📐 Galois Criterion for Solvability A polynomial $f \\in F[X]$ is solvable by radicals if and only if its Galois group $\\operatorname{Gal}(E/F)$ is a solvable group. From: gal-weintraub Learn more:

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📖 Affine Group mod $p$ The affine group modulo a prime $p$ consists of all permutations of $\\{0, 1, \\ldots, p-1\\}$ of the form $j \\mapsto rj + s \\pmod{p}$ where $r \\not\\equiv 0 \\pmod{p}$ and $s \\in \\{0, 1, \\ldots, p-1\\}$. This group has order $p(p-1)$ and is solvable. Its normal subgroup of translations ($r = 1$) is cyclic of order $p$, and the quotient is cyclic of order $p - 1$. From: gal-edwards Learn more:

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📖 Primitive Root of Unity An element $\\epsilon$ is a primitive $n$th root of unity if $\\epsilon^n = 1$ and $\\epsilon$ has order exactly $n$. The roots of $x^n - 1$ are $1, \\epsilon, \\epsilon^2, \\ldots, \\epsilon^{n-1}$. From: gal-artin Learn more:

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📖 Transitive Action A group $G$ acts transitively on a set if for any two elements $a$ and $b$ of the set, there is an element of $G$ that maps $a$ to $b$. A polynomial $f(x)$ with no repeated roots is irreducible over $K$ if and only if the Galois group acts transitively on the roots. From: gal-edwards Learn more:

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📖 Galois Extension A finite extension $K/F$ is **Galois** if $|\\operatorname{Aut}(K/F)| = [K:F]$. Equivalently, $K/F$ is both normal and separable. From: gal-morandi Learn more:

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📐 Fundamental Theorem (Infinite Case) Let $K/F$ be a (possibly infinite) Galois extension. The Galois correspondence gives an inclusion-reversing bijection between intermediate fields and **closed** subgroups of $\\operatorname{Gal}(K/F)$ (in the Krull topology). An intermediate extension $L/F$ is Galois if and only if the corresponding subgroup is normal and closed. From: gal-morandi Learn more:

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📐 Characterization of Formally Real Fields A field $K$ admits an ordering if and only if $-1$ is not a sum of squares in $K$. Such fields are called formally real. From: gal-jacobson Learn more:

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📐 Cayley-Hamilton Theorem If $T \\in \\mathcal{L}(V)$ and $q$ is the characteristic polynomial of $T$, then $q(T) = 0$. From: linalg-axler Learn more:

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📖 Linear Independence Vectors $A_1, \\ldots, A_n$ are independent if the only choice of scalars for which $x_1 A_1 + \\cdots + x_n A_n = 0$ is the trivial one $x_1 = \\cdots = x_n = 0$. From: gal-artin Learn more:

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📖 Extension Field If $E$ is a field and $F$ is a subset of $E$ which itself forms a field under the operations of $E$, then $F$ is a subfield of $E$ and $E$ is an extension of $F$. From: gal-artin Learn more:

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📖 Purely Inseparable Extension An algebraic extension $K/F$ is **purely inseparable** if, for every $\\alpha \\in K$, the minimal polynomial of $\\alpha$ over $F$ has only one distinct root. In characteristic $p$, this means $\\alpha^{p^n} \\in F$ for some $n$. From: gal-morandi Learn more:

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📖 Field Extension A field extension $E/F$ is a pair of fields with $F \\subseteq E$. The degree $[E:F]$ is the dimension of $E$ as an $F$-vector space. From: gal-weintraub Learn more:

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📐 Theorem 1 A system of $m$ homogeneous linear equations in $n$ unknowns over a field $F$, with $n > m$, always has a non-trivial solution. Proof: By induction on $m$. For $m = 0$, all unknowns are free. For the inductive step, use elimination: assuming $a_{11} \\neq 0$, form $m-1$ equations in $n-1 > m-1$ unknowns by subtracting multiples of the first equation. The inductive hypothesis gives a non-trivial solution, which extends to the ful... From: gal-artin Learn more:

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📐 Artin\u2013Schreier Extensions Let $K$ be a field of characteristic $p > 0$. Every cyclic extension of $K$ of degree $p$ has the form $K(\\alpha)$ where $\\alpha^p - \\alpha = a$ for some $a \\in K$. The polynomial $x^p - x - a$ is either irreducible or splits completely over $K$. From: gal-jacobson Learn more:

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🎮 Interactive: Fundamental Theorem of Calculus See how differentiation and integration are inverse operations. The most important theorem in calculus! From: Calculus I Try it:

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📐 Gauss If $g$ and $h$ are polynomials with rational coefficients and $gh$ has integer coefficients, then any coefficient of $g$ times any coefficient of $h$ is an integer. Equivalently: if $F(x)$ is a polynomial with integer coefficients that is irreducible over $\\mathbb{Z}$ and has positive degree, then $F(x)$ is irreducible over $\\mathbb{Q}$. From: gal-edwards Learn more:

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📐 Theorem 15 (Artin) $E$ is a normal extension of $F$ if and only if $E$ is the splitting field of a separable polynomial $p(x)$ in $F$. From: gal-artin Learn more:

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📐 Tower Law If $F \\subseteq K \\subseteq E$ are fields, then $[E:F] = [E:K][K:F]$. Proof: If $\\{\\alpha_i\\}$ is a basis for $E/K$ and $\\{\\beta_j\\}$ is a basis for $K/F$, then $\\{\\alpha_i \\beta_j\\}$ is a basis for $E/F$. The proof involves showing linear independence and spanning. From: gal-weintraub Learn more:

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📖 Symmetric Group The group of all permutations on $n$ objects is the symmetric group $S_n$. It has order $n!$. From: gal-artin Learn more:

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📖 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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