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Chapter 3: Problem 7

If \(f(x)\) is in \(F[x]\), where \(F\) is the field of integers mod \(p, p\) a prime, and \(f(x)\) is irreducible over \(F\) of degree \(n\) prove that \(F[x] /(f(x))\) is a field with \(p^{n}\) elements.

Short Answer

Expert verified
\(F[x]/(f(x))\) is a field due to \((f(x))\) being a maximal ideal, containing \(p^n\) elements because each has an \(n\)-degree polynomial with coefficients in \(F\).

Step by step solution

01

Understanding the Problem

We need to show that the quotient ring \(F[x]/(f(x))\), where \(F\) is the field of integers modulo a prime \(p\), and \(f(x)\) is an irreducible polynomial over \(F\) of degree \(n\), forms a field with \(p^n\) elements.
02

Irreducibility of the Polynomial

Since \(f(x)\) is irreducible over \(F\), it means that \(f(x)\) has no divisors other than itself and units in \(F[x]\). This makes the ideal \((f(x))\) maximal in \(F[x]\).
03

Properties of Maximal Ideals

A crucial fact about maximal ideals is that the quotient of a ring by a maximal ideal is a field. Because \((f(x))\) is a maximal ideal, \(F[x]/(f(x))\) forms a field.
04

Determine the Structure of the Field

Elements of \(F[x]/(f(x))\) are equivalence classes of polynomials in \(F[x]\), with remainders of degree less than \(n\) when divided by \(f(x)\). This implies each element can be uniquely represented as \(a_0 + a_1 x + a_2 x^2 + \cdots + a_{n-1} x^{n-1}\) with \(a_i \in F\).
05

Calculate Number of Elements

Since each \(a_i\) can be any element in \(F\), which has \(p\) elements, there are \(p^n\) possible polynomials of degree less than \(n\). Thus, \(F[x]/(f(x))\) contains \(p^n\) elements.

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Key Concepts

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

Irreducible Polynomial
In mathematics, a polynomial is said to be irreducible over a given field if it cannot be factored into the product of two non-constant polynomials with coefficients in that field. Think of it as a building block, like a prime number.
  • For example, over the field of integers modulo prime number, an irreducible polynomial doesn't split any further without leaving the field.
  • Having no divisors other than itself and the units in the field means it's quite "sturdy." However, it might factor if you switch fields.
Why does this matter? In field theory, irreducible polynomials help form extensions of fields, growing them in a controlled, well-understood manner. When dealing with an irreducible polynomial of degree \(n\), we can create new structures called fields that have interesting properties, such as specific numbers of elements.
Quotient Ring
The concept of a quotient ring is central when we talk about diving polynomial rings by polynomials.
  • Imagine having a giant cheese wheel, but you only want to keep a slice of it. The larger wheel is your polynomial ring, and the slice is the quotient ring.
  • Formed by dividing a polynomial ring \(F[x]\) by an ideal \((f(x))\), the result \(F[x]/(f(x))\) consists of all possible remainders when you divide by \(f(x)\).
  • This concept allows us to compress information, capturing key properties of polynomials.
Essentially, it gives us a way to "mod out" the polynomial, focusing only on what's left behind after accounting for a full cycle. The elements of a quotient ring are more manageable and helpful when we want to explore deeper properties like field structure.
Maximal Ideal
Maximal ideals are special subsets in rings, pivotal to forming fields in the quotient ring.
  • An ideal is maximal if no larger ideal contains it, except for the whole ring itself.
  • The importance? When you divide a ring by a maximal ideal, what's left is a field.
  • In our context, our polynomial ring \(F[x]\), divided by the maximal ideal \((f(x))\), is crucial for proving that the resulting structure is a field.
Why do maximal ideals matter so much? They simplify structures. By ensuring nothing "bigger" contains them, they ensure the clean breakdown of the original polynomial ring into something with field properties. This transformation is a neat trick of algebra that solves otherwise complex problems.
Finite Fields
A finite field is a set containing a finite number of elements, where mathematical operations like addition, subtraction, multiplication, and division (except by zero) are possible.
  • Important to know: finite fields have \(p^n\) elements, where \(p\) is a prime number and \(n\) is a positive integer.
  • They are sometimes called Galois fields, denoted as \(GF(p^n)\).
When dealing with finite fields, familiarity with simple math operations extends to interesting and useful cyclic behaviors. For example, all the calculations "wrap around" in a structured way. In the problem we've been discussing, the quotient ring \(F[x]/(f(x))\) becomes such a finite field. Its cardinality, or size, is driven by the number of different coefficients in polynomials of degree less than \(n\). These fields are fundamental in coding theory, cryptography, and many areas of algebra.

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Most popular questions from this chapter

Find the greatest common divisor of the following polynomials over \(F\), the field of rational numbers: (a) \(x^{3}-6 x^{2}+x+4\) and \(x^{3}-6 x+1\). (b) \(x^{2}+1\) and \(x^{0}+x^{3}+x+1\).

If \(R\) is the ring of integers, let \(U\) be the ideal consisting of all multiples of 17 . Prove that if \(V\) is an ideal of \(R\) and \(R \supset V \supset U\) then either \(V=R\) or \(V=U\). Generalize!

If \(F\) is a field prove its only ideals are \((0)\) and \(F\) itself.

If \(R\) is a commutative ring with unit element, prove that \(a_{0}+a_{1} x+\) \(\cdots+a_{n} x^{n}\) in \(R[x]\) has an inverse in \(R[x]\) (i.e., is a unit in \(\left.R[x]\right)\) if and only if \(a_{0}\) is a unit in \(R\) and \(a_{1}, \ldots, a_{n}\) are nilpotent elements in \(R\).

Find the greatest common divisor in \(J[i]\) of: (a) \(3+4 i\) and \(4-3 i\) (b) \(11+7 i\) and \(18-i\)
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