digits is also a prime number. In decimal representation, $197$ is an example because $197,971,719$ are all prime numbers. A multi-digit circular prime can contain only $1,3,7,9$: any other digit can be rotated into the units position and force divisibility by $2$ or $5$.

For a prime number $p$ and nonnegative integers $h+k=p-1$, Wilson theorem gives $(-1{)}^{k}h!k!\equiv -1(\mathrm{mod}p)$ by replacing the last $k$ factors of $(p-1)!$ with $-1,-2,\dots ,-k$. For odd $p$, $h,k$ have the same parity; for $p=2$, both signs have the same residue class. Thus $h!k!+(-1{)}^h\equiv 0(\mathrm{mod}p)$, including the endpoint cases $0!=1$.

An arithmetic function $f$ is a completely multiplicative function when $f(1)=1$ and $f(mn)=f(m)f(n)$ for every pair of positive integers $m,n$. In particular $f(p^a)=f(p{)}^a$ for every prime number $p$ and every nonnegative exponent $a$. Both the constant-one arithmetic function and the identity arithmetic function are completely multiplicative functions: $1(mn)=1=1\cdot 1$ and $\mathrm{id}(mn)=mn=\mathrm{id}(m)\mathrm{id}(n)$. The divisor function fails this stronger condition at $m=n=2$.