📖 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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📐 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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