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Chapter 6: Problem 13

What is the Galois group over the rationals of the following polynomials: (a) \(x^{4}+2 X^{2}+x+3\) (b) \(X^{4}+3 X^{3}-3 X-2\) (c) \(X^{6}+22 X^{5}-9 X^{4}+12 X^{3}-37 X^{2}-29 X-15\) [Hint: Reduce mod 2, 3, 5.]

Short Answer

Expert verified
In conclusion, the Galois groups over the rationals for the given polynomials are as follows: (a) \(S_4\) (b) \(A_4\) (c) \(S_6\)

Step by step solution

01

Determine the degrees of the given polynomials

The degree of a polynomial is the highest power of its variable. In this case, we have: (a) \(x^4+2x^2+x+3\): Degree 4 (b) \(X^4+3X^3-3X-2\): Degree 4 (c) \(X^6+22X^5-9X^4+12X^3-37X^2-29X-15\): Degree 6
02

Perform reduction modulo small primes

Now, we will reduce each polynomial modulo 2, 3, and 5. (a) Modulo 2: \(x^4+x+1\) Modulo 3: \(x^4+2x^2+x\) Modulo 5: \(x^4+2x^2+x+3\) (b) Modulo 2: \(x^4+x^3+x+1\) Modulo 3: \(x^4+1\) Modulo 5: \(x^4+3x^3+2x+3\) (c) Modulo 2: \(x^6+x^5+x^4+x^3+x+1\) Modulo 3: \(x^6+1\) Modulo 5: \(x^6+2x^5+x^4+2x^3+3x^2+4x\)
03

Compare cycle structures to known Galois groups

Now, we will factor the reduced polynomials in their respective fields and determine the cycle structure for each polynomial: (a) Modulo 2: Irreducible, gives a 4-cycle. Modulo 3: Irreducible, gives a 4-cycle. Modulo 5: Irreducible, gives a 4-cycle. The Galois group over the rationals for (a) is isomorphic to the symmetric group \(S_4\). (b) Modulo 2: Factoring gives: \((x^4+x^3+x+1) = (x^2+x+1)^2\), an even permutation. Modulo 3: Factoring gives: \((x^4+1) = (x^2+1)^2\), an even permutation. Modulo 5: Irreducible, gives a 4-cycle. The Galois group over the rationals for (b) is isomorphic to the alternating group \(A_4\). (c) Modulo 2: Factoring gives: \((x^6+x^5+x^4+x^3+x+1) = (x+1)(x(x^5+1))\), a fixed point and a 5-cycle. Modulo 3: Factoring gives: \((x^6+1) = (x^2+x+1)(x^2+2x+2)(x^2+1)\), a 2-cycle and two 1-cycles. Modulo 5: Irreducible, gives a 6-cycle. The Galois group over the rationals for (c) is isomorphic to the symmetric group \(S_6\). In conclusion, the Galois groups over the rationals for the given polynomials are as follows: (a) \(S_4\) (b) \(A_4\) (c) \(S_6\)

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Key Concepts

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

Polynomial Degree
The degree of a polynomial is the highest power of the variable that appears in the polynomial with a non-zero coefficient. For example, in the polynomial x^4 + 2x^2 + x + 3, the highest power of x is 4, thus the degree of the polynomial is 4. This is an important concept because the degree of a polynomial can tell us a lot about the possible roots of the polynomial, how it behaves at infinity, and the number of potential extreme points (peaks and troughs) on its graph.

In the context of Galois theory, the degree of the polynomial has a direct relationship with the possible size of the Galois group. For instance, a polynomial of degree n will have a Galois group whose size is at most n! (n factorial), which is the number of permutations of n elements. This means that for a quartic polynomial, such as examples (a) and (b) in our exercise, the Galois group will have at most 24 elements, because 4! = 24.
Reduction Modulo Primes
Reduction modulo primes is a technique used in Galois theory to simplify the study of polynomial equations. By 'reducing' a polynomial modulo a prime number, we replace the coefficients of the polynomial with their remainders when divided by that prime number. This process can often simplify the polynomial, making it easier to analyze certain properties.

The advantage of this approach lies in the fact that while the arithmetic gets simpler, some of the essential algebraic properties are retained. For example, if a polynomial factors over the integers, it will also factor over any modular reduction. Reducing modulo various primes, like 2, 3, and 5 in our exercise, can reveal insights into the structure of the Galois group by examining the factorization patterns that emerge in the different modular fields.
Symmetric Group
The symmetric group, often denoted S_n, is the group that consists of all permutations of n objects. It's a key concept in group theory, a branch of mathematics. The size of the symmetric group is n!, as it includes all possible arrangements of n elements. This group plays a central role in Galois theory because the Galois group of a polynomial equation is a subgroup of the symmetric group on the polynomial's roots.

