maths.free › Abstract Algebra › 23. Galois Theory › Applications

Applications

Throughout this section we shall assume that all fields have characteristic zero to ensure that irreducible polynomials do not have multiple roots.

Solvability by Radicals

Throughout this section we shall assume that all fields have characteristic zero to ensure that irreducible polynomials do not have multiple roots. The immediate goal of this section is to determine when the roots of a polynomial \(f(x)\) can be computed with a finite number of operations on the coefficients of \(f(x)\). The allowable operations are addition, subtraction, multiplication, division, and the extraction of \(n\)th roots. Certainly the solution to the quadratic equation, \(a x^2 + b x +c = 0\), illustrates this process: \[\begin{aligned}\end{aligned}\]. The only one of these operations that might demand a larger field is the taking of \(n\)th roots. We are led to the following definition.

An extension field \(E\) of a field \(F\) is an extension by radicals if there exists a chain of subfields \[\begin{aligned}\end{aligned}\] such for \(i = 1, 2, \ldots, r\), we have \(F_i = F_{i - 1}(\alpha_i)\) and \(\alpha_i^{n_i} \in F_{i-1}\) for some positive integer \(n_i\). A polynomial \(f(x)\) is solvable by radicals over \(F\) if the splitting field \(K\) of \(f(x)\) over \(F\) is contained in an extension of \(F\) by radicals. Our goal is to arrive at criteria that will tell us whether or not a polynomial \(f(x)\) is solvable by radicals by examining the Galois group \(f(x)\).

The easiest polynomial to solve by radicals is one of the form \(x^n - a\). As we discussed in , the roots of \(x^n - 1\) are called the nth roots of unity. These roots are a finite subgroup of the splitting field of \(x^n -1\). By , the \(n\)th roots of unity form a cyclic group. Any generator of this group is called a primitive nth root of unity.

Example

The polynomial \(x^n - 1\) is solvable by radicals over \({\mathbb Q}\). The roots of this polynomial are \(1, \omega, \omega^2, \ldots, \omega^{n - 1}\), where \[\begin{aligned}\end{aligned}\]. The splitting field of \(x^n - 1\) over \({\mathbb Q}\) is \({\mathbb Q}(\omega)\).

We will now prove the main theorem about solvability by radicals.

The converse of is also true. For a proof, see any of the references at the end of this chapter.

Condensed — the full section is in Judson, Abstract Algebra: Theory and Applications.

Insolvability of the Quintic

We are now in a position to find a fifth-degree polynomial that is not solvable by radicals. We merely need to find a polynomial whose Galois group is \(S_5\). We begin by proving a lemma.

Lemma

If \(p\) is prime, then any subgroup of \(S_p\) that contains a transposition and a cycle of length \(p\) must be all of \(S_p\).

Proof

Let \(G\) be a subgroup of \(S_p\) that contains a transposition \(\sigma\) and \(\tau\) a cycle of length \(p\). We may assume that \(\sigma = (1 \, 2)\). The order of \(\tau\) is \(p\) and \(\tau^n\) must be a cycle of length \(p\) for \(1 \leq n \lt p\). Therefore, we may assume that \(\mu = \tau^n = (1, 2, i_3, \ldots, i_p)\) for some \(n\), where \(1 \leq n \lt p\) (see in ). Noting that \((1 \, 2)(1, 2, i_3,\ldots, i_p) = (2, i_3, \ldots, i_p)\) and \((2,i_3, \ldots, i_p)^k(1 \, 2)(2,i_3, \ldots, i_p)^{-k} = (1 \, i_k)\), we can obtain all the transpositions of the form \((1n)\) for \(1 \leq n \lt p\). However, these transpositions generate all transpositions in \(S_p\), since \((1 \, j)(1 \, i)(1 \, j) = (i \, j)\). The transpositions generate \(S_p\).

Condensed — the full section is in Judson, Abstract Algebra: Theory and Applications.

The Fundamental Theorem of Algebra

It seems fitting that the last theorem that we will state and prove is the Fundamental Theorem of Algebra. This theorem was first proven by Gauss in his doctoral thesis. Prior to Gauss's proof, mathematicians suspected that there might exist polynomials over the real and complex numbers having no solutions. The Fundamental Theorem of Algebra states that every polynomial over the complex numbers factors into distinct linear factors.

Although our proof was strictly algebraic, we were forced to rely on results from calculus. It is necessary to assume the completeness axiom from analysis to show that every polynomial of odd degree has a real root and that every positive real number has a square root. It seems that there is no possible way to avoid this difficulty and formulate a purely algebraic argument. It is somewhat amazing that there are several elegant proofs of the Fundamental Theorem of Algebra that use complex analysis. It is also interesting to note that we can obtain a proof of such an important theorem from two very different fields of mathematics.

Fields, field extensions, roots of polynomials, and group theory Sage has it all, and so it is possible to carefully study very complicated examples from Galois Theory with Sage.

Condensed — the full section is in Judson, Abstract Algebra: Theory and Applications.

Symbols used here

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

\sigma,\ s,\ \sigma^2

standard deviation, sample s.d., variance
Typical distance from the mean; its square.

\bar{x},\ \mu

sample mean, population mean
Average of the data; average of the whole population.

imaginary unit
i² = −1.

\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.

\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.

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.

자신만의 길을 찾아보세요

Parts of this page are adapted from Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Condensed and re-explained here; errors are ours.

에 더 Abstract Algebra

Groups Subgroups, cosets and Lagrange's theorem Cyclic groups and permutation groups Homomorphisms, normal subgroups and quotient groups Rings and fields Galois theory: why the quintic has no formula