MagicInternetMath
Magic Internet Math

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Daily Insights from Magic Internet Math courses. Learn at https://mathacademy-cyan.vercel.app

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Last Notes

2026-09-03 21:53:37 UTC

📖 Algebraic Closure
An **algebraic closure** of $F$ is an algebraically closed field $\\overline{F}$ that is algebraic over $F$. It exists and is unique up to $F$-isomorphism. The **absolute Galois group** is $\\operatorname{Gal}(\\overline{F}/F)$.
From: gal-morandi
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2026-09-03 01:27:37 UTC

📖 Mining Pool
A mining pool aggregates many miners\
From: mining-pools
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2026-09-02 23:33:41 UTC

📖 PPS / FPPS
Pay-per-share: the pool pays a fixed price for every share submitted, block or no block — the operator absorbs all luck variance. FPPS additionally pays the expected transaction-fee revenue per share.
From: mining-pools
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2026-09-02 21:18:21 UTC

📐 Bonferroni Bound
Testing $m$ cells at once, requiring per-test $p < \\alpha/m$ keeps the probability of even one false positive below $\\alpha$. For the era: $m = 1440$, $\\alpha = 0.05 \\Rightarrow p < 3.47 \\times 10^{-5}$.
Proof: Union bound: $P(\\text{any false positive}) \\leq \\sum_{i=1}^{m} P(\\text{test } i \\text{ false positive}) = m \\cdot (\\alpha/m) = \\alpha$.
From: mining-pools
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2026-08-09 04:50:15 UTC

📖 Roots of Unity
The roots of $x^n - 1$ form a cyclic group under multiplication. A generator $\\epsilon$ (of order exactly $n$) is a primitive $n$-th root of unity.
From: gal-artin
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2026-08-08 23:04:18 UTC

📖 Resolvent Equation
The auxiliary equation of lower degree that arises in the process of solving a polynomial equation is called its resolvent equation (or simply resolvent). For the quartic, the resolvent is a cubic. This pattern -- solving an equation by reducing it to a resolvent of lower degree -- is the fundamental idea that Lagrange would later analyze in full generality.
From: gal-edwards
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2026-08-08 22:02:24 UTC

📐 Characteristic is Prime
The characteristic of a field is either $0$ or a prime number $p$.
Proof: If $\\operatorname{char}(F) = n = ab$ with $1 < a, b < n$, then $(a \\cdot 1)(b \\cdot 1) = n \\cdot 1 = 0$. Since $F$ is a field (hence an integral domain), either $a \\cdot 1 = 0$ or $b \\cdot 1 = 0$, contradicting the minimality of $n$.
From: gal-weintraub
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2026-08-08 17:09:21 UTC

📖 Linear Substitution Modulo q
A permutation $\\sigma$ of $\\{1, \\ldots, q\\}$ (with $q$ prime) is a linear substitution modulo $q$ if $\\sigma(i) \\equiv bi + c \\pmod{q}$ for integers $b \\not\\equiv 0$ and $c$. These form a group of order $q(q-1)$.
From: gal-artin
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2026-08-08 16:02:56 UTC

📖 Galois Group
The Galois group of the equation $f(x) = 0$ over the field $K$ is the group of substitutions of the roots $a, b, c, \\ldots$ presented by the table whose rows are $\\phi_a(t\
From: gal-edwards
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2026-08-08 15:04:03 UTC

📐 Gram-Schmidt Procedure
If $v_1, \\ldots, v_m$ is a linearly independent list in $V$, then there exists an orthonormal list $e_1, \\ldots, e_m$ such that $\\operatorname{span}(v_1, \\ldots, v_j) = \\operatorname{span}(e_1, \\ldots, e_j)$ for each $j$.
From: linalg-axler
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2026-08-08 14:18:17 UTC

🎮 Interactive: Determinant Visualizer
Understand the determinant as signed area/volume. See why det(AB) = det(A)det(B) and when matrices are invertible.
From: Linear Algebra
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2026-08-08 11:06:02 UTC

📐 Fundamental Theorem of Galois Theory
Let $E/K$ be a finite Galois extension with $G = \\mathrm{Gal}(E/K)$. There is an inclusion-reversing bijection between the set of intermediate fields $K \\subseteq F \\subseteq E$ and the set of subgroups $H \\leq G$, given by $F \\mapsto \\mathrm{Gal}(E/F)$ and $H \\mapsto E^H$. Moreover, $[E:F] = |\\mathrm{Gal}(E/F)|$ and $F/K$ is normal iff $\\mathrm{Gal}(E/F) \\trianglelefteq G$.
From: gal-jacobson
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2026-08-08 10:13:12 UTC

