Galois group of a polynomial - Revision history

R-a-jones at 09:43, 12 June 2024

2024-06-12T09:43:18Z

← Older revision		Revision as of 09:43, 12 June 2024
Line 2:		Line 2:

	{{termrelatedto\|Galois theory}}		{{termrelatedto\|Galois theory}}

			{{basicdef in\|Galois theory}}

	==Definition==		==Definition==

R-a-jones: /* Results helping with the computation of a Galois group */

2024-06-10T17:06:43Z

Results helping with the computation of a Galois group

← Older revision		Revision as of 17:06, 10 June 2024
Line 31:		Line 31:
	* It is natural to ask whether or not the Galois group is a subgroup of the [[alternating group]] <math>A_n</math>, i.e. whether it contains an odd permutation or not. The following result turns out to hold: suppose <math>f</math> is a separable polynomial of degree <math>n</math> in a field <math>K</math>, with the characteristic of <math>K</math> not equal to <math>2</math>. Then <math>\mathrm{Gal}(f/K)</math> is a subgroup of <math>A_n</math> if and only if the discriminant of <math>f</math> is a square in <math>K</math>.		* It is natural to ask whether or not the Galois group is a subgroup of the [[alternating group]] <math>A_n</math>, i.e. whether it contains an odd permutation or not. The following result turns out to hold: suppose <math>f</math> is a separable polynomial of degree <math>n</math> in a field <math>K</math>, with the characteristic of <math>K</math> not equal to <math>2</math>. Then <math>\mathrm{Gal}(f/K)</math> is a subgroup of <math>A_n</math> if and only if the discriminant of <math>f</math> is a square in <math>K</math>.

	* '''"Reduction mod p":''' Suppose we want to find the Galois group of a polynomial <math>f \in \mathbb{Z}[X]</math> over <math>\mathbb{Q}</math>. For <math>p</math> a prime number, we may define <math>\overline{f} \in \mathbb{F}_p[X]</math> to be the polynomial in the finite field with <math>p</math> elements such that <math>f \equiv \overline{f}</math> modulo <math>p</math>. Then, it turns out that <math>\mathrm{Gal}(\overline{f}/\mathbb{F}_p) \leq \mathrm{Gal}(f/\mathbb{Q})</math>. This turns out to be quite helpful - [[Galois group of a polynomial#Over finite fields\|as remarked in the examples section on this page]], the Galois group of a polynomial over the field <math>\mathbb{F}_p</math> has a particular structure (a cyclic group generated by a permutation with a cycle type corresponding to the irreducible factors' degrees) allowing for its computation.		* '''"Reduction mod p":''' Suppose we want to find the Galois group of a polynomial <math>f \in \mathbb{Z}[X]</math> over <math>\mathbb{Q}</math>. For <math>p</math> a prime number, we may define <math>\overline{f} \in \mathbb{F}_p[X]</math> to be the polynomial in the finite field with <math>p</math> elements such that <math>f \equiv \overline{f}</math> modulo <math>p</math>. Then, it turns out that <math>\mathrm{Gal}(\overline{f}/\mathbb{F}_p) \leq \mathrm{Gal}(f/\mathbb{Q})</math>. This turns out to be quite helpful - [[Galois group of a polynomial#Over finite fields\|as remarked in the examples section on this page]], the Galois group of a polynomial over the field <math>\mathbb{F}_p</math> has a particular structure (a cyclic group generated by a permutation with a cycle type corresponding to the irreducible factors' degrees) allowing for its computation. One may thus iterate through the primes to gain information on what cycle types are present in the Galois group over <math>\mathbb{Q}</math>, which may determine what the group is.

