1. Prove that the order of a finite field is never divisible by two distinct prime numbers.

Answer:

Introduction:

The problem here involves the field of mathematics known as “finite fields” or “Galois fields”. Finite fields are fields that contain a finite number of elements. In the context of this problem, we are asked to prove that the order (or number of elements) in a finite field can never be divisible by two distinct prime numbers.

Method/Approach:

We will approach this problem using the properties of finite fields. The main properties we will use are:

1. Every finite field has order \(p^n\) where \(p\) is a prime number and \(n\) is a positive integer.

2. Every finite field with \(p^n\) elements is isomorphic to the field \(GF(p^n)\) (the Galois Field).

3. If a finite field has \(p^n\) elements, where \(p\) and \(n\) are as above, then it has exactly \(p^n\) elements.

Work/Calculations:

Based on these properties, we can state that the order of a finite field is always in the form of a power of a prime number, say \(p^n\). If a number is divisible by two distinct primes, it cannot be written as \(p^n\), where \(p\) is a prime and \(n\) is a positive integer. This is because \(p^n\) is only divisible by the prime number \(p\) and its powers.

The only way a finite field can have a number of elements that is divisible by a prime number other than \(p\) is if that other prime divides \(n\). However, this isn’t possible, since \(n\) is the exponent in \(p^n\) and therefore must be an integer. A prime number cannot divide an integer and result in another integer unless it is equal to that integer, which contradicts our assumption that the two prime numbers are distinct.

So we can conclude that the order of a finite field can never be divisible by two distinct prime numbers.

Conclusion:

In conclusion, we have proven that the order of a finite field cannot be divisible by two distinct prime numbers. This is a consequence of the fact that the order of a finite field is always a power of a prime number, and a power of a prime number cannot be divisible by any other distinct prime number.

2. Prove that the 5th roots of unity form a cyclic group under multiplication.

Answer:

Introduction:

We’re asked to prove that the 5th roots of unity form a cyclic group under the operation of multiplication. The “5th roots of unity” are the complex numbers that solve the equation \(z^5=1\). In other words, they are the complex numbers \(z\) such that when raised to the 5th power, they give the result 1.

A “cyclic group” is a type of algebraic structure in which all the elements of the group can be generated by repeated application of a single element.

Method/Approach:

In this problem, we’ll prove that the 5th roots of unity form a cyclic group by demonstrating that they meet the four necessary conditions to be a group, and that they can be generated by a single element.

Work/Calculations:

1. Closure: The 5th roots of unity are the solutions to the equation \(z^5=1\), which can be written in the form \(e^{2\pi ik/5}\) for \(k=0, 1, 2, 3, 4\). If we multiply any two of these roots, we get another 5th root of unity, as \(e^{2\pi ik/5} \cdot e^{2\pi ij/5} = e^{2\pi i(k+j)/5}\), which is a 5th root of unity for \(k+j \mod 5\). Hence, the set is closed under multiplication.

2. Associativity: Multiplication of complex numbers is always associative; that is, for any complex numbers a, b, and c, \((ab)c = a(bc)\). Hence, the set of 5th roots of unity is associative under multiplication.

3. Identity: The complex number 1 is a 5th root of unity because \(1^5=1\), and for any 5th root of unity \(z\), we have \(z \cdot 1 = z\) and \(1 \cdot z = z\). Hence, 1 is the multiplicative identity in this group.

4. Inverses: For each 5th root of unity \(z = e^{2\pi ik/5}\), the complex conjugate \(z^{-1} = e^{-2\pi ik/5}\) is also a 5th root of unity, and \(z \cdot z^{-1} = 1\), so each element has a multiplicative inverse in the group.

Having shown that the 5th roots of unity satisfy the four conditions for a group under multiplication, we now show they form a cyclic group. A group is cyclic if there exists an element in the group such that the group can be generated by that element. The element \(z = e^{2\pi i/5}\) generates the 5th roots of unity because \((e^{2\pi i/5})^k\) for \(k=0, 1, 2, 3, 4\) gives all the 5th roots of unity, namely \(1, e^{2\pi i/5}, e^{4\pi i/5}, e^{6\pi i/5}\), and \(e^{8\pi i/5}\). Therefore, the 5th roots of unity form a cyclic group under multiplication.

Conclusion:

We have shown that the 5th roots of unity form a group under the operation of multiplication as they satisfy the group axioms: closure, associativity, identity, and inverses. Moreover, we have shown that they form a cyclic group, since all the 5th roots of unity can be generated by repeatedly multiplying a single root of unity. Thus, the 5th roots of unity form a cyclic group under multiplication.