📖 Affine Algebraic Variety
An **affine algebraic variety** over a field $k$ is the set $V(I) = \\{a \\in k^n : f(a) = 0 \\text{ for all } f \\in I\\}$ where $I$ is an ideal in $k[x_1, \\ldots, x_n]$. The **coordinate ring** is $k[V] = k[x_1, \\ldots, x_n]/I(V)$.
From: gal-morandi
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2026-08-08 09:22:15 UTC

📐 Fundamental Theorem of Symmetric Functions (Generalized)
$S = F$ and $(E/F) = n!$. Any polynomial in $x_1, \\ldots, x_n$ can be uniquely expressed as a linear combination of $x_1^{\\nu_1} \\cdots x_n^{\\nu_n}$ (with $\\nu_i \\leq i-1$) with coefficients that are polynomials in $a_1, \\ldots, a_n$.
From: gal-artin
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2026-08-08 05:26:06 UTC

📖 Fixed Point and Fixed Field
An element $a \\in E$ such that $\\sigma_1(a) = \\sigma_2(a) = \\cdots = \\sigma_n(a)$ is a fixed point. The set of all fixed points forms a subfield called the fixed field.
From: gal-artin
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2026-08-08 02:35:03 UTC

📐 Galois Groups of Cubics
An irreducible cubic $f(x) \\in \\mathbb{Q}[x]$ has Galois group $S_3$ if $\\operatorname{disc}(f)$ is not a perfect square in $\\mathbb{Q}$, and $A_3 \\cong \\mathbb{Z}/3\\mathbb{Z}$ if it is.
From: gal-morandi
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2026-08-07 23:17:26 UTC

📖 Algebraic Element
An element $\\alpha \\in E$ is algebraic over $F$ if it is a root of some nonzero polynomial in $F[X]$. Otherwise, $\\alpha$ is transcendental over $F$.
From: gal-weintraub
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2026-08-07 21:22:12 UTC

📐 Eigenvectors are Independent
Let $T \\in \\mathcal{L}(V)$. Eigenvectors of $T$ corresponding to distinct eigenvalues are linearly independent.
From: linalg-axler
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2026-08-07 19:39:00 UTC

📐 Resolvent is Solvable
The resolvent equation of the cubic, although of degree 6 in the variable $t$, is actually a quadratic equation in $t^3$. It can therefore be solved by first solving a quadratic (to find $t^3$) and then taking a cube root (to find $t$). Once $t$ is found, the original roots are determined.
From: gal-edwards
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2026-08-07 18:29:55 UTC

📐 Galois Group of the Cyclotomic Equation
The Galois group of $x^p - 1 = 0$ over $\\mathbb{Q}$ (equivalently, the Galois group of $\\Phi_p(x) = 0$ over $\\mathbb{Q}$) is the cyclic group of order $p - 1$, isomorphic to $(\\mathbb{Z}/p\\mathbb{Z})^*$. Each automorphism sends $a$ to some power $a^j$ where $j \\not\\equiv 0 \\pmod{p}$.
From: gal-edwards
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2026-08-07 16:34:03 UTC

💡 Proposition (Reduction by a Cyclic Extension)
Consider the Galois group of $f(x) = 0$ over a field $K$. Let $p$ be a prime, let $K$ contain primitive $p$th roots of unity, and let $K\
Proof: Let $H(X)$ be the irreducible factor of $F(X)$ over $K\
From: gal-edwards
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2026-08-07 15:36:47 UTC

📖 Perfect Field
A field $F$ is **perfect** if every irreducible polynomial in $F[x]$ is separable. Every field of characteristic $0$ is perfect, and a field of characteristic $p$ is perfect if and only if $F = F^p$ (i.e., the Frobenius is surjective).
From: gal-morandi
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2026-08-07 12:27:58 UTC

