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Galois theory: why the quintic has no formula

Field extensions, the Galois group, and solvability by radicals.

A polynomial's roots generate a field; the symmetries of that field fixing the base form the Galois group. A polynomial is solvable by radicals exactly when its Galois group is solvable — and S₅ is not. Picture it: the cubic's three roots permuted by S₃; the quintic's five roots have too much symmetry to untangle. Think it: Galois turned a question about formulas into a question about groups — the template for all of modern algebra.

İşlənmiş nümunə: x^3 + x + 1 = 0

Algebra · Polynomial-Equations

Solve x^3 + x + 1 = 0

x^{3} + x + 1 = 0

Addım-addım

  1. x^{3} + x + 1 = 0

    Start from the equation as given.

  2. x^{3} + x + 1 = 0

    The polynomial has degree 3 and no rational factors; the roots come from the general solution.

  3. x = \frac{1}{\left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{27}{2} + \frac{3 \sqrt{93}}{2}}} - \frac{\left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{27}{2} + \frac{3 \sqrt{93}}{2}}}{3} ,\; x = - \frac{\left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{27}{2} + \frac{3 \sqrt{93}}{2}}}{3} + \frac{1}{\left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{\frac{27}{2} + \frac{3 \sqrt{93}}{2}}} ,\; x = - \frac{\sqrt[3]{\frac{27}{2} + \frac{3 \sqrt{93}}{2}}}{3} + \frac{1}{\sqrt[3]{\frac{27}{2} + \frac{3 \sqrt{93}}{2}}} \approx -0.68233

    All 3 roots.

  4. x \approx 0.34116411 + 1.16154099 i ,\; x \approx 0.34116411 - 1.16154099 i ,\; x \approx -0.68232822

    The exact forms are unwieldy (Cardano-style radicals); here they are numerically.

Cavabı göstər
x \approx 0.34116411 + 1.16154099 i ,\; x \approx 0.34116411 - 1.16154099 i ,\; x \approx -0.68232822

Symbols used here

\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
i
imaginary unit
i² = −1.
\approx
approximately equal
Equal to the precision shown, not exactly.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}
number sets
Naturals, integers, rationals, reals, complex numbers.
x \in A,\ A \subseteq B
element of, subset
x belongs to A; every element of A is in B.
\blacksquare\ \text{or}\ \square
end of proof (halmos)
Marks the point where the statement has been established.
a \equiv b \pmod n
congruent modulo n
n divides a − b; a and b have the same remainder.
a \mid b,\ \gcd(a,b)
divides, greatest common divisor
b is a multiple of a; the largest number dividing both.
(G, \cdot),\ e,\ g^{-1}
group, identity, inverse
A set with an operation; the do-nothing element; the element that undoes g.
G \cong H,\ G / N
isomorphic, quotient group
Same structure; the group of cosets of a normal subgroup N.
\mathbb{Z}/n\mathbb{Z},\ \mathbb{Z}_n
integers modulo n
The remainders 0…n−1 with clock arithmetic.
\operatorname{Hom}(A, B),\ f \circ g
arrows from A to B, composition
The set of morphisms; do g then f.

How to: Galois theory: why the quintic has no formula

  1. Start from the equation as given.
  2. The polynomial has degree 3 and no rational factors; the roots come from the general solution.
  3. All 3 roots.
  4. The exact forms are unwieldy (Cardano-style radicals); here they are numerically.

Questions people ask

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group — that is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

Özün sına

Daha çox Abstract Algebra

GroupsSubgroups, cosets and Lagrange's theoremCyclic groups and permutation groupsHomomorphisms, normal subgroups and quotient groupsRings and fields