Problem 22 - Abstract Algebra: Theory and Applications

Question

Let \(F\) be a field of characteristic zero and let \(f(x) \in F[x]\) be a separable polynomial of degree \(n .\) If \(E\) is the splitting field of \(f(x),\) let \(\alpha_{1}, \ldots, \alpha_{n}\) be the roots of \(f(x)\) in \(E .\) Let \(\Delta=\prod_{i

Step-by-Step Solution

Answer

Question: Determine the discriminant \(\Delta^2\) for the quadratic polynomial \(f(x) = x^2 + bx + c\) and cubic polynomial \(f(x) = x^3 + px + q\). Show that \(\Delta^2\) is in the field \(F\) for any field \(F\). If \(\sigma\) is a transposition of the roots of a polynomial, show that \(\sigma(\Delta) = -\Delta\). If \(\sigma\) is an even permutation of the roots of a polynomial, show that \(\sigma(\Delta) = \Delta\). Prove that \(\Delta \in F\) if and only if the Galois group \(G(E/F)\) of the polynomial is a subgroup of the alternating group \(A_n\). Determine the Galois group of the cubic polynomials \(x^3+2x-4\) and \(x^3+x-3\). Answer: The discriminant \(\Delta^2\) for the quadratic polynomial \(f(x) = x^2 + bx + c\) is \(b^2 - 4c\), and the discriminant \(\Delta^2\) for the cubic polynomial \(f(x) = x^3 + px + q\) is \(-4p^3 - 27q^2\). We have shown that \(\Delta^2\) is in the field \(F\) for any field \(F\). When \(\sigma\) is a transposition of the roots, it holds that \(\sigma(\Delta) = -\Delta\), and when \(\sigma\) is an even permutation of the roots, it holds that \(\sigma(\Delta) = \Delta\). The isomorphism condition states that \(\Delta \in F\) if and only if the Galois group \(G(E/F)\) is a subgroup of the alternating group \(A_n\). For the cubic polynomials \(x^3+2x-4\) and \(x^3+x-3\), the Galois groups are both isomorphic to \(S_3\).

Step 1: Calculate the roots of the polynomial

We calculate the roots using the quadratic formula: \(\alpha_i = \frac{-b \pm \sqrt{b^2-4c}}{2}\). So, we have \(\alpha_1 = \frac{-b + \sqrt{b^2-4c}}{2}\) and \(\alpha_2 = \frac{-b - \sqrt{b^2-4c}}{2}\).

Step 2: Calculate \(\Delta\)

Now, according to the definition of \(\Delta\), we have \(\Delta = (\alpha_1 - \alpha_2) = \sqrt{b^2 - 4c}\).

Step 3: Calculate \(\Delta^2\)

Squaring \(\Delta\), we get \(\Delta^2 = (b^2 - 4c)\). Hence, we have shown that \(\Delta^2 = b^2 - 4c\) for the given quadratic polynomial. (b) Finding the discriminant \(\Delta^2\) for \(f(x) = x^3 + px + q\)

Step 4: Calculate the roots of the polynomial

Since the cubic polynomial has three roots, let them be \(\alpha_1\), \(\alpha_2\), and \(\alpha_3\). Note that we don't need to find their explicit expressions in this case.

Step 5: Calculate \(\Delta\)

From the definition of \(\Delta\), we form the product of differences of roots pairwise: \(\Delta = (\alpha_1-\alpha_2)(\alpha_1-\alpha_3)(\alpha_2-\alpha_3)\).

Step 6: Use Vieta's formulas

Using Vieta's formulas for the sum of roots, we have \(\alpha_1+\alpha_2+\alpha_3 = 0\). Additionally, we know that \(\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3 = p\) and \(\alpha_1\alpha_2\alpha_3 = -q\).

Step 7: Determine \(\Delta^2\)

Squaring \(\Delta\), we can rewrite it as \(\Delta^2 = \left[ (\alpha_1-\alpha_2)(\alpha_1-\alpha_3)(\alpha_2-\alpha_3)\right]^2\). Expanding and simplifying, and using Vieta's formulas, we find the final expression as \(\Delta^2 = -4p^3 - 27q^2\). Hence, we have shown that \(\Delta^2 = -4p^3 - 27q^2\) for the given cubic polynomial. (c) Proof that \(\Delta^2\) is in \(F\)

Step 8: Express \(\Delta^2\) in terms of symmetric functions

Note that in parts (a) and (b), we have expressed the discriminant \(\Delta^2\) as a polynomial in the coefficients of \(f(x)\). This is a general property of the discriminant: it can be written as a symmetric polynomial in the roots of \(f(x)\), and thus as a polynomial in the coefficients of \(f(x)\).