2026-08-07 11:31:37 UTC

📐 Division Algorithm
For any two polynomials $f(x)$ and $g(x)$ in $F$ with $g \\neq 0$: $f(x) = q(x) \\cdot g(x) + r(x)$ where $q(x)$ and $r(x)$ are unique and $\\deg(r) < \\deg(g)$.
Proof: Subtract suitable multiples of $g(x)$ from $f(x)$ to reduce the degree. Since the degree decreases at each step, the process terminates with $\\deg(r) < \\deg(g)$. Uniqueness: if $q_1 g + r_1 = q_2 g + r_2$, then $(q_1 - q_2)g = r_2 - r_1$, and the degree constraint forces $q_1 = q_2$, $r_1 = r_2$.
From: gal-artin
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2026-08-07 10:33:34 UTC

📖 Irreducible Polynomial
A polynomial $f \\in F[X]$ of degree $\\geq 1$ is irreducible over $F$ if it cannot be written as a product $f = gh$ with $\\deg(g), \\deg(h) \\geq 1$.
From: gal-weintraub
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2026-08-07 06:53:17 UTC

📐 Invertibility Characterization
A linear map $T \\in \\mathcal{L}(V, W)$ is invertible if and only if $T$ is injective and surjective.
From: linalg-axler
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2026-08-07 00:31:06 UTC

📖 Cyclic Extension
A field extension $K\
From: gal-edwards
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2026-08-06 13:19:48 UTC

2026-08-06 11:07:06 UTC

🎮 Interactive: Quaternion Rotation Demo
Visualize 3D rotations using quaternions. See why video game developers and aerospace engineers prefer quaternions over Euler angles.
From: Four Pillars of Geometry
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2026-08-06 08:06:27 UTC

📐 Galois
A polynomial equation with rational coefficients is solvable by radicals if and only if its Galois group is a solvable group. In particular, for equations of degree 5 or higher, the Galois group may fail to be solvable, which is why no general radical formula exists.
Proof: The proof of this theorem is the goal of the entire book. It requires developing the theory of Galois groups, the connection between field extensions and group theory, and the concept of solvable groups. The full proof appears in Part 6 of Edwards\
From: gal-edwards
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2026-08-06 04:48:53 UTC

🎯 Corollary (Isomorphism of Splitting Fields)
Any two splitting fields for $p(x)$ over $F$ are isomorphic.
Proof: Apply Theorem 10 with $F = F\
From: gal-artin
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2026-08-06 01:05:53 UTC

📐 Cauchy-Schwarz Inequality
If $u, v \\in V$ (inner product space), then $|\\langle u, v \\rangle| \\leq \\|u\\| \\|v\\|$. Equality holds iff one is a scalar multiple of the other.
Proof: If $v = 0$, both sides are 0. Otherwise, let $c = \\langle u, v \\rangle / \\|v\\|^2$. Then $0 \\leq \\|u - cv\\|^2 = \\|u\\|^2 - |\\langle u,v\\rangle|^2/\\|v\\|^2$, giving the result.
From: linalg-axler
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2026-08-05 21:56:05 UTC

📐 Newton
Any symmetric polynomial in the roots of an equation can be expressed in terms of the coefficients of that equation. That is, symmetric functions of the roots are computable without finding the roots themselves.
Proof: The formulas follow immediately from the identity $x^3 + bx^2 + cx + d = (x - r)(x - s)(x - t)$. When the right side is multiplied out and coefficients of like powers of $x$ are equated, one obtains: $r + s + t = -b$, $rs + rt + st = c$, $rst = -d$. Any other symmetric function can be built from ...
From: gal-edwards
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2026-08-05 20:14:30 UTC

📖 Normal Subgroup
A subgroup $H$ of a group $G$ is called normal if for every $S$ in $H$ and every $T$ in $G$, the conjugate $T^{-1}ST$ is also in $H$. Equivalently, $H$ is normal if the various coset presentations of the subgroup differ from one another by a single substitution.
From: gal-edwards
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2026-08-05 18:11:25 UTC

📖 Solution to Noether
A system $\\{x_\\sigma\\}$ indexed by elements of a group $G$ of automorphisms of $E$ is a solution to Noether\
From: gal-artin
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2026-08-05 04:48:34 UTC

📐 Insolvability of the General Quintic
The general polynomial of degree $n \\geq 5$ is not solvable by radicals. The symmetric group $S_n$ for $n \\geq 5$ is not solvable, since $A_n$ is simple for $n \\geq 5$.
From: gal-morandi
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2026-08-04 23:45:13 UTC

