📐 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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📖 Elementary Symmetric Functions For the polynomial $(t - x_1)(t - x_2) \\cdots (t - x_n) = t^n + a_1 t^{n-1} + \\cdots + a_n$, the coefficients $a_i = (-1)^i$ times the sum of all products of $i$ different variables are the elementary symmetric functions. From: gal-artin Learn more:

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📖 Solvable by Radicals A polynomial $f(x) \\in F[x]$ is **solvable by radicals** if there is a tower $F = F_0 \\subset F_1 \\subset \\cdots \\subset F_n$ with $F_n$ containing all roots of $f$, where each $F_{i+1} = F_i(\\alpha_i)$ with $\\alpha_i^{n_i} \\in F_i$ for some $n_i$. From: gal-morandi Learn more:

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📖 Dependence of Characters Characters $\\sigma_1, \\ldots, \\sigma_n$ are dependent if there exist $a_1, \\ldots, a_n \\in F$, not all zero, such that $a_1 \\sigma_1(x) + \\cdots + a_n \\sigma_n(x) = 0$ for each $x \\in G$. Otherwise they are independent. From: gal-artin Learn more:

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📖 Transcendental Extension An element $a$ is transcendental over a field $K$ if it satisfies no polynomial equation with coefficients in $K$. The extension $K(a)$ is then a transcendental extension. Polynomials with coefficients in $K(a)$ can be regarded as polynomials in two variables ($a$ and $x$) with coefficients in $K$. From: gal-edwards Learn more:

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🎮 Interactive: Isometry Group Demo Discover the isometries of 3D space: translations, rotations, reflections, and glide reflections. Every distance-preserving map is one of these! From: Four Pillars of Geometry Try it:

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📐 Kronecker Let $f(x)$ be a polynomial of degree $n > 0$ with integer coefficients. If $f(x)$ has a nontrivial factorization over $\\mathbb{Z}$, then one of the factors has degree $\\leq n/2$. A factor of degree $m$ is determined by its values at $m + 1$ points and must divide $f(j)$ at each integer $j$. This gives a finite search using Lagrange interpolation. From: gal-edwards Learn more:

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🔗 Hensel Let $K$ be a complete valued field with valuation ring $\\mathcal{O}$ and residue field $k$. If $f(x) \\in \\mathcal{O}[x]$ and $\\bar{f}(x) = \\bar{g}(x)\\bar{h}(x)$ in $k[x]$ with $\\gcd(\\bar{g}, \\bar{h}) = 1$, then there exist $g(x), h(x) \\in \\mathcal{O}[x]$ lifting $\\bar{g}, \\bar{h}$ with $f = gh$. From: gal-jacobson Learn more:

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📐 Division Algorithm for Polynomials If $f, g \\in F[X]$ with $g \\neq 0$, there exist unique $q, r \\in F[X]$ with $f = qg + r$ and $\\deg(r) < \\deg(g)$. From: gal-weintraub Learn more:

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📖 Sample Definition A function $f: A \\to B$ is a mapping from set $A$ to set $B$. From: gal-howie Learn more:

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🔗 Lemma (Irreducible Divisibility) If $f(x)$ is irreducible of degree $n$, there do not exist two polynomials each of degree less than $n$ whose product is divisible by $f(x)$. Proof: Suppose $g(x)h(x)$ is divisible by $f(x)$ with $\\deg(g), \\deg(h) < n$. Choose $g$ of minimal degree. Dividing $f$ by $g$ gives $f = qg + r$ with $0 < \\deg(r) < \\deg(g)$. Then $r \\cdot h$ is divisible by $f$, contradicting the minimality of $\\deg(g)$. From: gal-artin 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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📐 Theorem 1: Existence of Factorization Every polynomial with integer coefficients that is not a unit (not $\\pm 1$) can be written as a product of irreducible polynomials with integer coefficients. From: gal-edwards Learn more:

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📐 Lagrange In all the successful methods for solving equations of degree $n \\leq 4$ by radicals, the resolvent has the form of a Lagrange resolvent: $t = x_1 + \\alpha x_2 + \\cdots + \\alpha^{n-1} x_n$ where $\\alpha$ is an $n$th root of unity. From: gal-edwards Learn more:

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📐 Memorylessness For an exponential waiting time, $P(T > s + t \\mid T > s) = P(T > t)$: no matter how long you have waited, the distribution of the remaining wait is unchanged. Proof: $P(T > s+t \\mid T > s) = \\frac{P(T > s+t)}{P(T > s)} = \\frac{e^{-\\lambda(s+t)}}{e^{-\\lambda s}} = e^{-\\lambda t} = P(T > t)$. From: mining-pools 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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📖 Algebraic Function Field An **algebraic function field** $K/k$ in one variable is a finitely generated field extension of transcendence degree $1$. A **place** of $K/k$ is an equivalence class of valuations trivial on $k$. From: gal-morandi 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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📐 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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📖 Irreducible Element (in a Ring) An element $p$ of a ring is irreducible if it is not a unit (not $\\pm 1$ in $\\mathbb{Z}$) and if the only factorizations $p = ab$ have one of $a$, $b$ as a unit. In $\\mathbb{Z}$, irreducible elements are the prime numbers and their negatives. From: gal-edwards Learn more:

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