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(a) Let \(G\) be a finite abelian group. Prove that there exists an abelian extension of Q whose Galois group is \(G\). (b) Let \(k\) be a finite extension of \(\mathbf{Q}\), and let \(G\) be a finite abelian group. Prove that there exist infinitely many abelian extensions of \(k\) whose Galois group is \(G\).

Short Answer

Expert verified
To summarize, the existence of abelian extensions of Q and k whose Galois group is G can be shown as follows: (a) There exists an abelian extension of Q whose Galois group is G. This can be proven using the Kronecker-Weber theorem, which states that every finite abelian extension of Q is contained inside a cyclotomic extension of Q. (b) For a finite extension k of Q and a finite abelian group G, there exist infinitely many abelian extensions of k whose Galois group is G. This can be shown by considering the set of primes that are distinct from the primes dividing the exponent of G and the degree [K : k]. These primes are unramified in K, and we can construct distinct abelian extensions L_i of k for each prime p_i by adjoining the p_i-th root of unity to K.

Step by step solution

01

Review Definitions and Properties

Before we dive into the proofs, let's review some key definitions and properties related to abelian groups, abelian extensions, and Galois groups. An abelian group is a group where the group operation is commutative. In other words, if a and b are in the group, then a * b = b * a. An abelian extension K of a field F is a field extension such that the Galois group of K over F, Gal(K/F), is an abelian group. The Galois group of an extension K over F, Gal(K/F), is a group of automorphisms of K that fix F. It measures the degree of non-abelian-ness of the extension.
02

Prove Statement (a)

Now, let's prove statement (a): There exists an abelian extension of Q whose Galois group is G. We will use the Kronecker-Weber theorem to prove this statement. The Kronecker-Weber theorem states that every finite abelian extension of Q is contained inside a cyclotomic extension of Q. A cyclotomic extension of Q is an extension obtained by adjoining a root of unity. Let G be a finite abelian group and let n be the exponent of G. Then, by the Kronecker-Weber theorem, we can find a cyclotomic extension Q(zeta_n) of Q such that Gal(Q(zeta_n)/Q) is isomorphic to G. Hanence, there exists an abelian extension of Q whose Galois group is G.
03

Prove Statement (b)

Now, let's prove statement (b): For a finite extension k of Q and a finite abelian group G, there exist infinitely many abelian extensions of k whose Galois group is G. Let k be a finite extension of Q, and let K be an abelian extension of k whose Galois group is G. We want to show that there exist infinitely many such extensions. Consider the set of primes {p_1, p_2, ...} that are distinct from the primes dividing the exponent of G and the degree [K : k]. These primes are also unramified in K. Now, let L_i be the extension of K obtained by adjoining the p_i-th root of unity to K. Then, L_i is an abelian extension of k, and Gal(L_i/k) is isomorphic to G. Furthermore, for distinct primes p_i and p_j, L_i and L_j are distinct extensions of k. This is because they correspond to different primes, and the p_i-th root of unity generates a different field from the p_j-th root of unity. Therefore, there exist infinitely many abelian extensions of k whose Galois group is G, as desired.

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Key Concepts

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

Galois Group
Understanding the Galois group is essential for studying field extensions in algebra. It is defined as the group of field automorphisms (symmetries of a field) of a field extension that fix the base field. In simpler terms, it’s the set of all possible transformations that you can apply to the elements of the extended field without changing the base field's elements. For example, if we have a field extension of the rational numbers \(\mathbb{Q}\), the Galois group consists of automorphisms that leave the rational numbers unchanged but may permute the other elements of the extension.

When discussing abelian extensions, we’re focusing on the field extensions whose Galois group is abelian, meaning that any two of its automorphisms commute. This property simplifies many aspects of the theory because abelian groups are well-understood and can be broken down into simpler, cyclic components. As an analogy, think of an abelian Galois group as a well-organized team where all members work seamlessly together, allowing for predictable outcomes and analysis.
Kronecker-Weber Theorem
The Kronecker-Weber theorem is a fascinating result in the realm of number theory and algebra. It answers a very specific question: which abelian extensions of the rational numbers \(\mathbb{Q}\) can we obtain? The theorem states that all such finite extensions are found within cyclotomic fields—fields created by adding roots of unity to \(\mathbb{Q}\).