Step 9: Show that \(\Delta^2 \in F\)

Since \(\Delta^2\) is a polynomial in the coefficients of \(f(x)\), which are in \(F\), we can conclude that \(\Delta^2\) is an element of the field \(F\). Thus, \(\Delta^2 \in F\). (d) Proof that for a transposition \(\sigma\), \(\sigma(\Delta) = -\Delta\)

Step 10: Recall transpositions

A transposition is a permutation that involves swapping exactly two elements of a set. Here, \(\sigma\) is a transposition of two roots of \(f(x)\); assume that it swaps \(\alpha_i\) and \(\alpha_j\).

Step 11: Apply transposition to \(\Delta\)

We know that \(\Delta = \prod_{i

Step 12: Show that \(\sigma(\Delta)=-\Delta\)

Since \((\alpha_j - \alpha_i) = -(\alpha_i - \alpha_j)\), it follows that the difference of each pair of the roots is negated. Therefore, we have \(\sigma(\Delta) = -\Delta\). (e) Proof that for an even permutation \(\sigma\), \(\sigma(\Delta) = \Delta\)

Step 13: Express even permutation as a composition of transpositions

The permutation \(\sigma\) is called an even permutation if it can be expressed as a composition of an even number of transpositions, say \(2k\).

Step 14: Apply even permutation to \(\Delta\)

Applying the even permutation \(\sigma\) to \(\Delta\), we get \(\sigma(\Delta) = \tau_{2k} \circ \tau_{2k-1} \circ ... \circ \tau_2 \circ \tau_1 (\Delta)\).

Step 15: Apply transpositions successively

We apply the transpositions one by one to \(\Delta\). From part (d), we know that each transposition negates \(\Delta\). So, if we apply an even number of transpositions, the result will be the original \(\Delta\).

Step 16: Show that \(\sigma(\Delta)=\Delta\)

Since \(\sigma\) is an even permutation, we have \(\sigma(\Delta) = \Delta\). (f) Proof of the isomorphism condition

Step 17: Suppose \(\Delta \in F\)

In this case, from part (e), we know that all the even permutations in \(G(E/F)\) fix \(\Delta\). Thus, \(G(E/F)\) is a subgroup of the alternating group \(A_n\).

Step 18: Suppose \(G(E/F)\) is a subgroup of \(A_n\)

If \(G(E/F)\) is a subgroup of \(A_n\), then all the even permutations in \(G(E/F)\) fix \(\Delta\), so \(\Delta\) must be an element of \(F\). Therefore, \(\Delta \in F\). (g) Determining the Galois groups of \(x^3+2x-4\) and \(x^3+x-3\)

Step 19: Calculate \(\Delta^2\) for the first polynomial

For the polynomial \(f(x) = x^3+2x-4\), \(p = 2\) and \(q = -4\). Then using part (b), we find \(\Delta^2 = -4p^3 - 27q^2 = -64 - 432 = -496\).

Step 20: Calculate \(\Delta^2\) for the second polynomial

For the polynomial \(g(x) = x^3+x-3\), \(p = 1\) and \(q = -3\). Then using part (b), we find \(\Delta^2 = -4p^3 - 27q^2 = -4 - 243 = -247\).

Step 21: Determine the Galois groups

As both discriminants are non-square elements of the rationals, neither discriminant is an element of the base field. By part (f), the Galois group of both these polynomials is not a subgroup of \(A_3\). In fact, they are isomorphic to \(S_3\) since the degree of both polynomials is \(3\).

Key Concepts

Field Theory

In abstract algebra, a field is a set equipped with two operations that generalize the arithmetic of numbers: addition and multiplication. Field theory is the branch of mathematics that studies the properties of fields. Fields are fundamental in constructing algebraic numbers and manipulating polynomial equations.
In particular, fields provide a structure within which we can investigate the roots of polynomials. The polynomial roots can be elements of another field, called a field extension. For instance, a field extension of a field \( F \) might allow solutions to polynomial equations that have coefficients in \( F \). When working with polynomial equations, it is often useful to consider the minimal field extension that contains all roots of the polynomial. This specific extension is called the splitting field.
Field theory helps us understand the capacity of these extensions and the constraints in which arithmetic operations can be performed, laying the groundwork for deeper algebraic theories like Galois Theory.

Galois Theory

Galois theory provides a profound connection between field theory and group theory. Named after the mathematician Évariste Galois, this theory examines how the symmetries of the roots of polynomials relate to the symmetries of field extensions.
One of the key concepts in Galois theory is the Galois group. This is a group that consists of all automorphisms of a field extension that leave the base field unchanged. These automorphisms behave like permutations, acting on the roots of the polynomial.
Through this perspective, Galois theory answers fundamental questions about solvability, such as whether a polynomial of a given degree can be solved by radicals. By studying the structure of Galois groups, mathematicians can determine if the solution to a polynomial equation is expressible in terms of simple arithmetic operations and root extractions. This insight allows us to classify field extensions and provides tools to solve polynomial-related problems more effectively.

