📖 Determinant If $T \\in \\mathcal{L}(V)$ and $\\lambda_1, \\ldots, \\lambda_n$ are the eigenvalues of $T$ (counted with multiplicity), then $\\det T = \\lambda_1 \\cdots \\lambda_n$. From: linalg-axler Learn more:

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📐 Cramer If the determinant $D$ of the coefficient matrix is nonzero, the system $\\sum_j a_{ij}x_j = b_i$ has the unique solution $D \\cdot x_k = \\sum_i A_{ik} b_i$, where $A_{ik}$ are the cofactors. Proof: Multiply the $i$-th equation by $A_{ik}$ and sum over $i$. By the orthogonality relations of cofactors, $\\sum_i a_{ij} A_{ik} = D$ if $j=k$ and $0$ otherwise. This gives $D \\cdot x_k = \\sum_i A_{ik} b_i$. From: gal-artin 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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📖 Fixed Field For a group $G$ of automorphisms of $K$, the **fixed field** is $K^G = \\{a \\in K : \\sigma(a) = a \\text{ for all } \\sigma \\in G\\}$. From: gal-morandi Learn more:

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📖 Axiom 2: Vanishing for Equal Adjacent Columns The determinant is $0$ if two adjacent columns $A_k$ and $A_{k+1}$ are equal. From: gal-artin Learn more:

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📖 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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📐 Gauss Let $d$ and $D$ be divisors of $p - 1$, with $d | D$ and $D/d = q$. Then $K_D$ is a simple algebraic extension of $K_d$ of degree $q$. Any given element of $K_D$ can be expressed rationally in terms of elements of $K_d$, a $q$th root of an element of $K_d$, and a primitive $q$th root of unity. Proof: The Lagrange resolvent $t = \\gamma + \\beta \\cdot S^d\\gamma + \\beta^2 \\cdot S^{2d}\\gamma + \\cdots + \\beta^{q-1} \\cdot S^{(q-1)d}\\gamma$ where $\\beta$ is a primitive $q$th root of unity satisfies $t^q \\in K_d$, so the extension from $K_d$ to $K_D$ requires only a $q$th root. From: gal-edwards 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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📖 Normal Extension An algebraic extension $E/K$ is normal if every irreducible polynomial in $K[x]$ that has at least one root in $E$ splits completely in $E[x]$. Equivalently, $E$ is the splitting field of some family of polynomials over $K$. From: gal-jacobson Learn more:

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📐 Fundamental Theorem (Infinite Case) Let $E/K$ be a (possibly infinite) Galois extension. There is an inclusion-reversing bijection between intermediate fields $K \\subseteq F \\subseteq E$ and closed subgroups of $\\mathrm{Gal}(E/K)$ (in the Krull topology). Open subgroups correspond to finite extensions of $K$ within $E$. From: gal-jacobson Learn more:

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📐 Irreducible Equations of Prime Degree An irreducible equation of prime degree $p$ is solvable by radicals if and only if each of its roots can be expressed as a rational function (with coefficients in $K$) of any two of the roots. Proof: If solvable, the Galois group is a solvable transitive subgroup of $S_p$. Such a group must be contained in the affine group $x \\mapsto ax + b \\pmod{p}$, in which every element is determined by its effect on any two roots. Conversely, if every root is rational in any two others, the Galois grou... From: gal-edwards Learn more:

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🎯 Transcendence of e and pi The numbers $e$ and $\\pi$ are both transcendental over $\\mathbb{Q}$. In particular, $e$ is transcendental because $e^1 = e$ with $1$ algebraic, and $\\pi$ is transcendental because $e^{i\\pi} = -1$ would give an algebraic relation if $\\pi$ were algebraic. From: gal-morandi Learn more:

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📐 Insolvability of the General Quintic The general polynomial equation of degree $\\geq 5$ is not solvable by radicals. In particular, there exist degree 5 polynomials over $\\mathbb{Q}$ whose Galois group is $S_5$, which is not solvable. From: gal-weintraub Learn more:

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📖 Characteristic The characteristic of a field $F$ is the smallest positive integer $p$ such that $p \\cdot a = 0$ for all $a \\in F$. If no such integer exists, $F$ has characteristic 0. The characteristic is always prime. From: gal-artin Learn more:

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📖 Linearly Disjoint Extensions Subfields $K$ and $L$ of an extension $E/F$ are **linearly disjoint** over $F$ if every subset of $K$ that is $F$-linearly independent remains $L$-linearly independent in $E$ (and vice versa). Equivalently, the natural map $K \\otimes_F L \\to KL$ is injective. From: gal-morandi Learn more:

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📐 Steinitz Every field $K$ has an algebraic closure $\\overline{K}$. Any two algebraic closures of $K$ are $K$-isomorphic (though the isomorphism is not canonical). From: gal-jacobson Learn more:

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📐 Galois Group as Permutation Group If $f(x) \\in K[x]$ has splitting field $E$ over $K$ with roots $\\alpha_1, \\ldots, \\alpha_n$, then $\\mathrm{Gal}(E/K)$ acts faithfully on $\\{\\alpha_1, \\ldots, \\alpha_n\\}$, giving an embedding $\\mathrm{Gal}(E/K) \\hookrightarrow S_n$. The action is transitive iff $f$ is irreducible. From: gal-jacobson Learn more:

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📖 Trace If $T \\in \\mathcal{L}(V)$ and $\\lambda_1, \\ldots, \\lambda_n$ are the eigenvalues of $T$ (counted with multiplicity), then $\\operatorname{trace} T = \\lambda_1 + \\cdots + \\lambda_n$. From: linalg-axler Learn more:

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📐 Artin\u2013Schreier Theorem If $K$ is a field that is not algebraically closed, but whose algebraic closure $\\overline{K}$ is a finite extension of $K$, then $[\\overline{K}:K] = 2$, $K$ is real closed, and $\\overline{K} = K(\\sqrt{-1})$. From: gal-jacobson Learn more:

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📐 Theorem 16: The Fundamental Theorem of Galois Theory For a separable polynomial with splitting field $E$ and Galois group $G$: (1) Each intermediate field $B$ is the fixed field of a subgroup $G_B$ of $G$. (2) $B$ is normal over $F$ iff $G_B$ is normal in $G$, with $\\mathrm{Gal}(B/F) \\cong G/G_B$. (3) $(B/F) = [G : G_B]$ and $(E/B) = |G_B|$. From: gal-artin Learn more:

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