	==Applications==		==Applications==

R-a-jones: /* Results helping with the computation of a Galois group */

2024-06-10T17:05:16Z

Results helping with the computation of a Galois group

← Older revision		Revision as of 17:05, 10 June 2024
Line 31:		Line 31:
	* It is natural to ask whether or not the Galois group is a subgroup of the [[alternating group]] <math>A_n</math>, i.e. whether it contains an odd permutation or not. The following result turns out to hold: suppose <math>f</math> is a separable polynomial of degree <math>n</math> in a field <math>K</math>, with the characteristic of <math>K</math> not equal to <math>2</math>. Then <math>\mathrm{Gal}(f/K)</math> is a subgroup of <math>A_n</math> if and only if the discriminant of <math>f</math> is a square in <math>K</math>.		* It is natural to ask whether or not the Galois group is a subgroup of the [[alternating group]] <math>A_n</math>, i.e. whether it contains an odd permutation or not. The following result turns out to hold: suppose <math>f</math> is a separable polynomial of degree <math>n</math> in a field <math>K</math>, with the characteristic of <math>K</math> not equal to <math>2</math>. Then <math>\mathrm{Gal}(f/K)</math> is a subgroup of <math>A_n</math> if and only if the discriminant of <math>f</math> is a square in <math>K</math>.

	* '''"Reduction mod p":''' Suppose we want to find the Galois group of a polynomial <math>f \in \mathbb{Z}[X]</math> over <math>\mathbb{Q}</math>. For <math>p</math> a prime number, we may define <math>\overline{f} \in \mathbb{F}_p[X]</math> to be the polynomial in the finite field with <math>p</math> elements such that <math>f \equiv \overline{f}</math> modulo <math>p</math>. Then, it turns out that <math>\mathrm{Gal}(\overline{f}/\mathbb{F}_p) \leq \mathrm{Gal}(f/\mathbb{Q})</math>. This turns out to be quite helpful - as remarked in the examples section on this page, the Galois group of a polynomial over the field <math>\mathbb{F}_p</math> has a particular structure (a cyclic group generated by a permutation with a cycle type corresponding to the irreducible factors' degrees) allowing for its computation.		* '''"Reduction mod p":''' Suppose we want to find the Galois group of a polynomial <math>f \in \mathbb{Z}[X]</math> over <math>\mathbb{Q}</math>. For <math>p</math> a prime number, we may define <math>\overline{f} \in \mathbb{F}_p[X]</math> to be the polynomial in the finite field with <math>p</math> elements such that <math>f \equiv \overline{f}</math> modulo <math>p</math>. Then, it turns out that <math>\mathrm{Gal}(\overline{f}/\mathbb{F}_p) \leq \mathrm{Gal}(f/\mathbb{Q})</math>. This turns out to be quite helpful - [[Galois group of a polynomial#Over finite fields\|as remarked in the examples section on this page]], the Galois group of a polynomial over the field <math>\mathbb{F}_p</math> has a particular structure (a cyclic group generated by a permutation with a cycle type corresponding to the irreducible factors' degrees) allowing for its computation.

	==Applications==		==Applications==

R-a-jones at 17:04, 10 June 2024

2024-06-10T17:04:13Z

← Older revision		Revision as of 17:04, 10 June 2024
Line 10:		Line 10:

	==Examples==		==Examples==

			===Particular examples===

	* The polynomial <math>f(X) = (X^2-2)(X^2-3) \in \Q[X]</math> has splitting field <math>\Q(\sqrt{2}, \sqrt{3})</math> over <math>\Q</math>. It can be shown that this field extension has Galois group isomorphic to the [[Klein four-group]].		* The polynomial <math>f(X) = (X^2-2)(X^2-3) \in \Q[X]</math> has splitting field <math>\Q(\sqrt{2}, \sqrt{3})</math> over <math>\Q</math>. It can be shown that this field extension has Galois group isomorphic to the [[Klein four-group]].

	* The polynomial <math>X^n-X-1 \in \Q[X]</math> has Galois group the [[symmetric group]] on <math>n</math> letters.		* The polynomial <math>X^n-X-1 \in \Q[X]</math> has Galois group the [[symmetric group]] on <math>n</math> letters.

