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Chapter 16: Problem 49

Show that the polynomial \(X^{4}+1\) is irreducible in \(\mathbb{Q}[X]\).

Short Answer

Expert verified
To show that the polynomial \(X^{4}+1\) is irreducible in \(\mathbb{Q}[X]\), we substituted \(X\) with \(Y+1\), resulting in the polynomial \(Y^4 + 4Y^3 + 6Y^2 + 4Y + 2\). By applying Eisenstein's Criterion with \(p=2\), we proved that the transformed polynomial is irreducible in \(\mathbb{Q}[Y]\). Consequently, we concluded that the original polynomial \(X^{4}+1\) is also irreducible in \(\mathbb{Q}[X]\).

Step by step solution

01

Substitute \(X\) with \(Y+1\)

Start by substituting \(X\) with \(Y+1\) to transform the polynomial into a possible polynomial with integer coefficients: $$(Y+1)^{4}+1 = Y^4 + 4Y^3 + 6Y^2 + 4Y + 2$$
02

Apply Eisenstein's Criterion

Now, to prove that the polynomial \(Y^4 + 4Y^3 + 6Y^2 + 4Y + 2\) is irreducible in \(\mathbb{Q}[Y]\), we can apply Eisenstein's Criterion with the prime number \(p=2\). Eisenstein's Criterion states that if: 1. All coefficients of a polynomial with integer coefficients other than the leading coefficient are divisible by a prime number \(p\). 2. The constant term must not be divisible by \(p^2\). 3. The leading coefficient is not divisible by \(p\). Then, the polynomial is irreducible over \(\mathbb{Q}\). The polynomial \(Y^4 + 4Y^3 + 6Y^2 + 4Y + 2\) satisfies all the above conditions for \(p=2\). All coefficients except the leading coefficient are divisible by \(2\), the constant term is not divisible by \(2^2=4\), and the leading coefficient is not divisible by \(2\). Therefore, we can apply Eisenstein's criterion and state that \(Y^4 + 4Y^3 + 6Y^2 + 4Y + 2\) is irreducible in \(\mathbb{Q}[Y]\).
03

Conclude irreducibility in \(\mathbb{Q}[X]\)

Since we have shown that the transformed polynomial \(Y^4 + 4Y^3 + 6Y^2 + 4Y + 2\) is irreducible in \(\mathbb{Q}[Y]\), we can now conclude that the original polynomial \(X^{4}+1\) is also irreducible in \(\mathbb{Q}[X]\) due to the reverse change of variables (replacing \(Y+1\) with \(X\)). Therefore, the polynomial \(X^{4}+1\) is irreducible in \(\mathbb{Q}[X]\).

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Key Concepts

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

Eisenstein's Criterion
Eisenstein's Criterion is a valuable method for determining whether a polynomial is irreducible over the field of rationals, denoted as \(\mathbb{Q}\). Irreducibility is an important concept in algebra which indicates that a polynomial cannot be factored into simpler non-constant polynomials with rational coefficients. To apply Eisenstein's Criterion, a polynomial with integer coefficients needs to satisfy certain conditions with respect to a chosen prime number \(p\).

Here are the criteria that need to be met:
  • All coefficients of the polynomial, except the leading one (the coefficient of the highest degree term), must be divisible by \(p\).
  • The constant term (the term without any variables) should not be divisible by \(p^2\).
  • The leading coefficient must not be divisible by \(p\).
If a polynomial satisfies these conditions, then it is considered irreducible over \(\mathbb{Q}\). This criterion is often used because it provides a straightforward way to ascertain irreducibility, especially when direct factorization is complex.
Polynomials
Polynomials are foundational elements in algebra, consisting of variables and coefficients, all combined using addition, subtraction, multiplication, and non-negative integer exponents. A polynomial generally has the form \(a_nX^n + a_{n-1}X^{n-1} + \ldots + a_1X + a_0\), where \(a_n, a_{n-1},...,a_0\) are coefficients, and \(X\) is a variable.

