📖 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 Learn more:

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📖 Axiom 3: Normalization The determinant is $1$ when the columns are the standard unit vectors $U_1, U_2, \\ldots, U_n$. From: gal-artin Learn more:

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📐 Theorem 5 A system of $n$ non-homogeneous linear equations in $n$ unknowns has a unique solution if and only if the corresponding homogeneous system has only the trivial solution. Proof: If the non-homogeneous system has two solutions, their difference solves the homogeneous system non-trivially. Conversely, if the homogeneous system has only the trivial solution, the column vectors are independent and form a generating system, so the non-homogeneous system has a unique solution. From: gal-artin Learn more:

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📖 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 Learn more:

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📐 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 Learn more:

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🎮 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 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: Matrix Transformation Demo Watch how matrices transform the plane. Rotations, reflections, shears, and stretches are all matrix multiplications! From: Linear Algebra Try it:

Introduction to Linear Algebra | Math Academy

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📖 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 Learn more:

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📐 Theorem 11 (Unique Factorization of Polynomials) If $p(x) = p_1(x) \\cdots p_r(x) = q_1(x) \\cdots q_s(x)$ are two factorizations into irreducible polynomials, then $r = s$ and after reordering, $p_i(x) = c_i q_i(x)$ for constants $c_i \\in F$. Proof: Let $\\alpha$ be a root of $p_1$ in an extension (Kronecker). Then $q_j(\\alpha) = 0$ for some $j$, so $q_j$ and $p_1$ share a root and both are irreducible and monic, forcing $p_1 = q_1$ (after reindexing). Cancel and repeat. From: gal-artin Learn more:

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📖 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 Learn more:

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📖 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 Learn more:

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📐 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 Learn more:

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🎮 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 Try it:

Introduction to Linear Algebra | Math Academy

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📐 Galois When the function $t$ has been chosen so that it has $n!$ different values when the roots are permuted, it has the property that all the roots of the equation can be expressed rationally in terms of $t$. From: gal-edwards Learn more:

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📖 Separable Polynomial A polynomial $f \\in F[X]$ is separable if it has no repeated roots in any extension of $F$. An algebraic element is separable if its minimal polynomial is separable. From: gal-weintraub Learn more:

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📐 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 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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📐 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 Learn more:

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📐 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 Learn more:

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📖 The Resolvent Degree Problem When $n = 3$, the resolvent equation has degree $3! = 6$ but actually has degree $2! = 2$ in $X^3$ and is therefore solvable. When $n = 4$, it has degree $4! = 24$ but actually has degree $3! = 6$ in $X^4$. But when $n = 5$, the resolvent is a polynomial of degree 24 in $X^5$ -- harder than the original equation. From: gal-edwards Learn more:

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