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๐Ÿ“ 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: Explore all courses:
๐Ÿ“– Solvable Group (Preview) A group $G$ is solvable if there exists a chain of subgroups $\\{e\\} = G_0 \\triangleleft G_1 \\triangleleft G_2 \\triangleleft \\cdots \\triangleleft G_k = G$ where each $G_i$ is normal in $G_{i+1}$ and each quotient $G_{i+1}/G_i$ is cyclic of prime order. $S_3$ and $S_4$ are solvable, but $S_5$ is not. From: gal-edwards Learn more: Explore all courses:
๐Ÿ“ Division Algorithm For any two polynomials $f(x)$ and $g(x)$ in $F$ with $g \\neq 0$: $f(x) = q(x) \\cdot g(x) + r(x)$ where $q(x)$ and $r(x)$ are unique and $\\deg(r) < \\deg(g)$. Proof: Subtract suitable multiples of $g(x)$ from $f(x)$ to reduce the degree. Since the degree decreases at each step, the process terminates with $\\deg(r) < \\deg(g)$. Uniqueness: if $q_1 g + r_1 = q_2 g + r_2$, then $(q_1 - q_2)g = r_2 - r_1$, and the degree constraint forces $q_1 = q_2$, $r_1 = r_2$. From: gal-artin Learn more: Explore all courses:
๐Ÿ“ Dedekind Distinct characters $\\chi_1, \\ldots, \\chi_n: G \\to K^\\times$ of a group $G$ are linearly independent over $K$. That is, if $\\sum a_i \\chi_i = 0$ for $a_i \\in K$, then all $a_i = 0$. From: gal-jacobson Learn more: Explore all courses:
๐Ÿ“ Existence and Uniqueness of Splitting Fields For any polynomial $f(x) \\in K[x]$ of degree $n \\geq 1$, there exists a splitting field $E$ of $f$ over $K$. Any two splitting fields of $f$ over $K$ are $K$-isomorphic. Moreover, $[E:K] \\leq n!$. From: gal-jacobson Learn more: Explore all courses:
๐Ÿ“ Abel There is no formula that expresses the roots of the general fifth-degree polynomial equation in terms of the coefficients using only addition, subtraction, multiplication, division, and the extraction of $n$th roots. That is, the general quintic is not solvable by radicals. From: gal-edwards Learn more: Explore all courses:
๐Ÿ“ Krull Topology and Infinite Galois Theory If $E/F$ is an algebraic Galois extension (possibly infinite), the Galois group $\\operatorname{Gal}(E/F)$ is a profinite group under the Krull topology. The Fundamental Theorem holds: closed subgroups correspond bijectively to intermediate fields. From: gal-weintraub Learn more: Explore all courses:
๐Ÿ“ Lindemann-Weierstrass Theorem If $\\alpha_1, \\ldots, \\alpha_n$ are algebraic numbers that are linearly independent over $\\mathbb{Q}$, then $e^{\\alpha_1}, \\ldots, e^{\\alpha_n}$ are algebraically independent over $\\mathbb{Q}$. From: gal-morandi Learn more: Explore all courses:
๐Ÿ“ Primitive Element Theorem If $E/K$ is a finite separable extension, then $E = K(\\theta)$ for some $\\theta \\in E$. That is, every finite separable extension is simple. From: gal-jacobson Learn more: Explore all courses:
๐Ÿ“ Kummer Theory Let $F$ be a field containing a primitive $n$-th root of unity. The abelian extensions $E/F$ of exponent dividing $n$ are in bijection with subgroups of $F^\\times / (F^\\times)^n$. Explicitly, $E = F(\\sqrt[n]{a_1}, \\ldots, \\sqrt[n]{a_r})$. From: gal-weintraub Learn more: Explore all courses:
๐Ÿ“– Simple Root of $F(X)$ The polynomial $t$ in the roots is a simple root of its equation $F(X)$ if the polynomials $t, \\phi_1 t, \\phi_2 t, \\ldots$ obtained by distinct permutations of the roots are all numerically distinct (when actual root values are substituted). In modern language, this means $F(X)$ has no repeated roots. From: gal-edwards Learn more: Explore all courses:
๐Ÿ“– Reducible and Irreducible Polynomials A polynomial in $F$ is reducible if it equals the product of two polynomials in $F$ each of degree at least one. Polynomials that are not reducible are called irreducible. From: gal-artin Learn more: Explore all courses:
๐Ÿ“– Symmetric Polynomial in Roots Let $r, s, t$ be the three roots of a cubic equation $x^3 + bx^2 + cx + d = 0$. The elementary symmetric polynomials are: $r + s + t = -b$, $rs + rt + st = c$, and $rst = -d$. Any symmetric polynomial in the roots can be expressed in terms of these. From: gal-edwards Learn more: Explore all courses:
๐Ÿ“ Characteristic is Prime The characteristic of a field is either $0$ or a prime number $p$. Proof: If $\\operatorname{char}(F) = n = ab$ with $1 < a, b < n$, then $(a \\cdot 1)(b \\cdot 1) = n \\cdot 1 = 0$. Since $F$ is a field (hence an integral domain), either $a \\cdot 1 = 0$ or $b \\cdot 1 = 0$, contradicting the minimality of $n$. From: gal-weintraub Learn more: Explore all courses:
๐Ÿ“ Genus and the Riemann-Roch Space For an algebraic function field $K/k$, the **genus** $g$ is defined such that for any divisor $D$, the Riemann-Roch space $L(D)$ is a finite-dimensional $k$-vector space with $\\dim L(D) \\geq \\deg D - g + 1$. From: gal-morandi Learn more: Explore all courses:
๐Ÿ“ Characterization of Normal Subgroups A subgroup $H$ of a group $G$ is normal if and only if for every $S \\in H$ and every $T \\in G$, the element $T^{-1}ST$ is in $H$. Proof: A subgroup $H$ has the property that every presentation differs from the first by a single substitution if and only if for every $T$ in $G$ and every $S$ in $H$, $T^{-1}ST$ is in $H$. In other words, the subgroup is invariant under conjugation by elements of $G$. From: gal-edwards Learn more: Explore all courses:
๐Ÿ“ Fundamental Theorem on Symmetric Functions Every symmetric polynomial in $x_1, \\ldots, x_n$ over a field $F$ can be uniquely expressed as a polynomial in the elementary symmetric polynomials $e_1, \\ldots, e_n$. From: gal-weintraub Learn more: Explore all courses:
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