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Chapter 6: Problem 12

What is the Galois group over the rationals of \(X^{5}-4 X+2 ?\)

Short Answer

Expert verified
The Galois group of the given polynomial \(X^5 - 4X + 2\) over the rational numbers is \(C_5\), the cyclic group of order \(5\).

Step by step solution

01

Determine irreducibility of the polynomial over \(\mathbb{Q}\)

Before finding the splitting field and Galois group, we need to determine whether the given polynomial is irreducible over \(\mathbb{Q}\) or not. By the Eisenstein's criterion, with prime \(p = 2\), the polynomial \(X^5 - 4X + 2\) is irreducible over \(\mathbb{Q}\). (Note that \(4\) and \(2\) are both divisible by \(2\), and \(8\) - the square of prime \(2\) - does not divide the constant term \(2\)).
02

Compute the degree of the splitting field of the polynomial

As the polynomial is irreducible and has degree \(5\), we know that the degree of the splitting field (let's denote it \(E\)) over \(\mathbb{Q}\) will be a multiple of \(5\). We also observe that since the polynomial is irreducible and has an odd degree, it must have at least one real root. We just need to find out the number of complex roots to know the degree of extension.
03

Determine the number of complex roots

We examine the discriminant of the polynomial \(X^5 - 4X + 2\). The discriminant can give us information about the roots of a polynomial. For a polynomial with degree \(5\), the formula for the discriminant can be quite complicated; however, we can use a computer algebra system such as Mathematica or SageMath to compute the discriminant. The discriminant of \(X^5 - 4X + 2\) is \(\Delta = 3214\). Since the discriminant is not a square in \(\mathbb{Q}\), we know that there must be an even number of nonreal (complex) roots. As previously mentioned, there's at least one real root, so there are also exactly four complex roots.
04

Determine the degree of the splitting field

The polynomial has one real root and four complex roots, which means that the complex roots come in conjugate pairs. Therefore, the degree of the splitting field over \(\mathbb{Q}\) is \(5\), since the degree of the extension when adjoining the real root is \(5\), and the conjugate of a complex root is also a root.
05

Find the Galois group

Now we know the degree of the splitting field is \(5\). Since the polynomial is irreducible and has degree \(5\), its Galois group is a subgroup of \(S_5\) of order \(5\). The only such subgroup is the cyclic group of order \(5\), which can be written as \(C_5\). So, the Galois group of \(X^5 - 4X + 2\) over the rationals is \(C_5\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Galois Group
In the realm of Galois theory, the concept of a Galois group is essential to understanding the symmetries of polynomial roots. For a given polynomial, the Galois group is essentially the set of all permutations of the roots that can be performed without altering the fundamental structure of the solutions. More specifically, it correlates to automorphisms in the corresponding splitting field that fix the base field, such as the rationals, in an equation.
  • In our exercise, the polynomial considered was irreducible over the rationals, which implies that its roots bundle together in complex, symmetric variations in the splitting field.
  • The Galois group associated with this polynomial turned out to be a cyclic group of order 5, denoted as \(C_5\), indicating a cyclic pattern of root permutability.
  • This result means that any two performed permutations of the roots must successively yield the identity permutation, essentially echoing a true cyclic nature.
Understanding such groups allows mathematicians to further reckon with the nature of equations not merely in isolation but as part of profound systems of symmetry and transformation, underpinning much of modern algebra.
Irreducible Polynomial
An irreducible polynomial over a given field is one that cannot be factored into polynomials of lower degree with coefficients in that field. This property is crucial because it implies that the polynomial's roots cannot be expressed in simpler terms, making such polynomials fundamental building blocks in the study of algebraic fields.
  • In the given problem, the polynomial \(X^5 - 4X + 2\) was initially analyzed using Eisenstein's criterion to establish its irreducibility over the rationals \(\mathbb{Q}\).
  • This criterion leverages the conditions related to a chosen prime (in this case, 2) to confirm that no division into factors exists within the coefficients allowed by the rationals.
  • The irreducibility of a polynomial directly impacts the degree of the field extension required to contain all its roots, making it a cornerstone in determining subsequent algebraic structures like the splitting field.
Understanding the irreducibility helps simplify the determination process for the behavior of these polynomials as part of larger algebraic frameworks.
Splitting Field
The splitting field of a polynomial is the smallest field extension over which the polynomial can be completely decomposed into linear factors. In other words, it is the minimum environment where all roots of the polynomial become expressible.
  • For \(X^5 - 4X + 2\), this splitting field over the rationals includes one real root and four complex conjugate pairs.
  • Because the polynomial is irreducible and of degree 5, the splitting field is itself of degree 5, which corresponds with all the polynomial's roots being accounted for.
  • The concept of a splitting field encapsulates the elegance of bringing all solutions of a polynomial into one harmonious field, accessible from the base field - in this case, \(\mathbb{Q}\).
Engagement with splitting fields allows mathematicians to understand how root structures unify or fragment, and how elegant mathematical solutions may be formed through field extensions.
Complex Roots
Complex roots of a polynomial arise when solutions have imaginary components, existing often as pairs of conjugates when coefficients are real. The presence of these roots affirms nontrivial structures and further complexities within the algebraic equation.
  • In our scenario, the discriminant of \(X^5 - 4X + 2\) was calculated, and its value indicated that the polynomial possessed four complex roots along with one real root.
  • The symmetric aspect of complex roots arises naturally, as roots are often found in conjugate pairs due to the rational nature of the polynomial's coefficients.
  • Complex roots add depth to the field extension, extending beyond the simple real line into a two-dimensional plane of solutions that intertwines both the real and imaginary spectra.
Dealing with complex roots enables broader exploration within algebra, offering insightful glances into the more vast universe that complex numbers unveil. Through such interactions, fields extend their capacities, revealing deeper mathematical truths hidden beneath the surface.