2026-08-04 22:44:38 UTC

📐 Tower Law
If $K \\subseteq L \\subseteq M$ are fields, then $[M:K] = [M:L] \\cdot [L:K]$. In particular, $[M:K]$ is finite if and only if both $[M:L]$ and $[L:K]$ are finite.
Proof: Let $\\{e_1, \\ldots, e_m\\}$ be a basis for $M/L$ and $\\{f_1, \\ldots, f_n\\}$ a basis for $L/K$. Then $\\{e_i f_j\\}$ is a basis for $M/K$, giving $[M:K] = mn = [M:L][L:K]$.
From: gal-jacobson
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2026-08-04 20:55:43 UTC

📐 Theorem 19 (Artin)
A polynomial $f$ has repeated roots if and only if $f$ and $f\
From: gal-artin
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2026-08-04 19:14:27 UTC

2026-08-04 17:21:08 UTC

📐 Newton\
The power sums $s_k = r_1^k + r_2^k + \\cdots + r_n^k$ satisfy the recurrence relation: $s_k - s_{k-1}\\sigma_1 + s_{k-2}\\sigma_2 - \\cdots + (-1)^{k-1}s_1\\sigma_{k-1} + (-1)^k k\\sigma_k = 0$ for $k = 1, 2, 3, \\ldots$, where $\\sigma_j = 0$ for $j > n$.
Proof: This recurrence follows from the identity $r_i^n - \\sigma_1 r_i^{n-1} + \\sigma_2 r_i^{n-2} - \\cdots \\pm \\sigma_n = 0$, which holds for each root $r_i$. Summing over $i$ and using the definition of the power sums gives $s_n - \\sigma_1 s_{n-1} + \\sigma_2 s_{n-2} - \\cdots \\pm n\\sigma_n = 0...
From: gal-edwards
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2026-08-04 12:29:35 UTC

📖 Krull Topology
The **Krull topology** on $\\operatorname{Gal}(K/F)$ for an (infinite) Galois extension has as basic open sets the cosets $\\sigma \\cdot \\operatorname{Gal}(K/L)$ where $L/F$ is a finite Galois sub-extension. With this topology, $\\operatorname{Gal}(K/F)$ is a profinite group.
From: gal-morandi
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2026-08-04 10:18:51 UTC

🎮 Interactive: RSA Encryption Demo
See how RSA public-key encryption works. The security relies on the difficulty of factoring large numbers!
From: Cryptography Math
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2026-08-04 03:54:34 UTC

2026-08-03 23:48:19 UTC

📐 Primitive Element Theorem
If $E/F$ is a finite separable extension, then $E = F(\\alpha)$ for some $\\alpha \\in E$.
Proof: If $F$ is infinite, let $E = F(\\alpha, \\beta)$. Choose $c \\in F$ so that $\\gamma = \\alpha + c\\beta$ is a primitive element. This is possible since only finitely many values of $c$ fail. The key step is showing $\\beta \\in F(\\gamma)$ by analyzing gcd arguments with minimal polynomials.
From: gal-weintraub
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2026-08-03 20:47:12 UTC

📖 Transcendence Base and Degree
A subset $S$ of $K$ is **algebraically independent** over $F$ if no element of $S$ is algebraic over $F(S \\setminus \\{s\\})$ for any $s \\in S$. A **transcendence base** is a maximal algebraically independent set. The **transcendence degree** $\\operatorname{tr.deg}(K/F)$ is the cardinality of any transcendence base.
From: gal-morandi
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2026-08-03 19:10:40 UTC

📐 Theorem 7 (Kronecker)
If $f(x)$ is a polynomial in a field $F$, there exists an extension field $E$ of $F$ in which $f(x)$ has a root.
Proof: Factor $f(x)$ into irreducible factors. For an irreducible factor of degree $n$, construct $E_1 = F[\\xi]/(f(\\xi))$, the set of formal polynomials in a symbol $\\xi$ of degree $< n$, with multiplication defined modulo $f(\\xi)$. The irreducibility of $f$ guarantees that $E_1$ is a field in which...
From: gal-artin
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2026-08-03 17:28:34 UTC

🔗 Lemma 1 (LCM of Orders)
In an abelian group, if $A$ and $B$ have orders $a$ and $b$ with $\\mathrm{lcm}(a,b) = c$, then there exists an element of order $c$.
From: gal-artin
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2026-08-03 15:32:41 UTC

