📖 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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📖 Resolvent Equation The auxiliary equation of lower degree that arises in the process of solving a polynomial equation is called its resolvent equation (or simply resolvent). For the quartic, the resolvent is a cubic. This pattern -- solving an equation by reducing it to a resolvent of lower degree -- is the fundamental idea that Lagrange would later analyze in full generality. From: gal-edwards Learn more:

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📖 Eigenvalue A scalar $\\lambda \\in \\mathbf{F}$ is an eigenvalue of $T \\in \\mathcal{L}(V)$ if there exists $v \\in V$ with $v \\neq 0$ such that $Tv = \\lambda v$. From: linalg-axler Learn more:

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📖 Galois Group (Restated) The Galois group of $f(x) = 0$ over $K$ is the group $G$ of all automorphisms of the splitting field $K(a, b, c, \\ldots)$ that leave every element of $K$ fixed. In terms of the roots, it is the group of all permutations of $a, b, c, \\ldots$ that can be extended to automorphisms of the splitting field over $K$. From: gal-edwards Learn more:

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📐 Theorem 8 (Extension of Isomorphisms) An isomorphism $\\sigma: F \\to F\ Proof: Both $F(\\beta)$ and $F\ From: gal-artin Learn more:

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📖 Algebraic and Transcendental Elements An element $\\alpha \\in K$ is **algebraic** over $F$ if it is a root of some nonzero polynomial in $F[x]$. Otherwise, $\\alpha$ is **transcendental** over $F$. The **minimal polynomial** of an algebraic element is the unique monic irreducible polynomial $\\min(F, \\alpha) \\in F[x]$ having $\\alpha$ as a root. From: gal-morandi Learn more:

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📖 Normal Extension A finite extension $E/F$ is normal if the group $G$ of automorphisms of $E$ fixing $F$ has $F$ as its fixed field. Equivalently, $F$ is exactly the set of elements fixed by all automorphisms leaving $F$ fixed. From: gal-artin Learn more:

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📐 Complex Spectral Theorem An operator $T$ on a finite-dimensional complex inner product space is normal if and only if $V$ has an orthonormal basis consisting of eigenvectors of $T$. From: linalg-axler Learn more:

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🎮 Interactive: Caesar Cipher Demo Encrypt messages like Julius Caesar did! Shift letters by a fixed amount. Try breaking it with frequency analysis. From: Cryptography Math Try it:

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📖 Trivial and Non-trivial Solutions A solution $x_1, x_2, \\ldots, x_n$ of a homogeneous system is called non-trivial if not all of the elements are zero. Otherwise it is called trivial. From: gal-artin Learn more:

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📖 Subspace A subset of a vector space is a subspace if it is a subgroup of the additive group and is closed under scalar multiplication by any element of the field. From: gal-artin Learn more:

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📐 Fundamental Theorem of Symmetric Functions (Generalized) $S = F$ and $(E/F) = n!$. Any polynomial in $x_1, \\ldots, x_n$ can be uniquely expressed as a linear combination of $x_1^{\\nu_1} \\cdots x_n^{\\nu_n}$ (with $\\nu_i \\leq i-1$) with coefficients that are polynomials in $a_1, \\ldots, a_n$. From: gal-artin Learn more:

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📐 Cauchy-Schwarz Inequality If $u, v \\in V$ (inner product space), then $|\\langle u, v \\rangle| \\leq \\|u\\| \\|v\\|$. Equality holds iff one is a scalar multiple of the other. Proof: If $v = 0$, both sides are 0. Otherwise, let $c = \\langle u, v \\rangle / \\|v\\|^2$. Then $0 \\leq \\|u - cv\\|^2 = \\|u\\|^2 - |\\langle u,v\\rangle|^2/\\|v\\|^2$, giving the result. From: linalg-axler Learn more:

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📖 The Resolvent Degree Problem When $n = 3$, the resolvent equation has degree $3! = 6$ but actually has degree $2! = 2$ in $X^3$ and is therefore solvable. When $n = 4$, it has degree $4! = 24$ but actually has degree $3! = 6$ in $X^4$. But when $n = 5$, the resolvent is a polynomial of degree 24 in $X^5$ -- harder than the original equation. From: gal-edwards Learn more:

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📐 Tower Law If $K \\subseteq L \\subseteq M$ are fields, then $[M:K] = [M:L] \\cdot [L:K]$. In particular, $[M:K]$ is finite if and only if both $[M:L]$ and $[L:K]$ are finite. Proof: Let $\\{e_1, \\ldots, e_m\\}$ be a basis for $M/L$ and $\\{f_1, \\ldots, f_n\\}$ a basis for $L/K$. Then $\\{e_i f_j\\}$ is a basis for $M/K$, giving $[M:K] = mn = [M:L][L:K]$. From: gal-jacobson 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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🎮 Interactive: Tree Explorer Discover trees: connected graphs with no cycles. Every tree with n vertices has exactly n-1 edges! From: Introduction to Graph Theory Try it:

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📐 Orthogonal Decomposition If $U$ is a finite-dimensional subspace of $V$, then $V = U \\oplus U^\\perp$. From: linalg-axler Learn more:

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📖 Field A field $F$ is a set which, under addition, forms an additive abelian group, and for which the nonzero elements form a multiplicative group. The two operations are connected by the distributive law $a(b+c) = ab + ac$ and $(b+c)a = ba + ca$. From: gal-artin Learn more:

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📐 Non-Solvability of the General $n$th Degree Equation The general $n$th degree equation is solvable by radicals for $n \\leq 4$ and is not solvable by radicals for $n \\geq 5$. From: gal-edwards Learn more:

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