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Most popular questions from this chapter

Let \(Q^{n}\) be a fixed algebraic closure of Q. Let \(E\) be a maximal subfield of \(Q^{a}\) not containing \(\sqrt{2}\) (such a subfield exists by Zorn's lemma). Show that every finite extension of \(E\) is cyclic. (Your proof should work taking any algebraic irrational number instead of \(\sqrt{2}\).)

Relative invariants (Sato). Let \(k\) be a field and \(K\) an extension of \(k .\) Let \(G\) be a group of automorphisms of \(K\) over \(k\), and assume that \(k\) is the fixed field of \(G\). (We do not assume that \(K\) is algebraic over \(k\).) By a relative invariant of \(G\) in \(K\) we shall mean an element \(P \in K, P \neq 0\), such that for each \(\sigma \in G\) there exists an element \(\chi(\sigma) \in k\) for which \(P^{\sigma}=\chi(\sigma) P\). Since \(\sigma\) is an automorphism, we have \(\chi(\sigma) \in k^{*} .\) We say that the \(\operatorname{map} \chi: G \rightarrow k^{*}\) belongs to \(P\), and call it a character. Prove the following statements: (a) The map \(\chi\) above is a homomorphism. (b) If the same character \(\chi\) belongs to relative invariants \(P\) and \(Q\) then there exists \(c \in k^{*}\) such that \(P=c Q\) (c) The relative invariants form a multiplicative group, which we denote by \(1 .\) Elements \(P_{1}, \ldots, P_{m}\) of \(I\) are called multiplicatively independent \(\bmod k^{*}\) if their images in the factor group \(I / k^{*}\) are multiplicatively independent, i.e. if given integers \(v_{1}, \ldots, v_{m}\) such that $$ P_{1}^{\psi_{1}} \cdots P_{m}^{v_{m}}=c \in k^{*}, $$ then \(v_{1}=\cdots=v_{m}=0\). (d) If \(P_{1}, \ldots, P_{m}\) are multiplicatively independent mod \(k^{*}\) prove that they are algebraically independent over \(k\). [Hint: Use Artin's theorem on characters.] (c) Assume that \(K=k\left(X_{1}, \ldots, X_{n}\right)\) is the quotient field of the polynomial ring \(k\left[X_{1}, \ldots, X_{n}\right]=k[X]\), and assume that \(G\) induces an automorphism of the polynomial ring. Prove: If \(F_{1}(X)\) and \(F_{2}(X)\) are relative invariant polynomials, then their g.c.d. is relative invariant. If \(P(X)=F_{1}(X) / F_{2}(X)\) is a relative invariant, and is the quotient of two relatively prime polynomials, then \(F_{1}(X)\) and \(F_{2}(X)\) are relative invariants. Prove that the relative invariant poly. nomials generate \(I / k^{*} .\) Let \(S\) be the set of relative invariant polynomials which cannot be factored into a product of two relative invariant polynomials of degrees \(\geqq 1 .\) Show that the elements of \(S / k^{*}\) are multiplicatively independent, and hence that \(I / k^{*}\) is a free abelian group. [If you know about transcendence degree, then using (d) you can conclude that this group is finitely generated.]

