📐 Factorization over Transcendental Extensions If there is a factorization method for polynomials with coefficients in a field $K$, then there is also a factorization method for polynomials with coefficients in $K(a)$ where $a$ is transcendental over $K$. The method uses the substitution $x \\mapsto x + ua$ to reduce to factoring polynomials in two variables. From: gal-edwards Learn more:

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📖 Solvable by Radicals A polynomial $f(x)$ in $F$ is solvable by radicals if its splitting field lies in an extension of $F$ by radicals. From: gal-artin Learn more:

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📐 Galois Criterion for Solvability A polynomial $f \\in F[X]$ is solvable by radicals if and only if its Galois group $\\operatorname{Gal}(E/F)$ is a solvable group. From: gal-weintraub Learn more:

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📐 Kummer Theory Let $F$ be a field containing a primitive $n$-th root of unity. The abelian extensions $E/F$ of exponent dividing $n$ are in bijection with subgroups of $F^\\times / (F^\\times)^n$. Explicitly, $E = F(\\sqrt[n]{a_1}, \\ldots, \\sqrt[n]{a_r})$. From: gal-weintraub Learn more:

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📖 Division with Remainder Given polynomials $a(X)$ and $b(X) \\neq 0$ with coefficients in a field $K$, there exist unique polynomials $q(X)$ (the quotient) and $r(X)$ (the remainder) such that $a(X) = q(X) b(X) + r(X)$ and either $r(X) = 0$ or $\\deg r < \\deg b$. From: gal-edwards Learn more:

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🎮 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 Try it:

The Four Pillars of Geometry | Magic Internet Math

Explore geometry from four perspectives: Euclidean constructions, coordinates, projective geometry, and transformations. Based on John Stillwell

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🎮 Interactive: Quadratic Residue Checker Test which numbers are perfect squares modulo a prime. Visualize the pattern that led Gauss to call quadratic reciprocity the "golden theorem." From: Disquisitiones Arithmeticae Try it:

Disquisitiones Arithmeticae | Math Academy

Gauss

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📐 Galois Group of $x^n - 1 = 0$ over $\\mathbb{Q}$ The splitting field of $x^n - 1$ over $\\mathbb{Q}$ is $\\mathbb{Q}(a)$ where $a$ is a primitive $n$th root of unity. The Galois group is $\\text{Gal}(x^n - 1 = 0 / \\mathbb{Q}) \\cong (\\mathbb{Z}/n\\mathbb{Z})^*$, the multiplicative group of integers modulo $n$. Each automorphism sends $a \\mapsto a^j$ where $\\gcd(j, n) = 1$. From: gal-edwards Learn more:

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📐 Automorphism Group Bound For any finite extension $K/F$, we have $|\\operatorname{Aut}(K/F)| \\leq [K:F]$. From: gal-morandi Learn more:

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🎮 Interactive: Diffie-Hellman Key Exchange Watch two parties create a shared secret over a public channel. The magic of discrete logarithms makes this possible! From: Cryptography Math Try it:

Cryptography Mathematics | Magic Internet Math

Learn the mathematics that powers modern encryption, from modular arithmetic to elliptic curves.

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📐 Theorem 19 (Artin) A polynomial $f$ has repeated roots if and only if $f$ and $f\ From: gal-artin Learn more:

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📐 Determination of the Fixed Field For the six automorphisms of $k(x)$ given by the cross-ratio substitutions, $F = S = k(I)$ where $I$ is the cross-ratio invariant, and $(E/F) = 6$. From: gal-artin Learn more:

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📖 p-Basis In characteristic $p$, a **$p$-basis** of $K/F$ is a subset $B \\subseteq K$ such that $K = F(K^p)(B)$ and $B$ is $p$-independent over $F(K^p)$. The cardinality of a $p$-basis equals $\\dim_K \\Omega_{K/F}$. From: gal-morandi Learn more:

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🎮 Interactive: Matching Visualizer Find maximum matchings in bipartite graphs. This algorithm solves the stable marriage problem and job assignment! From: Introduction to Graph Theory Try it:

Introduction to Graph Theory | Math Academy

Interactive course based on Douglas B. West

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📖 Primitive Root of Unity An $n$th root of unity $\\alpha$ is called primitive if $\\alpha^k \\neq 1$ for $0 < k < n$. Equivalently, $\\alpha$ is primitive if the full list $\\alpha, \\alpha^2, \\ldots, \\alpha^n = 1$ gives all $n$ roots of unity. When $n$ is prime, every root of unity other than 1 is primitive. From: gal-edwards Learn more:

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🔗 Lemma 2 (Independence of Roots) Let $p$ be a prime, let $\\alpha$ be a primitive $p$th root of unity, and let $\\beta$ be a primitive $(p-1)$st root of unity. If $P_1(\\beta)\\alpha + P_2(\\beta)\\alpha^2 + \\cdots + P_{p-1}(\\beta)\\alpha^{p-1} = 0$ where the $P_i$ are polynomials with rational coefficients, then $P_1(\\beta) = P_2(\\beta) = \\cdots = P_{p-1}(\\beta) = 0$. Proof: The fact that changing $\\alpha$ to $\\alpha^2$ changes $t$ to $\\beta^{-1}t$ and therefore leaves $t^{p-1}$ unchanged shows that all coefficients $p_1(\\beta), p_2(\\beta), \\ldots$ are equal. Call this common value $p(\\beta)$. Then $t^{p-1} = p_0(\\beta) + p(\\beta)(\\alpha + \\alpha^2 + \\cdo... From: gal-edwards Learn more:

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📖 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 Learn more:

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📖 Division with Remainder Given polynomials $a(X)$ and $b(X) \\neq 0$ with coefficients in a field $K$, there exist unique polynomials $q(X)$ (the quotient) and $r(X)$ (the remainder) such that $a(X) = q(X) b(X) + r(X)$ and either $r(X) = 0$ or $\\deg r < \\deg b$. From: gal-edwards Learn more:

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📖 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 Learn more:

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📖 Permutation A permutation is a one-to-one mapping of a set of $n$ objects onto itself. The set of all such mappings forms a group under composition. From: gal-artin Learn more:

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