📐 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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📖 Basis A basis of $V$ is a list of vectors in $V$ that is linearly independent and spans $V$. From: linalg-axler Learn more:

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📐 Structure of Roots of Unity To find primitive $n$th roots of unity for an arbitrary positive integer $n$, it suffices to be able to find primitive $p$th roots of unity for each prime $p$. This is because any $n$ can be written as a product of prime powers, and the $p^{n-1}$st root of a primitive $p$th root of unity is a primitive $p^n$th root. Proof: When $n = jk$ where $j$ and $k$ are relatively prime, the product of a primitive $j$th root and a primitive $k$th root is a primitive $n$th root. More generally, if $n = j \\cdot k \\cdots m$ with pairwise coprime factors, the primitive $n$th roots of unity are products of primitive roots of the ... From: gal-edwards Learn more:

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📖 Character of a Group A homomorphism $\\sigma$ from a multiplicative group $G$ into a field $F$ (with $\\sigma(\\alpha) \\neq 0$ for all $\\alpha$) is called a character of $G$ in $F$. From: gal-artin Learn more:

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📐 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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🔗 Lemma 2 (Maximal Order Divides All) If $C$ has maximal order $c$ in an abelian group, then $c$ is divisible by the order of every element, so $x^c = 1$ for all elements. From: gal-artin Learn more:

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📖 Intermediate Fields $K_d$ For each divisor $d$ of $p - 1$, let $K_d$ denote the set of all elements of $\\mathbb{Q}(\\alpha)$ invariant under $S^d$ (replacing $\\alpha$ by $\\alpha^{g^d}$). Then $K_1 = \\mathbb{Q}$ and $K_{p-1} = \\mathbb{Q}(\\alpha)$. The intermediate fields form a chain $\\mathbb{Q} = K_1 \\subset K_d \\subset K_D \\subset K_{p-1} = \\mathbb{Q}(\\alpha)$ whenever $d | D | (p-1)$. From: gal-edwards Learn more:

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📐 Theorem 14 (Artin) If $\\sigma_1, \\ldots, \\sigma_n$ form a group of automorphisms of $E$ with fixed field $F$, then $(E/F) = n$. From: gal-artin Learn more:

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📖 Krull Topology The **Krull topology** on $\\operatorname{Gal}(K/F)$ for an (infinite) Galois extension has as basic open sets the cosets $\\sigma \\cdot \\operatorname{Gal}(K/L)$ where $L/F$ is a finite Galois sub-extension. With this topology, $\\operatorname{Gal}(K/F)$ is a profinite group. From: gal-morandi Learn more:

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📐 Fundamental Theorem of Algebra Every nonconstant polynomial with complex coefficients has a zero in $\\mathbf{C}$. From: linalg-axler Learn more:

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📖 Primitive Root of Unity An $n$th root of unity is a number $\\alpha$ satisfying $\\alpha^n = 1$. It is primitive if no smaller positive power of $\\alpha$ equals 1. For $n = 3$, the primitive cube roots of unity are $\\alpha = (-1 \\pm \\sqrt{-3})/2$. From: gal-edwards Learn more:

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