📖 Solving by Radicals
An equation is said to be solvable by radicals if its roots can be expressed in terms of its coefficients using only the operations of addition, subtraction, multiplication, division, and the extraction of $n$th roots for various $n$.
From: gal-edwards
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2026-08-03 08:51:10 UTC

📐 Dedekind
Distinct characters $\\chi_1, \\ldots, \\chi_n: G \\to K^\\times$ of a group $G$ are linearly independent over $K$. That is, if $\\sum a_i \\chi_i = 0$ for $a_i \\in K$, then all $a_i = 0$.
From: gal-jacobson
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2026-08-03 05:22:27 UTC

📐 Characterization of Cyclic Extensions
Let $F$ contain a primitive $n$th root of unity. Then $K/F$ is cyclic of degree $n$ if and only if $K = F(\\alpha)$ where $\\alpha^n \\in F$.
From: gal-morandi
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2026-08-02 23:42:23 UTC

📐 Frobenius Automorphism
The Galois group $\\operatorname{Gal}(\\mathbb{F}_{p^n}/\\mathbb{F}_p)$ is cyclic of order $n$, generated by the Frobenius automorphism $\\phi: a \\mapsto a^p$.
From: gal-morandi
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2026-08-02 21:37:28 UTC

2026-08-02 17:44:45 UTC

📐 Simple Algebraic Extensions
Let $K$ be a given field and let $G(X)$ be an irreducible polynomial with coefficients in $K$. Then one can construct a field $K(t)$ such that: (1) $K(t)$ contains $K$, (2) $K(t)$ contains an element $t$ with $G(t) = 0$, and (3) every element of $K(t)$ can be expressed as a polynomial $b_0 + b_1 t + \\cdots + b_\\nu t^\\nu$ where $\\nu < \\deg G$. Moreover, any two such fields are naturally iso...
Proof: Let $R$ be the set of all polynomials in $X$ with coefficients in $K$. Two elements are congruent mod $G$ if their difference is divisible by $G(X)$. The quotient $L$ is a ring. The mapping $k \\mapsto$ [class of constant $k$] embeds $K$ into $L$. The class of $X$ is a root of $G$ in $L$. The Euc...
From: gal-edwards
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2026-08-02 16:39:31 UTC

📐 Abel\u2013Ruffini Theorem
The general polynomial equation of degree $n \\geq 5$ is not solvable by radicals. This follows from the fact that $S_n$ is not solvable for $n \\geq 5$.
From: gal-jacobson
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2026-08-02 15:45:01 UTC

📐 Sufficient Condition for Diagonalizability
If $T \\in \\mathcal{L}(V)$ has $\\dim V$ distinct eigenvalues, then $T$ is diagonalizable.
From: linalg-axler
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2026-08-02 12:43:52 UTC

📖 Derivation
A **derivation** from a ring $R$ to an $R$-module $M$ is a map $D: R \\to M$ satisfying $D(ab) = aD(b) + bD(a)$ (the Leibniz rule). The module of **K\\u00E4hler differentials** $\\Omega_{K/F}$ is the universal target for $F$-derivations from $K$.
From: gal-morandi
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2026-08-01 23:41:28 UTC

📐 Solution by Radicals (Full Version)
Let $f(x) = 0$ be an equation with coefficients in a field $K$ (obtained from $\\mathbb{Q}$ by a finite number of adjunctions). A solution by radicals is a sequence of field extensions $K \\subset K_1 \\subset \\cdots \\subset K_\\mu$ where each $K_i$ is obtained by adjoining a $p_i$th root of an element of $K_{i-1}$ (with $p_i$th roots of unity present in $K_{i-1}$). Such a solution exists if ...
From: gal-edwards
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2026-08-01 22:39:08 UTC

📐 Gauss
Let $d$ and $D$ be divisors of $p - 1$, with $d | D$ and $D/d = q$. Then $K_D$ is a simple algebraic extension of $K_d$ of degree $q$. Any given element of $K_D$ can be expressed rationally in terms of elements of $K_d$, a $q$th root of an element of $K_d$, and a primitive $q$th root of unity.
Proof: The Lagrange resolvent $t = \\gamma + \\beta \\cdot S^d\\gamma + \\beta^2 \\cdot S^{2d}\\gamma + \\cdots + \\beta^{q-1} \\cdot S^{(q-1)d}\\gamma$ where $\\beta$ is a primitive $q$th root of unity satisfies $t^q \\in K_d$, so the extension from $K_d$ to $K_D$ requires only a $q$th root.
From: gal-edwards
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2026-08-01 21:37:18 UTC