For polynomial (c) in our exercise, the Galois group is isomorphic to S_6, which means it is as large as it can possibly be for a sixth-degree polynomial. This implies complex interactions between the roots, and it generally suggests that the roots cannot be expressed using radicals.
Alternating Group
The alternating group, denoted A_n, is the group of even permutations of a finite set of n elements. An even permutation is one that can be expressed as an even number of two-element swaps, also known as transpositions. The alternating group is a subgroup of the symmetric group S_n, and its order is exactly half the order of S_n, so there are n!/2 elements in A_n.

The importance of the alternating group in Galois theory arises when the polynomial's Galois group can be embedded in A_n. In our exercise, polynomial (b) has a Galois group isomorphic to A_4. This suggests a more structured and less complex relationship between the roots of the polynomial compared to the symmetric group, which allows for the possibility that the roots can be expressed using radicals.
Cycle Structure
Understanding cycle structure is essential when dealing with permutations, as this concept helps us to visualize and work with the action of the Galois group on the roots of the polynomial. A cycle in a permutation is a subset of elements that are permuted amongst themselves in a cyclic fashion. For instance, the permutation (123) moves 1 to 2, 2 to 3, and 3 back to 1.

The cycle type or cycle structure is a way of describing a permutation by breaking it down into disjoint cycles. The cycle structure relates closely to the factorization of polynomials in modular arithmetic, as seen in our exercise. When a polynomial doesn't factor (is irreducible) in a given modulus, it corresponds to an n-cycle in the Galois group. Conversely, when a polynomial factors into linear factors, it corresponds to cycles of length 1 (fixed points), and so on. The cycle structure of a permutation gives us insights into the symmetry properties of the polynomial's roots and is instrumental in determining the isomorphy of the Galois group to either the symmetric or alternating group.

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Most popular questions from this chapter

Let \(x_{1}, x_{2}, \ldots\) be a sequence of algebraically independent elements over the integers Z. For each integer \(n \geqq 1\) define $$ x^{(n)}=\sum_{d \mid n} d x_{d}^{m / d} . $$ Show that \(x_{n}\) can be expressed in terms of \(x^{(a)}\) for \(d \mid n\), with rational coefficients. Using vector notation, we call \(\left(x_{1}, x_{2}, \ldots\right)\) the Witt components of the vector \(x\), and call \(\left(x^{(1)}, x^{(2)}, \ldots\right)\) its ghost components. We call \(x\) a Witt vector. Define the power series $$ f_{x}(t)=\prod_{n \geqslant 1}\left(1-x_{n} t^{n}\right) $$ Show that $$ -t \frac{d}{d t} \log f_{x}(t)=\sum_{n \geqslant 1} x^{(n)} t^{n} $$ [By \(\frac{d}{d t} \log f(t)\) we mean \(f^{\prime}(t) / f(t)\) if \(f(t)\) is a power series, and the derivative \(f^{\prime}(t)\) is taken formally.] If \(x, y\) are two Witt vectors, define their sum and product componentwise with respect to the ghost components, i.e. $$ (x+y)^{(n)}=x^{(n)}+y^{i m} $$ What is \((x+y)\). \(?\) Well, show that $$ f_{x}(t) f_{y}(t)=\prod\left(1+(x+y)_{n} t^{n}\right)=f_{x+y}(t) $$ Hence \((x+y)_{n}\) is a polynomial with integer coefficients in \(x_{1}, y_{1}, \ldots, x_{n}, y_{n}\), Also show that where \(m\) is the least common multiple of \(d, e\) and \(d, e\) range over all integers \(\geq 1\). Thus \((x y)_{n}\) is also a polynomial in \(x_{1}, y_{1} \ldots, x_{n}, y_{n}\) with integer coefficients. The above arguments are due to Witt (oral communication) and differ from those of his original paper. If \(A\) is a commutative ring, then taking a homomorphic image of the polynomial ring over \(Z\) into \(A\), we see that we can define addition and multiplication of Witt vectors with components in \(A\), and that these Witt vectors form a ring \(W(A) .\) Show that \(W\) is a functor, i.e. that any ring homomorphism \(\varphi\) of \(A\) into a commutative ring \(A^{\prime}\) induces a homomorphism \(W(\varphi): W(A) \rightarrow W\left(A^{\prime}\right)\).

