📐 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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📐 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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📖 Product and Inverse of Automorphisms The product $\\sigma\\tau$ maps $x \\mapsto \\sigma(\\tau(x))$. The inverse $\\sigma^{-1}$ maps $y$ to $x$ where $\\sigma(x) = y$. The identity $1(x) = x$ is the unit automorphism. From: gal-artin Learn more:

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🔗 Linear Dependence Lemma If $v_1, \\ldots, v_m$ is linearly dependent in $V$ and $v_1 \\neq 0$, then there exists $j \\in \\{2, \\ldots, m\\}$ such that $v_j \\in \\operatorname{span}(v_1, \\ldots, v_{j-1})$ and removing $v_j$ does not change the span. From: linalg-axler Learn more:

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📖 Generating System A system $A_1, \\ldots, A_m$ of elements of $V$ is a generating system if each element $A \\in V$ can be expressed as $A = \\sum a_i A_i$ for suitable $a_i \\in F$. From: gal-artin Learn more:

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📐 Kummer Theory Let $F$ contain a primitive $n$th root of unity. There is a bijection between abelian extensions of $F$ of exponent dividing $n$ and subgroups of $F^\\times/(F^\\times)^n$. Specifically, if $K/F$ is abelian of exponent $n$, then $K = F(\\sqrt[n]{a_1}, \\ldots, \\sqrt[n]{a_r})$ for suitable $a_i \\in F^\\times$. From: gal-morandi 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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📖 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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📖 Group of an Equation If $f(x)$ is a polynomial in $F$ and $E$ is its splitting field, the group of automorphisms of $E$ over $F$ is the group of the equation $f(x) = 0$. 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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