📖 Normal Extension
A finite extension $E/F$ is normal if the group $G$ of automorphisms of $E$ fixing $F$ has $F$ as its fixed field. Equivalently, $F$ is exactly the set of elements fixed by all automorphisms leaving $F$ fixed.
From: gal-artin
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2026-08-01 18:42:03 UTC

📐 Existence of Splitting Fields
Every polynomial $f \\in F[X]$ of degree $\\geq 1$ has a splitting field, and any two splitting fields of $f$ over $F$ are isomorphic via an isomorphism fixing $F$.
From: gal-weintraub
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2026-08-01 17:42:37 UTC

📐 Classical Impossibility Results
It is impossible to (1) double the cube ($\\sqrt[3]{2}$ is not constructible since $[\\mathbb{Q}(\\sqrt[3]{2}):\\mathbb{Q}] = 3$), (2) trisect a general angle, or (3) square the circle ($\\pi$ is transcendental).
From: gal-morandi
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2026-08-01 07:51:37 UTC

🎮 Interactive: Angle Preservation in Hyperbolic Geometry
See how the hyperbolic plane preserves angles but distorts distances. Maps that preserve angles are called conformal.
From: Four Pillars of Geometry
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2026-08-01 05:00:09 UTC

2026-08-01 01:15:38 UTC

📐 Existence of Galois Resolvents
For any polynomial equation of degree $n$ with distinct roots, there exist integers $A, B, C, \\ldots$ such that $t = Aa + Bb + Cc + \\cdots$ has $n!$ distinct values under all $n!$ permutations of the roots.
From: gal-edwards
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2026-07-10 04:39:54 UTC

📐 Unique Factorization for Polynomials over $\\mathbb{Z}$
A representation of a polynomial with integer coefficients as a product of irreducibles is unique up to the order of the factors and their signs. That is, if $F_1 F_2 \\cdots F_\\mu = G_1 G_2 \\cdots G_\\nu$ where all factors are irreducible, then $\\mu = \\nu$ and the $G_j$\
From: gal-edwards
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2026-07-09 23:50:29 UTC

📖 Elementary Symmetric Polynomials
The $k$th elementary symmetric polynomial $\\sigma_k$ in variables $r_1, r_2, \\ldots, r_n$ is the sum of all products of $k$ distinct variables chosen from the $r_i$. The relationships $\\sigma_k = (-1)^k b_k$ connect the elementary symmetric polynomials to the coefficients of the equation.
From: gal-edwards
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2026-07-09 22:08:21 UTC

📐 Theorem 10 (Uniqueness of Splitting Fields)
An isomorphism $\\sigma: F \\to F\
Proof: Induction on the number of roots outside $F$. Use Theorem 8 to extend $\\sigma$ to $F(\\alpha) \\to F\
From: gal-artin
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2026-07-09 20:28:53 UTC

📖 The Babylonian Normal Form
The Babylonians commonly solved quadratic equations by reducing them to a normal form: given two numbers $p$ and $s$, find two numbers $x$ and $y$ such that $xy = p$ and $x + y = s$. This is equivalent to solving $x^2 - sx + p = 0$.
From: gal-edwards
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2026-07-09 18:27:12 UTC

📐 Theorem 4
The right column rank, left column rank, right row rank, and left row rank of a matrix are all equal.
Proof: Show $c \\leq r$ by truncating to the first $r$ independent rows (the column rank does not change). Applying the same argument to the transpose gives $r \\leq c$, hence $r = c$. The same reasoning equates all four rank notions.
From: gal-artin
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2026-07-09 16:16:01 UTC

🎮 Interactive: Non-Euclidean Lines Demo
Explore straight lines in hyperbolic geometry. In the Poincare disk, geodesics appear as circular arcs perpendicular to the boundary.
From: Four Pillars of Geometry
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2026-07-09 13:36:24 UTC

📐 Trace is Basis-Independent
$\\operatorname{trace} T = \\operatorname{trace} \\mathcal{M}(T)$ for any basis, where the trace of a matrix is the sum of its diagonal entries.
From: linalg-axler
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2026-07-09 10:14:29 UTC