It's crucial to understand polynomials because they play a critical role in various applications, such as solving equations, modeling real-world situations, and understanding continuity and differentiability in calculus.

Polynomials can be classified by their degree, which is determined by the highest exponent of the variable. For example, a polynomial like \(3X^2 + 2X + 1\) is quadratic because its highest degree is 2. Knowing the degree helps in recognizing patterns and estimating the number of roots the polynomial might have.
Field of Rationals
The field of rationals, denoted as \(\mathbb{Q}\), is the set of all numbers that can be expressed as the quotient \(\frac{a}{b}\) of two integers, where \(a\) and \(b\) are integers and \(beq0\). This field forms a major part of mathematical study and is the setting for many algebraic concepts, including polynomial irreducibility.

The field of rationals is characterized by its straightforward arithmetic properties such as:
  • Closure under addition, subtraction, multiplication, and division (except division by zero).
  • A unique additive identity, which is 0.
  • A unique multiplicative identity, which is 1.
  • The existence of additive inverses (opposites) and multiplicative inverses (reciprocals) for every element, except for zero in the case of multiplication.
Understanding \(\mathbb{Q}\) is essential for exploring deeper algebraic structures, as it serves as a building block for more complex fields and spaces.

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Most popular questions from this chapter

Let \(F\) be a field, and consider the ring of bivariate polynomials \(F[X, Y]\). Show that in this ring, the polynomial \(X^{2}+Y^{2}-1\) is irreducible, provided \(F\) does not have characteristic 2 . What happens if \(F\) has characteristic \(2 ?\)

Let \(E\) be an \(R\) -algebra. For \(\alpha \in E,\) consider the \(\alpha\) -multiplication map on \(E\), which sends \(\beta \in E\) to \(\alpha \beta \in E\). Show that this map is an \(R\) -linear map.

In Exercise 16.41, we saw that 2 factors as \(-i(1+i)^{2}\) in \(\mathbb{Z}[i]\), 16.9 Unique factorization domains (*) where \(1+i\) is irreducible. This exercise examines the factorization in \(\mathbb{Z}[i]\) of prime numbers \(p>2\). Show that: (a) for every irreducible \(\pi \in \mathbb{Z}[i],\) there exists a unique prime number \(p\) such that \(\pi\) divides \(p\); (b) for all prime numbers \(p \equiv 1(\bmod 4),\) we have \(p=\pi \bar{\pi},\) where \(\pi \in \mathbb{Z}[i]\) is irreducible, and the complex conjugate \(\bar{\pi}\) of \(\pi\) is also irreducible and not associate to \(\pi\); (c) all prime numbers \(p \equiv 3(\bmod 4)\) are irreducible in \(\mathbb{Z}[i]\).

Let \(g \in R[X]\) with \(\operatorname{deg}(g)=k \geq 0,\) and let \(x \in R .\) Show that if we evaluate \(g\) at \(X+x,\) writing $$ g(X+x)=\sum_{i=0}^{k} b_{i} X^{i} $$ with \(b_{0}, \ldots, b_{k} \in R,\) then we have $$ i ! \cdot b_{i}=\left(\mathbf{D}^{i}(g)\right)(x) \text { for } i=0, \ldots, k $$

Show that if the factorization of \(f\) over \(F[X]\) into monic irreducibles is \(f=f_{1}^{e_{1}} \cdots f_{r}^{e_{r}},\) and if \(\alpha=[h]_{f} \in F[X] /(f),\) then the minimal polynomial \(\phi\) of \(\alpha\) over \(F\) is \(\operatorname{lcm}\left(\phi_{1}, \ldots, \phi_{r}\right),\) where each \(\phi_{i}\) is the minimal polynomial of \([h]_{f_{i}^{e_{i}}} \in F[X] /\left(f_{i}^{e_{i}}\right)\) over \(F\)
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