📖 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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📐 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 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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📖 Normal Extension A finite extension $E/F$ is normal if every irreducible polynomial in $F[X]$ having a root in $E$ splits completely in $E[X]$. From: gal-weintraub 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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🎮 Interactive: Derivative Visualizer See how the derivative measures instantaneous rate of change. Watch the tangent line move along any curve. From: Calculus I Try it:

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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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💡 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 Learn more:

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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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📐 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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📖 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 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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