📖 Symmetric Polynomial in Roots
Let $r, s, t$ be the three roots of a cubic equation $x^3 + bx^2 + cx + d = 0$. The elementary symmetric polynomials are: $r + s + t = -b$, $rs + rt + st = c$, and $rst = -d$. Any symmetric polynomial in the roots can be expressed in terms of these.
From: gal-edwards
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2026-07-09 05:53:17 UTC

📐 Theorem 1
A system of $m$ homogeneous linear equations in $n$ unknowns over a field $F$, with $n > m$, always has a non-trivial solution.
Proof: By induction on $m$. For $m = 0$, all unknowns are free. For the inductive step, use elimination: assuming $a_{11} \\neq 0$, form $m-1$ equations in $n-1 > m-1$ unknowns by subtracting multiples of the first equation. The inductive hypothesis gives a non-trivial solution, which extends to the ful...
From: gal-artin
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2026-07-09 01:20:07 UTC

📖 Adjoint
If $T \\in \\mathcal{L}(V, W)$, the adjoint $T^* \\in \\mathcal{L}(W, V)$ is the unique operator such that $\\langle Tv, w \\rangle = \\langle v, T^*w \\rangle$ for all $v \\in V$, $w \\in W$.
From: linalg-axler
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2026-07-08 23:45:25 UTC

📖 Extension Field
If $E$ is a field and $F$ is a subset of $E$ which itself forms a field under the operations of $E$, then $F$ is a subfield of $E$ and $E$ is an extension of $F$.
From: gal-artin
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2026-07-08 21:48:19 UTC

📐 Galois
Let $f(x) = 0$ be an equation with distinct roots whose Galois group over $K$ is $G$. Then $f(x) = 0$ can be solved by radicals if and only if $G$ is solvable -- that is, has a composition series $G \\supset G_1 \\supset G_2 \\supset \\cdots \\supset G_\\nu = \\{e\\}$ in which each $G_i$ is a normal subgroup of prime index in its predecessor.
Proof: Necessity: If solvable by radicals, the tower of field extensions reduces the Galois group at each step to a normal subgroup of prime index (by the proposition of \u00a744). Taking only steps where the group decreases gives the composition series. Sufficiency: If $G$ is solvable, the proposition ...
From: gal-edwards
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2026-07-08 20:09:20 UTC

📖 Separable Polynomial and Separable Extension
A polynomial is separable if its irreducible factors have no repeated roots. An element is separable if it is a root of a separable polynomial. The extension $E/F$ is separable if every element of $E$ is separable over $F$.
From: gal-artin
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2026-07-08 18:06:33 UTC

2026-07-08 15:53:44 UTC

2026-07-08 13:15:57 UTC

📖 Lagrange Resolvent for the Cyclotomic Equation
For the equation $x^{p} - 1 = 0$ (with $p$ prime), let $\\beta$ be a primitive $(p-1)$st root of unity. The Lagrange resolvent is $t = \\alpha^j + \\beta \\alpha^k + \\beta^2 \\alpha^m + \\cdots$ where the roots are listed in the order determined by a primitive root modulo $p$. The quantity $t^{p-1}$ is a known quantity, expressible in terms of $\\beta$ alone.
From: gal-edwards
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2026-07-08 10:49:18 UTC

📖 Primitive Root of Unity
An $n$th root of unity is a number $\\alpha$ satisfying $\\alpha^n = 1$. It is primitive if no smaller positive power of $\\alpha$ equals 1. For $n = 3$, the primitive cube roots of unity are $\\alpha = (-1 \\pm \\sqrt{-3})/2$.
From: gal-edwards
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2026-07-08 07:58:40 UTC

📐 Norm and Trace via Galois Group
If $K/F$ is Galois with group $G$, then $N_{K/F}(\\alpha) = \\prod_{\\sigma \\in G} \\sigma(\\alpha)$ and $T_{K/F}(\\alpha) = \\sum_{\\sigma \\in G} \\sigma(\\alpha)$.
From: gal-morandi
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2026-07-08 01:11:20 UTC

📐 Kummer Theory
Let $K$ be a field containing a primitive $n$-th root of unity $\\zeta_n$ (with $\\mathrm{char}(K) \\nmid n$). There is a bijection between cyclic extensions $L/K$ of degree dividing $n$ and subgroups of $K^\\times/(K^\\times)^n$, given by $L = K(a^{1/n})$ for $a \\in K^\\times$.
From: gal-jacobson
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2026-07-07 23:36:45 UTC