Roots of unity are complex numbers that satisfy the equation \(z^n = 1\), and a cyclotomic field is what you get when you include these roots into \(\mathbb{Q}\). So, if you have an abelian group \(G\) and you're looking for an extension of \(\mathbb{Q}\) with \(G\) as its Galois group, simply turn to the cyclotomic fields. This theorem is a powerful tool, enabling mathematicians to understand the structure and properties of abelian extensions of \(\mathbb{Q}\) by using the well-paved roads laid down by cyclotomic fields.
Cyclotomic Extension
Cyclotomic extensions are essentially musical chords in the symphony of number theory, created by including the \(n\)-th roots of unity into a base field. Roots of unity are solutions to the polynomial equation \(x^n - 1 = 0\), and they form a cyclic group under multiplication. Imagine these roots sitting on the complex plane like evenly spaced dots around a circle.

When you adjoin an \(n\)-th root of unity to the rational numbers, you create a cyclotomic extension. The beauty of these extensions is their predictable structure. Their Galois groups are not only abelian but also have a direct tie to the geometry of those dots on the circle—the symmetries that rotate the circle without disturbing the arrangement of those dots. Cyclotomic extensions are central to many results in number theory, and their well-understood nature makes them ideal for constructing other abelian extensions, as the Kronecker-Weber theorem suggests.

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Most popular questions from this chapter

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 numériques invariables des fonctions rationnelles entières, Mémoires 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 hypothèses concernant les nombres premiers, Acta Arith. \(4(1958)\) pp. \(185-208\)

(a) Let \(a\) be a non-zero integer, \(p\) a prime, \(n\) a positive integer, and \(p \times n\). Prove that \(p \mid \Phi_{n}(a)\) if and only if \(a\) has period \(n\) in \((\mathbf{Z} / p \mathbf{Z})^{*} .\) (b) Again assume \(p X n\) Prove that \(p \mid \Phi_{n}(a)\) for some \(a \in \mathbf{Z}\) if and only if \(p=1\) mod \(n\). Deduce from this that there are infinitely many primes \(\equiv 1\) mod \(n\), a special case of Dirichlet's theorem for the existence of primes in an arithmetic progression.

Let \(F=\mathbf{F}_{p}\) be the prime field of characteristic \(p .\) Let \(K\) be the field obtained from \(F\) by adjoining all primitive \(l\) -th roots of unity, for all prime numbers \(l \neq p\). Prove that \(K\) is algebraically closed. [Hint: Show that if \(q\) is a prime number, and \(r\) an integer \(\geqq 1\), there exists a prime \(l\) such that the period of \(p \bmod l\) is \(q^{r}\), by using the following old trick of Van der Waerden: Let \(l\) be a prime dividing the number $$ b=\frac{p^{r}-1}{p^{r^{-1}}-1}=\left(p^{\boldsymbol{r}^{-1}}-1\right)^{q-1}+q\left(p^{y^{r-1}}-1\right)^{e^{-2}}+\cdots+q $$ If \(I\) does not divide \(p^{\sigma^{-1}}-1\), we are done. Otherwise, \(l=q .\) But in that case \(q^{2}\) does not divide \(b\), and hence there exists a prime \(l \neq q\) such that \(I\) divides \(b\). Then the degree of \(F\left(\zeta_{1}\right)\) over \(F\) is \(q^{\prime}\), so \(K\) contains subfields of arbitrary degree over \(\left.F .\right]\)

Let \(f(X) \in \mathrm{Q}[X]\) be a polynomial of degree \(n\), and let \(K\) be a splitting field of \(f\) over \(Q\). Suppose that \(\operatorname{Gal}(K / Q)\) is the symmetric group \(S_{n}\) with \(n>2\). (a) Show that \(f\) is irreducible over \(\mathbf{Q}\). (b) If \(\alpha\) is a root of \(f\), show that the only automorphism of \(\mathbf{Q}(\alpha)\) is the identity. (c) If \(n \geqq 4\), show that \(\alpha^{n} \notin Q\).

A polynomial \(f(X)\) is said to be reciprocal if whenever \(\alpha\) is a root, then \(1 / x\) is also a root. We suppose that \(f\) has coefficients in a real subfield \(k\) of the complex numbers. If \(f\) is irreducible over \(k\), and has a nonreal root of absolute value 1 , show that \(/\) is reciprocal of even degree

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