📖 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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📖 Splitting Field Let $K$ be a field and let $f(x) = 0$ be a polynomial with coefficients in $K$. A splitting field of $f$ over $K$ is a field $K(a, b, c, \\ldots)$ containing $K$ and a complete set of roots $a, b, c, \\ldots$ of $f(x) = 0$. By the Corollary, the splitting field equals $K(t)$ for a single Galois resolvent $t$. From: gal-edwards Learn more:

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📖 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 Learn more:

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📐 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 Learn more:

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📖 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 Learn more:

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📐 Eigenvectors are Independent Let $T \\in \\mathcal{L}(V)$. Eigenvectors of $T$ corresponding to distinct eigenvalues are linearly independent. From: linalg-axler Learn more:

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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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📐 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 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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📐 Theorem 20 (Artin) If $E$ is generated from $F$ by a primitive $n$-th root of unity, the Galois group $G$ of $E/F$ is abelian for any $n$, and cyclic if $n$ is prime. From: gal-artin Learn more:

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📐 Gauss The regular $p$-gon (for $p$ prime) can be constructed with ruler and compass if and only if $p - 1$ is a power of 2. That is, if and only if $p$ is a Fermat prime: a prime of the form $2^{2^k} + 1$. Proof: The regular $p$-gon is constructible if and only if $\\cos(2\\pi/p)$ can be expressed in terms of square roots alone. Since $\\cos(2\\pi/p) = (\\alpha + \\alpha^{-1})/2$, it suffices that $\\alpha + \\alpha^{-1}$ be constructible, which holds precisely when the chain of fields from $\\mathbb{Q}$ ... From: gal-edwards Learn more:

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📖 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 Learn more:

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📐 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 Learn more:

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📖 Cofactor The cofactor of $a_{ik}$ is $A_{ik} = (-1)^{i+k}$ times the determinant of the $(n-1) \\times (n-1)$ matrix obtained by deleting the $i$-th row and $k$-th column. From: gal-artin Learn more:

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📖 Transcendental Extension An element $a$ is transcendental over a field $K$ if it satisfies no polynomial equation with coefficients in $K$. The extension $K(a)$ is then a transcendental extension. Polynomials with coefficients in $K(a)$ can be regarded as polynomials in two variables ($a$ and $x$) with coefficients in $K$. From: gal-edwards Learn more:

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📖 Cyclic Extension A field extension $K\ From: gal-edwards Learn more:

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📐 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 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:

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📐 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 Learn more:

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🎯 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 Learn more:

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