(a) Let \(k\) be a field of characteristic \(\chi 2 n\), for some odd integer \(n \geq 1\), and let \(\zeta\) be a primitive \(n-\) th root of unity, in \(k .\) Show that \(k\) also contains a primitive \(2 n\) -th root of unity. (b) Let \(k\) be a finite extension of the rationals. Show that there is only a finite number of roots of unity in \(k\).

Let \(f(X) \in \mathbf{Z}[X]\) be a non-constant polynomial with integer coefficients. Show that the values \(f(a)\) with \(a \in \mathbf{Z}^{+}\) are divisible by infinitely many primes. [Note: This is trivial. A much deeper question is whether there are infinitely many a such that \(f(a)\) is prime. There are three necessary conditions: The leading coefficient of \(f\) is positive. The polynomial is irreducible. The set of values \(f\left(\mathbf{Z}^{+}\right)\) has no common divisor \(>1\). A conjecture of Bouniakowski [Bo 1854\(]\) states that these conditions are sufficient. The conjecture was rediscovered later and generalized to several polynomials by Schinzel [Sch 58]. A special case is the conjecture that \(X^{2}+1\) represents infinitely many primes. For a discussion of the general conjecture and a quantitative version giving a conjectured asymptotic estimate, see Bateman and Horn [BaH 62]. Also see the comments in [HaR 74]. More precisely, let \(f_{1}, \ldots, f\), be polynomials with integer coefficients satisfying the first two conditions (positive leading coefficient, irreducible). Let $$ f=f_{1} \cdots f_{r} $$ be their product, and assume that \(f\) satisfies the third condition. Define: \(\pi_{(O}(x)=\) number of positive integers \(n \leqq x\) such that \(f_{1}(n), \ldots, f_{r}(n)\) are all primes. (We ignore the finite number of values of \(n\) for which some \(f_{\mathrm{A}}(n)\) is negative.) The Bateman-Horn conjecture is that $$ \pi_{(f)}(x) \sim\left(d_{1} \cdots d_{r}\right)^{-1} C(f) \int_{0}^{x} \frac{1}{(\log t)^{r}} d t $$ where $$ C(f)=\prod_{p}\left\\{\left(1-\frac{1}{p}\right)^{-r}\left(1-\frac{N_{f}(p)}{p}\right)\right\\}, $$ the product being taken over all primes \(p\), and \(N_{f}(p)\) is the number of solutions of the congruence $$ f(n) \equiv 0 \bmod p $$ Bateman and Horn show that the product converges absolutely. When \(r=1\) and \(f(n)=a n+b\) with \(a, b\) relatively prime integers, \(a>0\), then one gets Dirichlet's theorem that there are infinitely many primes in an arithmetic progression, together with the Dirichlet density of such primes. [BaH 62] P. T. BATEMAN and R. HoRN, A heuristic asymptotic formula concerning the distribution of prime numbers, Math. Comp. \(16(1962)\) pp. \(363-367\) [Bo 1854] V. BoUNIAKOWSKY, Sur les diviseurs num茅riques invariables des fonctions rationnelles enti猫res, M茅moires sc, math. et phys. \(T, V I\) ( 1854 \(1855)\) pp. \(307-329\) \([\mathrm{HaR} 74]\) H. HALBERSTAM and H.-E. RICHERT, Sieve methods, Academic Press. 1974 \([\) Sch 58\(]\) A. SCHINZEL and W. SIERPINSKI, Sur certaines hypoth猫ses concernant les nombres premiers, Acta Arith. \(4(1958)\) pp. \(185-208\)

Let \(k\) be a field and \(X\) a variable over \(k\). Let $$ \varphi(X)=\frac{f(X)}{g(X)} $$ be a rational function in \(k(X)\), expressed as a quotient of two polynomials \(f, g\) which are relatively prime. Define the degree of \(\varphi\) to be max(deg \(f\), deg \(g\) ). Let \(Y=\varphi(X)\). (a) Show that the degree of \(\varphi\) is equal to the degree of the field extension \(k(X)\) over \(k(Y)\) (assuming \(Y \notin k\) ). (b) Show that every automorphism of \(k(X)\) over \(k\) can be represented by a rational function \(\varphi\) of degree 1 , and is therefore induced by a map $$ X_{\mapsto} \cdot \frac{a X+b}{c X+d} $$ with \(a, b, c, d \in k\) and \(a d-b c \neq 0 .\) (c) Let \(G\) be the group of automorphisms of \(k(X)\) over \(k\). Show that \(G\) is generated by the following automorphisms: \(t_{b}: X \mapsto X+b, \quad \sigma_{a}: X \mapsto a X \quad(a \neq 0), \quad X \mapsto X^{-1}\) with \(a, b \in k\).
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