			===Over finite fields===

	* Let <math>p</math> be a prime, and <math>\mathbb{F}_p</math> denote the [[finite field]] with <math>p</math> elements. Let <math>f \in \mathbb{F}_p[X]</math> be a separable polynomial that factors into irreducible factors of degrees <math>n_1, \dots n_r</math>, and denote the degree of <math>f</math> as <math>n := \sum_i n_i</math>. Then the Galois group of <math>f</math> over <math>\mathbb{F}_p</math> viewed as a subgroup of the symmetric group <math>S_n</math> is generated by a single permutation with cycle type <math>(n_1, \dots, n_r)</math>. Thus, it is [[cyclic group\|cyclic]] of order <math>\mathrm{lcm}(n_1, \dots, n_r)</math>.		* Let <math>p</math> be a prime, and <math>\mathbb{F}_p</math> denote the [[finite field]] with <math>p</math> elements. Let <math>f \in \mathbb{F}_p[X]</math> be a separable polynomial that factors into irreducible factors of degrees <math>n_1, \dots n_r</math>, and denote the degree of <math>f</math> as <math>n := \sum_i n_i</math>. Then the Galois group of <math>f</math> over <math>\mathbb{F}_p</math> viewed as a subgroup of the symmetric group <math>S_n</math> is generated by a single permutation with cycle type <math>(n_1, \dots, n_r)</math>. Thus, it is [[cyclic group\|cyclic]] of order <math>\mathrm{lcm}(n_1, \dots, n_r)</math>.
Line 19:		Line 23:
	{{fillin}}		{{fillin}}

	==Applications ~~and results~~==		==Results helping with the computation of a Galois group==

			One has that a separable polynomial of degree <math>n</math> has its Galois group a subgroup of the [[symmetric group]] <math>S_n</math>.

			* The Galois group of a polynomial is a [[Transitive subgroup of symmetric group\|transitive subgroup]] of <math>S_n</math> if and only if the polynomial is irreducible.

			* It is natural to ask whether or not the Galois group is a subgroup of the [[alternating group]] <math>A_n</math>, i.e. whether it contains an odd permutation or not. The following result turns out to hold: suppose <math>f</math> is a separable polynomial of degree <math>n</math> in a field <math>K</math>, with the characteristic of <math>K</math> not equal to <math>2</math>. Then <math>\mathrm{Gal}(f/K)</math> is a subgroup of <math>A_n</math> if and only if the discriminant of <math>f</math> is a square in <math>K</math>.

			* '''"Reduction mod p":''' Suppose we want to find the Galois group of a polynomial <math>f \in \mathbb{Z}[X]</math> over <math>\mathbb{Q}</math>. For <math>p</math> a prime number, we may define <math>\overline{f} \in \mathbb{F}_p[X]</math> to be the polynomial in the finite field with <math>p</math> elements such that <math>f \equiv \overline{f}</math> modulo <math>p</math>. Then, it turns out that <math>\mathrm{Gal}(\overline{f}/\mathbb{F}_p) \leq \mathrm{Gal}(f/\mathbb{Q})</math>. This turns out to be quite helpful - as remarked in the examples section on this page, the Galois group of a polynomial over the field <math>\mathbb{F}_p</math> has a particular structure (a cyclic group generated by a permutation with a cycle type corresponding to the irreducible factors' degrees) allowing for its computation.

			==Applications==

	===Insolvability of the quintic===		===Insolvability of the quintic===

R-a-jones at 11:49, 9 December 2023

2023-12-09T11:49:11Z

← Older revision		Revision as of 11:49, 9 December 2023
Line 28:		Line 28:

	Every subgroup of [[symmetric group:S4]] is solvable, which tells us that the quadratic formula, the cubic formula and the quartic formula do indeed exist, i.e. such polynomials are always solvable in radicals.		Every subgroup of [[symmetric group:S4]] is solvable, which tells us that the quadratic formula, the cubic formula and the quartic formula do indeed exist, i.e. such polynomials are always solvable in radicals.

			==See also==

			* [[Galois group]]

R-a-jones: /* Applications and results */

2023-12-09T11:48:31Z

Applications and results

← Older revision		Revision as of 11:48, 9 December 2023
Line 24:		Line 24:

	It can be shown that an irreducible polynomial <math>f</math> is solvable in terms of radicals if and only if its Galois group is a [[solvable group]]. But there exist degree five polynomials with Galois group [[alternating group:A5]] and [[symmetric group:S5]], which are not solvable groups. Thus in general, the quintic is not solvable in radicals.		It can be shown that an irreducible polynomial <math>f</math> is solvable in terms of radicals if and only if its Galois group is a [[solvable group]]. But there exist degree five polynomials with Galois group [[alternating group:A5]] and [[symmetric group:S5]], which are not solvable groups. Thus in general, the quintic is not solvable in radicals.