3. Prove that cartesian product of two solvable groups is again solvable.

Answer:

Introduction:

A group is called solvable if it can be decomposed into a series of abelian or commutative subgroups. The Cartesian product of two groups, often denoted as \(G \times H\), is the set of ordered pairs \((g,h)\) where \(g\) is in \(G\) and \(h\) is in \(H\), with group operations defined elementwise. The problem here is to prove that the Cartesian product of two solvable groups is also solvable.

Method/Approach:

To prove this, we will utilize the definition of solvability and the structure of the Cartesian product of two groups. Specifically, we will construct a series of subgroups for \(G \times H\) from the series of subgroups of \(G\) and \(H\).

Work/Calculations:

Let \(G\) and \(H\) be two solvable groups. By definition of solvability, there exist a chain of subgroups for each group:

For \(G\):

\[ \{e_G\} = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_n = G\]

For \(H\):

\[ \{e_H\} = H_0 \trianglelefteq H_1 \trianglelefteq \cdots \trianglelefteq H_m = H\]

where each \(G_i\) is a normal subgroup of \(G_{i+1}\) and each \(H_i\) is a normal subgroup of \(H_{i+1}\), and the quotient groups \(G_{i+1}/G_i\) and \(H_{i+1}/H_i\) are abelian.

We construct a chain of subgroups for \(G \times H\) as follows:

\[ \{(e_G,e_H)\} = (G_0,H_0) \trianglelefteq (G_1,H_0) \trianglelefteq \cdots \trianglelefteq (G_n,H_0) \trianglelefteq (G_n,H_1) \trianglelefteq \cdots \trianglelefteq (G_n,H_m) = (G,H)\]

Each group \((G_i,H_j)\) is clearly a subgroup of \((G_{i+1},H_j)\) and \((G_i,H_{j+1})\), and it is also a normal subgroup because the operations are defined elementwise and \(G_i\) is normal in \(G_{i+1}\) and \(H_j\) is normal in \(H_{j+1}\).

The quotient group \((G_{i+1},H_j)/(G_i,H_j) \cong G_{i+1}/G_i\) and the quotient group \((G_i,H_{j+1})/(G_i,H_j) \cong H_{j+1}/H_j\), both of which are abelian by the assumption that \(G\) and \(H\) are solvable.

Conclusion:

We have shown that if \(G\) and \(H\) are solvable groups, then their Cartesian product \(G \times H\) is also solvable. We achieved this by constructing a chain of subgroups for \(G \times H\) such that each subgroup is normal in the next, and the corresponding quotient groups are abelian. This satisfies the definition of a solvable group, so \(G \times H\) is solvable.

4. Find the smallest extension of Q having a root of x²-2 belongs to

Q[ x].

Answer:

Introduction:

The question is asking us to find the smallest extension of the rational numbers (\(\mathbb{Q}\)) that contains a root of the polynomial \(x^2 – 2\). This root does not exist in \(\mathbb{Q}\) itself, so we need to extend \(\mathbb{Q}\) to include it.

Method/Approach:

We’ll solve this problem using the concept of a field extension. A field extension is a larger field that contains a smaller field as a subfield. We’ll use the quadratic equation to find the roots of the given polynomial, and then form a field extension of \(\mathbb{Q}\) to include the roots.

Work/Calculations:

The roots of the polynomial \(x^2 – 2\) can be found using the quadratic formula. The roots are \(\sqrt{2}\) and \(-\sqrt{2}\). These numbers are not rational, so they are not in \(\mathbb{Q}\).

The smallest field extension of \(\mathbb{Q}\) that contains a root of the polynomial is \(\mathbb{Q}(\sqrt{2})\). This field includes all numbers of the form \(a + b\sqrt{2}\), where \(a\) and \(b\) are rational numbers. It’s clear that \(\mathbb{Q}(\sqrt{2})\) is a subset of \(\mathbb{Q}[x]\), since \(\mathbb{Q}[x]\) includes all polynomials with coefficients in \(\mathbb{Q}\), and the elements of \(\mathbb{Q}(\sqrt{2})\) can be thought of as polynomials of degree less than 2.

Conclusion:

The smallest extension of \(\mathbb{Q}\) that includes a root of the polynomial \(x^2 – 2\) and is a subset of \(\mathbb{Q}[x]\) is \(\mathbb{Q}(\sqrt{2})\). This extension includes all numbers of the form \(a + b\sqrt{2}\), where \(a\) and \(b\) are rational numbers.