📖 Separable Polynomial and Separable Extension A polynomial is separable if its irreducible factors have no repeated roots. An element is separable if it is a root of a separable polynomial. The extension $E/F$ is separable if every element of $E$ is separable over $F$. From: gal-artin Learn more:

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📐 Theorem 4 The right column rank, left column rank, right row rank, and left row rank of a matrix are all equal. Proof: Show $c \\leq r$ by truncating to the first $r$ independent rows (the column rank does not change). Applying the same argument to the transpose gives $r \\leq c$, hence $r = c$. The same reasoning equates all four rank notions. From: gal-artin Learn more:

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📐 Trace is Basis-Independent $\\operatorname{trace} T = \\operatorname{trace} \\mathcal{M}(T)$ for any basis, where the trace of a matrix is the sum of its diagonal entries. From: linalg-axler Learn more:

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📖 Lagrange Resolvent for the Cyclotomic Equation For the equation $x^{p} - 1 = 0$ (with $p$ prime), let $\\beta$ be a primitive $(p-1)$st root of unity. The Lagrange resolvent is $t = \\alpha^j + \\beta \\alpha^k + \\beta^2 \\alpha^m + \\cdots$ where the roots are listed in the order determined by a primitive root modulo $p$. The quantity $t^{p-1}$ is a known quantity, expressible in terms of $\\beta$ alone. From: gal-edwards 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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📐 Norm and Trace via Galois Group If $K/F$ is Galois with group $G$, then $N_{K/F}(\\alpha) = \\prod_{\\sigma \\in G} \\sigma(\\alpha)$ and $T_{K/F}(\\alpha) = \\sum_{\\sigma \\in G} \\sigma(\\alpha)$. From: gal-morandi Learn more:

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📐 Galois An irreducible equation of prime degree $p$ is solvable by radicals if and only if all its roots can be expressed as rational functions of any two of them. Equivalently: the Galois group is solvable if and only if it can be represented as affine permutations $j \\mapsto rj + s \\pmod{p}$ where $r \\not\\equiv 0$. From: gal-edwards Learn more:

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📐 Kummer Theory Let $K$ be a field containing a primitive $n$-th root of unity $\\zeta_n$ (with $\\mathrm{char}(K) \\nmid n$). There is a bijection between cyclic extensions $L/K$ of degree dividing $n$ and subgroups of $K^\\times/(K^\\times)^n$, given by $L = K(a^{1/n})$ for $a \\in K^\\times$. From: gal-jacobson 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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📐 Trace is Basis-Independent $\\operatorname{trace} T = \\operatorname{trace} \\mathcal{M}(T)$ for any basis, where the trace of a matrix is the sum of its diagonal entries. From: linalg-axler 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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📖 Dimension The dimension of a finite-dimensional vector space is the length of any basis of the vector space, denoted $\\dim V$. From: linalg-axler Learn more:

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📐 Fundamental Theorem on Symmetric Polynomials Every symmetric polynomial in $r_1, r_2, \\ldots, r_n$ can be expressed as a polynomial in the elementary symmetric polynomials $\\sigma_1, \\sigma_2, \\ldots, \\sigma_n$. Moreover, a symmetric polynomial with integer coefficients can be expressed as a polynomial in $\\sigma_1, \\sigma_2, \\ldots, \\sigma_n$ with integer coefficients. Proof: The proof proceeds by induction on the number of variables. Assume the theorem for $n-1$ variables. Let $G(r_1, \\ldots, r_n)$ be symmetric. Gather terms by powers of $r_n$. Each coefficient $G_i$ is symmetric in $r_1, \\ldots, r_{n-1}$ and can be expressed in terms of the elementary symmetric fu... From: gal-edwards Learn more:

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📖 Lagrange Resolvent Given an equation of degree $n$ with roots $x_1, x_2, \\ldots, x_n$, a Lagrange resolvent is a quantity of the form $t = x_1 + \\alpha x_2 + \\alpha^2 x_3 + \\cdots + \\alpha^{n-1} x_n$ where $\\alpha$ is an $n$th root of unity (not necessarily primitive). From: gal-edwards Learn more:

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📖 Normal Extension An algebraic extension $K/F$ is **normal** if every irreducible polynomial in $F[x]$ that has a root in $K$ splits completely in $K[x]$. Equivalently, $K$ is a splitting field of a family of polynomials over $F$. From: gal-morandi Learn more:

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📐 Simple Algebraic Extensions Let $K$ be a given field and let $G(X)$ be an irreducible polynomial with coefficients in $K$. Then one can construct a field $K(t)$ such that: (1) $K(t)$ contains $K$, (2) $K(t)$ contains an element $t$ with $G(t) = 0$, and (3) every element of $K(t)$ can be expressed as a polynomial $b_0 + b_1 t + \\cdots + b_\\nu t^\\nu$ where $\\nu < \\deg G$. Moreover, any two such fields are naturally iso... Proof: Let $R$ be the set of all polynomials in $X$ with coefficients in $K$. Two elements are congruent mod $G$ if their difference is divisible by $G(X)$. The quotient $L$ is a ring. The mapping $k \\mapsto$ [class of constant $k$] embeds $K$ into $L$. The class of $X$ is a root of $G$ in $L$. The Euc... 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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📖 Splitting Field A splitting field of $f \\in F[X]$ is an extension $E/F$ such that $f$ splits completely in $E[X]$ and $E$ is generated over $F$ by the roots of $f$. From: gal-weintraub 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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📐 Gram-Schmidt Procedure If $v_1, \\ldots, v_m$ is a linearly independent list in $V$, then there exists an orthonormal list $e_1, \\ldots, e_m$ such that $\\operatorname{span}(v_1, \\ldots, v_j) = \\operatorname{span}(e_1, \\ldots, e_j)$ for each $j$. From: linalg-axler Learn more:

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