			A particular polynomial over <math>\Q</math> that is not solvable in terms of radicals is <math>X^5-X+1</math>, which has Galois group [[symmetric group:S5]].

	Every subgroup of [[symmetric group:S4]] is solvable, which tells us that the quadratic formula, the cubic formula and the quartic formula do indeed exist, i.e. such polynomials are always solvable in radicals.		Every subgroup of [[symmetric group:S4]] is solvable, which tells us that the quadratic formula, the cubic formula and the quartic formula do indeed exist, i.e. such polynomials are always solvable in radicals.

R-a-jones: /* Applications and results */

2023-12-09T11:46:52Z

Applications and results

← Older revision		Revision as of 11:46, 9 December 2023
Line 25:		Line 25:
	It can be shown that an irreducible polynomial <math>f</math> is solvable in terms of radicals if and only if its Galois group is a [[solvable group]]. But there exist degree five polynomials with Galois group [[alternating group:A5]] and [[symmetric group:S5]], which are not solvable groups. Thus in general, the quintic is not solvable in radicals.		It can be shown that an irreducible polynomial <math>f</math> is solvable in terms of radicals if and only if its Galois group is a [[solvable group]]. But there exist degree five polynomials with Galois group [[alternating group:A5]] and [[symmetric group:S5]], which are not solvable groups. Thus in general, the quintic is not solvable in radicals.

	Every subgroup of [[symmetric group:S4]] is solvable, which tells us that the quadratic formula, the cubic formula and the quartic formula do indeed exist.		Every subgroup of [[symmetric group:S4]] is solvable, which tells us that the quadratic formula, the cubic formula and the quartic formula do indeed exist, i.e. such polynomials are always solvable in radicals.

R-a-jones: /* Applications and results */

2023-12-09T11:46:34Z

Applications and results

← Older revision		Revision as of 11:46, 9 December 2023
Line 24:		Line 24:

	It can be shown that an irreducible polynomial <math>f</math> is solvable in terms of radicals if and only if its Galois group is a [[solvable group]]. But there exist degree five polynomials with Galois group [[alternating group:A5]] and [[symmetric group:S5]], which are not solvable groups. Thus in general, the quintic is not solvable in radicals.		It can be shown that an irreducible polynomial <math>f</math> is solvable in terms of radicals if and only if its Galois group is a [[solvable group]]. But there exist degree five polynomials with Galois group [[alternating group:A5]] and [[symmetric group:S5]], which are not solvable groups. Thus in general, the quintic is not solvable in radicals.

			Every subgroup of [[symmetric group:S4]] is solvable, which tells us that the quadratic formula, the cubic formula and the quartic formula do indeed exist.

R-a-jones: /* Insolvability of the quintic */

2023-12-09T11:44:19Z

Insolvability of the quintic

← Older revision		Revision as of 11:44, 9 December 2023
Line 23:		Line 23:
	===Insolvability of the quintic===		===Insolvability of the quintic===

	It can be shown that an irreducible polynomial <math>f</math> is solvable in terms of radicals if and only if its Galois group is a [[solvable group]]. But there exist degree five polynomials with Galois group ~~<math>A_5</math>~~ and ~~<math>S_5</math>~~, which are not solvable groups. Thus in general, the quintic is not solvable in radicals.		It can be shown that an irreducible polynomial <math>f</math> is solvable in terms of radicals if and only if its Galois group is a [[solvable group]]. But there exist degree five polynomials with Galois group [[alternating group:A5]] and [[symmetric group:S5]], which are not solvable groups. Thus in general, the quintic is not solvable in radicals.

R-a-jones at 11:43, 9 December 2023

2023-12-09T11:43:35Z

← Older revision		Revision as of 11:43, 9 December 2023
Line 18:		Line 18:

	{{fillin}}		{{fillin}}

			==Applications and results==

			===Insolvability of the quintic===

			It can be shown that an irreducible polynomial <math>f</math> is solvable in terms of radicals if and only if its Galois group is a [[solvable group]]. But there exist degree five polynomials with Galois group <math>A_5</math> and <math>S_5</math>, which are not solvable groups. Thus in general, the quintic is not solvable in radicals.