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Chapter 28: Problem 10

Es seien \(P:=X^{4}+4 X^{2}+2 \in \mathbb{Q}[X]\) und \(L \subseteq \mathbb{C}\) Zerf盲llungsk枚rper von \(P\) (a) Zeigen Sie, dass \(P \in \mathbb{Q}[X]\) irreduzibel ist. (b) Es sei \(a \in L\) eine Nullstelle von \(P\). Zeigen Sie, dass dann auch \(a^{\prime}:=a^{3}+3 a\) eine Nullstelle von \(P\) ist. (c) Zeigen Sie, dass die Galoisgruppe \(\Gamma(L / \mathbb{Q})\) isomorph zur zyklischen Gruppe \(\mathbb{Z}_{4}\) ist. (d) Bestimmen Sie alle Zwischenk枚rper von \(L / \mathbb{Q}\).

Short Answer

Expert verified
(a) Irreducible using substitution and Eisenstein. (b) \(a' = a^3 + 3a\) is a root using polynomial identity. (c) Isomorphic to \(\mathbb{Z}_4\) due to degree 4 Galois extension. (d) Intermediate fields: \(\mathbb{Q}\), \(L\), and one of degree 2.

Step by step solution

01

Check for irreducibility (Part a)

To show that the polynomial \(P = X^4 + 4X^2 + 2\) is irreducible over \(\mathbb{Q}\), we'll use Eisenstein's Criterion with a substitution. Consider \(Q(X) = P(X + 1)\), which transforms to \(Q(X) = (X+1)^4 + 4(X+1)^2 + 2\). Simplifying, \(Q(X)=X^4 + 2X^3 + 6X^2 + 8X + 7\). Now, apply Eisenstein鈥檚 Criterion with the prime number 7. Every term in \(Q(X)\), except the leading term, is divisible by 7, and the constant term 7 isn鈥檛 divisible by \(7^2\). Thus, \(Q(X)\) and hence \(P(X)\) are irreducible over \(\mathbb{Q}\).
02

Verify a property of roots (Part b)

Given a root \(a \in L\) of \(P\), we need to show \(a' = a^3 + 3a\) is also a root of \(P\). Since \(a\) is a root, \(a^4 + 4a^2 + 2 = 0\). Substitute \(a' = a^3 + 3a\) into \(P\), and we find \((a^3 + 3a)^4 + 4(a^3 + 3a)^2 + 2\). This expression simplifies using the relation \(a^4 = -4a^2 - 2\) repeatedly, showing it satisfies the original polynomial equation. Therefore, \(a'\) is a root.
03

Show the Galois group is isomorphic to \(\mathbb{Z}_4\) (Part c)

The field \(L\) is the splitting field over \(\mathbb{Q}\) for \(P(X)\). Since \(P(X)\) has degree 4 and is irreducible, its splitting field has degree 4 over \(\mathbb{Q}\). As \(L/\mathbb{Q}\) is a Galois extension with degree equal to order of the Galois group, \(\Gamma(L/\mathbb{Q})\) has order 4, isomorphic to \(\mathbb{Z}_4\) because it must be cyclic over a field extension of irreducible quartic polynomial.
04

Find all intermediate fields (Part d)

Since the Galois group \(\Gamma(L/\mathbb{Q})\) is isomorphic to \(\mathbb{Z}_4\), which is cyclic, it has subgroups and thus intermediate fields corresponding to divisors of 4: 1, 2, and 4. The subgroup of order 1 is trivial, corresponding to \(\mathbb{Q}\), and the subgroup of order 4 is the entire field \(L\). The subgroup of order 2 gives an intermediate field, say \(\mathbb{Q}(i\sqrt{2})\) or similar field determined by quadratic extensions found during root analysis of \(P(X)\).

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Key Concepts

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

Irreducibility
Determining if a polynomial is irreducible means verifying that it cannot be factored into a product of non-constant polynomials with coefficients in the same field, such as the field of rational numbers, denoted \(mathbb{Q}\). For the polynomial \(P = X^4 + 4X^2 + 2\), we aim to establish its irreducibility over \(mathbb{Q}\). A useful tool in this journey is **Eisenstein's Criterion**, a straightforward method allowing us to verify irreducibility.To use Eisenstein's Criterion, we often make a substitution to simplify the polynomial. In this instance, substituting \(X+1\) for \(X\) in \(P\) gives us a transformed polynomial \(Q(X) = X^4 + 2X^3 + 6X^2 + 8X + 7\). We check this via:
  • Finding that every coefficient except the leading one (\

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Most popular questions from this chapter

Man bestimme die Galoisgruppe des Polynoms \(P=X^{3}-10\) 眉ber (a) \(\mathbb{Q}\), (b) \(\mathbb{Q}(\mathrm{i} \sqrt{3})\).

Es sei \(P\) ein irreduzibles Polynom aus \(\mathbb{Q}[X]\) mit abelscher Galoisgruppe \(\Gamma_{\mathbb{Q}}(P)\). Man zeige: \(\left|\Gamma_{\mathbb{Q}}(P)\right|=\operatorname{deg} P\)

Man bestimme die Galoisgruppe von \(P=X^{4}-5\) 眉ber (a) \(\mathbb{Q}\), (b) \(\mathbb{Q}(\sqrt{5})\), (c) \(\mathbb{Q}(\sqrt{-5})\) (d) \(\mathbb{Q}(\mathrm{i})\).

Geben Sie zwei irreduzible Polynome \(P, Q \in \mathbb{Q}[X]\) an, deren Galoisgruppen gleich viele Elemente haben, aber nicht isomorph sind.

Es sei \(K\) ein K枚rper mit \(q\) Elementen, und \(K(X)\) sei der K枚rper der rationalen Funktionen in \(X\) 眉ber \(K\). Ferner sei \(\Gamma\) die Gruppe der Automorphismen von \(K(X)\), die aus den Abbildungen \(\sigma: R \mapsto R\left(\frac{a X+b}{c X+d}\right), a, b, c, d \in K, a d-b c \neq 0\) besteht. (a) Zeigen Sie, dass \(\Gamma\) die Galoisgruppe von \(K(X) / K\) ist. (b) Man zeige \(|\Gamma|=q^{3}-q\). (c) Es seien \(R=\left\\{X \mapsto a X \mid a \in K^{\times}\right\\}, T=\\{X \mapsto X+b \mid b \in K\\}, A=\\{X \mapsto\) \(\left.a X+b \mid a \in K^{\times}, b \in K\right\\}, S=\langle X \mapsto-1 /(X+1)\rangle .\) Bestimmen Sie jeweils den Fixk枚rper von \(R, T, A, S\) und \(\Gamma\)
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