🔗 Lemma 2 (Maximal Order Divides All)
If $C$ has maximal order $c$ in an abelian group, then $c$ is divisible by the order of every element, so $x^c = 1$ for all elements.
From: gal-artin
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2026-07-07 22:00:59 UTC

2026-07-07 20:31:38 UTC

📖 Trace
If $T \\in \\mathcal{L}(V)$ and $\\lambda_1, \\ldots, \\lambda_n$ are the eigenvalues of $T$ (counted with multiplicity), then $\\operatorname{trace} T = \\lambda_1 + \\cdots + \\lambda_n$.
From: linalg-axler
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2026-07-07 18:36:08 UTC

📖 Dimension
The dimension of a finite-dimensional vector space is the length of any basis of the vector space, denoted $\\dim V$.
From: linalg-axler
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2026-07-07 13:05:04 UTC

📖 Lagrange Resolvent
Given an equation of degree $n$ with roots $x_1, x_2, \\ldots, x_n$, a Lagrange resolvent is a quantity of the form $t = x_1 + \\alpha x_2 + \\alpha^2 x_3 + \\cdots + \\alpha^{n-1} x_n$ where $\\alpha$ is an $n$th root of unity (not necessarily primitive).
From: gal-edwards
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2026-07-07 10:14:57 UTC

📖 Normal Extension
An algebraic extension $K/F$ is **normal** if every irreducible polynomial in $F[x]$ that has a root in $K$ splits completely in $K[x]$. Equivalently, $K$ is a splitting field of a family of polynomials over $F$.
From: gal-morandi
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2026-07-07 05:51:16 UTC

2026-07-07 01:24:13 UTC

📖 Character of a Group
A homomorphism $\\sigma$ from a multiplicative group $G$ into a field $F$ (with $\\sigma(\\alpha) \\neq 0$ for all $\\alpha$) is called a character of $G$ in $F$.
From: gal-artin
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2026-07-06 22:47:30 UTC

📖 Splitting Field
A splitting field of $f \\in F[X]$ is an extension $E/F$ such that $f$ splits completely in $E[X]$ and $E$ is generated over $F$ by the roots of $f$.
From: gal-weintraub
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2026-07-06 21:14:36 UTC

📐 Division Algorithm for Polynomials
If $f, g \\in F[X]$ with $g \\neq 0$, there exist unique $q, r \\in F[X]$ with $f = qg + r$ and $\\deg(r) < \\deg(g)$.
From: gal-weintraub
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2026-07-06 19:23:15 UTC

2026-07-06 17:30:39 UTC

📖 Norm and Trace
For a finite extension $K/F$ and $\\alpha \\in K$, the **norm** $N_{K/F}(\\alpha) = \\det(L_\\alpha)$ and the **trace** $T_{K/F}(\\alpha) = \\operatorname{tr}(L_\\alpha)$, where $L_\\alpha: K \\to K$ is left multiplication by $\\alpha$.
From: gal-morandi
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2026-07-06 11:16:45 UTC

📖 Simple Group
A group $G$ is called simple if it has no normal subgroups other than $\\{e\\}$ and $G$ itself. A simple group of order greater than 1 is an obstruction to solvability: if it appears as a quotient in any composition series, and it is not of prime order, then the group is not solvable. The alternating group $A_5$ (with 60 elements) is the smallest non-abelian simple group.
From: gal-edwards
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2026-07-06 06:10:22 UTC

🎮 Interactive: Riemann Sum Visualizer
Approximate area under a curve with rectangles. Watch how the limit of Riemann sums defines the integral.
From: Calculus I
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2026-07-06 01:26:41 UTC

🎮 Interactive: Eigenvalue Explorer
Find vectors that only get scaled by a matrix. Eigenvalues reveal the fundamental structure of linear transformations.
From: Linear Algebra
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2026-07-05 23:44:33 UTC

📐 Sample Theorem
If $A \\subseteq B$ and $B \\subseteq A$, then $A = B$
Proof: Let $x \\in A$. Since $A \\subseteq B$, we have $x \\in B$ by definition of subset.
Therefore, every element of $A$ is in $B$.
Now, let $y \\in B$. Since $B \\subseteq A$, we have $y \\in A$ by definition.
Therefore, every element of $B$ is in $A$.
Since $A \\subseteq B$ and $B \\subseteq A...
From: gal-howie
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