📐 Galois If an element of the splitting field $K(a, b, c, \\ldots)$ is left fixed by all the automorphisms of the Galois group, then it is in $K$. Equivalently: the elements of $K$ are precisely those elements of the splitting field that are fixed by every element of the Galois group. From: gal-edwards Learn more:

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📖 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 Learn more:

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🔗 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 Learn more:

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📐 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 Learn more:

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📖 Solvable Group A group $G$ is said to be solvable if there exists a sequence of subgroups $G = G_0 \\supset G_1 \\supset G_2 \\supset \\cdots \\supset G_\\nu = \\{e\\}$ in which each $G_i$ is a normal subgroup of $G_{i-1}$ of prime index, and the final subgroup $G_\\nu$ contains only the identity. Such a sequence is called a composition series (with prime factors). From: gal-edwards Learn more:

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📐 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 Learn more:

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📐 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 Learn more:

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📐 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 Learn more:

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📐 Constructibility Criterion A length $\\alpha$ is constructible by straightedge and compass if and only if $\\alpha$ lies in a field extension of $\\mathbb{Q}$ of degree $2^n$ for some $n \\geq 0$. From: gal-weintraub Learn more:

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🎮 Interactive: Linear Regression Fitter Fit a line to data points using least squares. See how regression minimizes the sum of squared residuals. From: Intro to Statistical Learning Try it:

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📐 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 Learn more:

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📐 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 Learn more:

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🎮 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 Try it:

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📐 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 Learn more:

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📖 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 Learn more:

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📐 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 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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📖 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 Learn more:

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📐 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 Learn more:

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📖 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 Learn more:

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