Polynomial Discriminant

The discriminant of a polynomial is a key concept that measures the nature of the roots without finding them explicitly. It is a certain type of symmetric polynomial derived from the roots of the polynomial.
For a quadratic equation \( ax^2 + bx + c \), the discriminant is given by \( Δ = b^2 - 4ac \). This expression can quickly tell us whether the roots are real or complex and whether they are distinct or repeated.
In higher degree polynomials, the discriminant generalizes as a more complex expression that remains symmetric in the roots. For instance, in the cubic polynomial \( x^3 + px + q \), the discriminant helps determine the root behavior. It becomes a crucial tool in assessing the Galois structure by indicating the nature of field extensions.
Understanding the discriminant allows mathematicians to infer the critical properties of the polynomial's roots, such as multiplicity and separability, merely from the coefficients.

Permutation Groups

Permutation groups are mathematical representations of the way objects can be arranged or permuted. In abstract algebra, these groups capture the essence of different symmetric arrangements.
In the context of polynomials, permutation groups arise naturally when considering the rearrangements of the roots of a polynomial. For instance, the symmetric group \( S_n \) on a set of \( n \) elements captures all possible permutations of these elements. An interesting subgroup is the alternating group \( A_n \), consisting of even permutations, or those permutations attainable by an even number of swaps.
These groups, especially the symmetric groups, play a crucial role in Galois theory, where the Galois group acts as permutations of the polynomial's roots. By analyzing these groups, one can deduce properties of solvability and symmetry in the solutions to polynomial equations. The interplay of permutations can also classify complex field extensions and ascertain the solvability of polynomial equations by radicals.

Previous

Problem 21

Other exercises in this chapter

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Problem 21

Let \(F\) be a finite field or a field of characteristic zero. Let \(E\) be a finite normal extension of \(F\) with Galois group \(G(E / F) .\) Prove that \(F \

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Problem 19

Let \(K\) be the splitting field of a polynomial over \(F\). If \(E\) is a field extension of \(F\) contained in \(K\) and \([E: F]=2,\) then \(E\) is the split

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Key Concepts

Field Theory

In abstract algebra, a field is a set equipped with two operations that generalize the arithmetic of numbers: addition and multiplication. Field theory is the branch of mathematics that studies the properties of fields. Fields are fundamental in constructing algebraic numbers and manipulating polynomial equations.
In particular, fields provide a structure within which we can investigate the roots of polynomials. The polynomial roots can be elements of another field, called a field extension. For instance, a field extension of a field \( F \) might allow solutions to polynomial equations that have coefficients in \( F \). When working with polynomial equations, it is often useful to consider the minimal field extension that contains all roots of the polynomial. This specific extension is called the splitting field.
Field theory helps us understand the capacity of these extensions and the constraints in which arithmetic operations can be performed, laying the groundwork for deeper algebraic theories like Galois Theory.

Galois Theory

Galois theory provides a profound connection between field theory and group theory. Named after the mathematician Évariste Galois, this theory examines how the symmetries of the roots of polynomials relate to the symmetries of field extensions.
One of the key concepts in Galois theory is the Galois group. This is a group that consists of all automorphisms of a field extension that leave the base field unchanged. These automorphisms behave like permutations, acting on the roots of the polynomial.
Through this perspective, Galois theory answers fundamental questions about solvability, such as whether a polynomial of a given degree can be solved by radicals. By studying the structure of Galois groups, mathematicians can determine if the solution to a polynomial equation is expressible in terms of simple arithmetic operations and root extractions. This insight allows us to classify field extensions and provides tools to solve polynomial-related problems more effectively.

Polynomial Discriminant

The discriminant of a polynomial is a key concept that measures the nature of the roots without finding them explicitly. It is a certain type of symmetric polynomial derived from the roots of the polynomial.
For a quadratic equation \( ax^2 + bx + c \), the discriminant is given by \( Δ = b^2 - 4ac \). This expression can quickly tell us whether the roots are real or complex and whether they are distinct or repeated.
In higher degree polynomials, the discriminant generalizes as a more complex expression that remains symmetric in the roots. For instance, in the cubic polynomial \( x^3 + px + q \), the discriminant helps determine the root behavior. It becomes a crucial tool in assessing the Galois structure by indicating the nature of field extensions.
Understanding the discriminant allows mathematicians to infer the critical properties of the polynomial's roots, such as multiplicity and separability, merely from the coefficients.

Permutation Groups

Permutation groups are mathematical representations of the way objects can be arranged or permuted. In abstract algebra, these groups capture the essence of different symmetric arrangements.
In the context of polynomials, permutation groups arise naturally when considering the rearrangements of the roots of a polynomial. For instance, the symmetric group \( S_n \) on a set of \( n \) elements captures all possible permutations of these elements. An interesting subgroup is the alternating group \( A_n \), consisting of even permutations, or those permutations attainable by an even number of swaps.
These groups, especially the symmetric groups, play a crucial role in Galois theory, where the Galois group acts as permutations of the polynomial's roots. By analyzing these groups, one can deduce properties of solvability and symmetry in the solutions to polynomial equations. The interplay of permutations can also classify complex field extensions and ascertain the solvability of polynomial equations by radicals.