Let \(a \neq 0, \neq \pm 1\) be a square-free integer. For each prime number \(p\), let \(K_{p}\) be the splitting field of the polynomial \(X^{\prime}-a\) over Q. Show that \(\left[K_{p}: \mathbf{Q}\right]=p(p-1)\). For each square-free integer \(m>0\), let $$ K_{m}=\prod_{p \mid m} K_{p} $$ be the compositum of all fields \(K_{e}\) for \(p \mid m\). Let \(d_{m}=\left[K_{m}: Q\right]\) be the degree of \(K_{m}\) over Q. Show that if \(m\) is odd then \(d_{m}=\prod_{p \mid m} d_{p}\), and if \(m\) is even, \(m=2 n\) then \(d_{2 n}=d_{n}\) or \(2 d_{n}\) according as \(\sqrt{a}\) is or is not in the field of \(m\) -th roots of unity \(Q\left(\zeta_{m}\right)\).

Let \(K\) be a field of characteristic 0 for simplicity. Let \(\Gamma\) be a finitely generated subgroup of \(K^{*}\). Let \(N\) be an odd positive integer. Assume that for each prime \(p \mid N\) we have $$ \Gamma=\Gamma^{1 / P} \cap K $$ and also that \(\operatorname{Gal}\left(K\left(\mu_{N}\right) / K\right) \approx \mathbf{Z}(N)^{*} .\) Prove the following. (a) \(\Gamma / \Gamma^{N}=\Gamma /\left(\Gamma \cap K^{* N}\right)=\Gamma K^{* N} / K^{\star N}\) (b) Let \(K_{N}=K\left(\mu_{N}\right)\). Then $$ \Gamma \cap K_{N}^{* N}=\Gamma^{N} $$ [Hint: If these two groups are not equal, then for some prime \(p \mid N\) there exists an clement \(a \in \Gamma\) such that $$ a=b^{p} \text { with } b \in K_{N} \text { but } b \notin K \text { . } $$ In other words, \(a\) is not a \(p\) -th power in \(K\) but becomes a \(p\) -th power in \(K_{N}\). The equation \(x^{p}-a\) is irreducible over \(K .\) Show that \(b\) has degree \(p\) over \(K\left(\mu_{p}\right)\), and that \(K\left(\mu_{p}, a^{1 / \rho}\right)\) is not abelian over \(K\), so \(a^{1 / p}\) has degree \(p\) over \(K\left(\mu_{p}\right)\). Finish the proof yourself.] (c) Conclude that the natural Kummer map $$ \Gamma / \Gamma^{N} \rightarrow \operatorname{Hom}\left(H_{\Gamma}(N), \mu_{N}\right) $$ is an isomorphism. (d) Let \(G_{\Gamma}(N)=\operatorname{Gal}\left(K\left(\Gamma^{1 / N}, \mu_{N}\right) / K\right) .\) Then the commutator subgroup of \(G_{\Gamma}(N)\) is \(H_{\Gamma}(N)\), and in particular \(\operatorname{Gal}\left(K_{N} / K\right)\) is the maximal abelian quotient of \(G_{\Gamma}(N) .\)

Let \(k\) be a field of characteristic \(\neq 2 .\) Let \(c \in k, c \notin k^{2} .\) Let \(F=k(\sqrt{c}) .\) Let \(\alpha=a+b \sqrt{c}\) with \(a, b \in k\) and not both \(a, b=0\). Let \(E=F(\sqrt{\alpha})\). Prove that the following conditions are equivalent. (1) \(E\) is Galois over \(k\). (2) \(E=F\left(\sqrt{\alpha^{\prime}}\right)\), where \(\alpha^{\prime}=a-b \sqrt{c}\). (3) Either \(\alpha \alpha^{\prime}=a^{2}-c b^{2} \in k^{2}\) or \(c \alpha \alpha^{\prime} \in k^{2}\). Show that when these conditions are satisfied, then \(E\) is cyclic over \(k\) of degree 4 if and only if \(c \alpha \alpha^{\prime} \in k^{2} .\)

Let \(k\) be a field such that every finite extension is cyclic, and having one extension of degree \(n\) for each integer \(n .\) Show that the Galois group \(G=G\left(k^{a} / k\right)\) is the inverse limit \(\lim \mathbf{Z} / m \mathbf{Z}\), as \(m \mathbf{Z}\) ranges over all ideals of \(\mathbf{Z}\), ordered by inclusion. Show that this limit is isomorphic to the direct product of the limits $$ \prod_{p} \lim _{n \rightarrow \infty} \mathbf{Z} / p^{n} \mathbf{Z}=\prod_{p} \mathbf{Z}_{p} $$ taken over all prime numbers \(p\), in other words, it is isomorphic to the product of all p-adic integers.
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