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Let \(\alpha\) be algebraic over the field \(F\), and let $$ D(t)=t^{*}+a_{n-1} t^{n-1}+\cdots+a_{0} $$ be the irreducible polynomial of \(\alpha\) over \(F\). Show that $$ N(\alpha)=(-1)^{n} a_{0} \quad \text { and } \quad \operatorname{Tr}(\alpha)=-a_{\mathrm{e}-1} . $$ (The norm and trace are taken from \(F(\alpha)\) to \(F .\) )

Short Answer

Expert verified
The Norm and Trace of the algebraic number \(\alpha\) with irreducible polynomial \(D(t)\) are given by \(\operatorname{N}(\alpha)=(-1)^n a_0\) and \(\operatorname{Tr}(\alpha)=-a_{n-1}\), where \(a_0\) and \(a_{n-1}\) are the constant term and the coefficient of \(t^{n-1}\) in \(D(t)\), respectively.

Step by step solution

01

Find the embeddings of \(F(\alpha)\)

To solve this problem, we need to find the distinct embeddings \(\sigma_i\) of the field extension \(F(\alpha)\) into its algebraic closure. Since \(\alpha\) is algebraic over \(F\), the degree of the field extension is equal to the degree of the irreducible polynomial, i.e., \(n = \deg(D(t))\). Since \(D(t)\) is irreducible, the embeddings are the \(n\) distinct roots of \(D(t)\), say \(\alpha_1, \alpha_2, \ldots, \alpha_n\). In other words, \(\sigma_i(\alpha) = \alpha_i\) for \(i = 1, 2, \ldots, n\).
02

Express the Norm and Trace using the embeddings

Using the embeddings \(\sigma_i(\alpha) = \alpha_i\), we can now express the Norm and Trace of \(\alpha\) as follows: $$ \operatorname{N}(\alpha) = \prod_{i=1}^n \alpha_i $$ $$ \operatorname{Tr}(\alpha) = \sum_{i=1}^n \alpha_i $$
03

Relate the Norm and Trace to the coefficients of \(D(t)\)

We know that \(D(t)\) has the roots \(\alpha_1, \alpha_2, \ldots, \alpha_n\) by definition. So, we can write \(D(t)\) as: $$ D(t) = (t - \alpha_1)(t - \alpha_2) \cdots (t - \alpha_n) $$ Expanding this product and comparing the coefficients, we can establish the following relationships: For the constant term, \(a_0 = (-1)^n \prod_{i=1}^n \alpha_i = (-1)^n \operatorname{N}(\alpha)\). For the coefficient of \(t^{n-1}\), \(a_{n-1} = -\sum_{i=1}^n \alpha_i = -\operatorname{Tr}(\alpha)\).
04

Conclusion

We have derived the expressions for the Norm and Trace of \(\alpha\) in terms of the coefficients of its irreducible polynomial \(D(t)\). Namely: $$ \operatorname{N}(\alpha) = (-1)^n a_0 $$ $$ \operatorname{Tr}(\alpha) = -a_{n-1} $$ These are the desired results.

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Key Concepts

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

Algebraic Extension
An algebraic extension is a type of field extension where each element in the larger field satisfies a polynomial equation with coefficients from the smaller field. This means that every element in the extended field, say \(F(\alpha)\), is "algebraic" over the base field \(F\).

To put it simply, an algebraic extension allows us to include additional numbers like \(\alpha\) that weren't originally in \(F\), but are solutions to polynomial equations with coefficients from \(F\).

Here's how it usually works:
  • We start with a base field \(F\), such as the set of rational numbers \(\mathbb{Q}\).
  • Select a number \(\alpha\) that solves a polynomial equation with rational coefficients but isn’t a rational number itself.
  • By adjoining \(\alpha\) to \(F\), we form the field extension \(F(\alpha)\).
In field theory, this is crucial because it helps understand how fields can be expanded. It allows mathematicians to study properties of numbers and their relations in much greater detail.
Irreducible Polynomial
An irreducible polynomial is one that cannot be factored into simpler polynomials over its coefficient field. This is a foundational concept in algebra, especially in understanding field extensions.

Consider a polynomial \(D(t) = t^n + a_{n-1} t^{n-1} + \cdots + a_0\). This polynomial is said to be irreducible over a field \(F\) if:
  • Its degree is greater than zero.
  • It has no roots in \(F\), meaning it cannot be written as a product of two non-constant polynomials with coefficients in \(F\).
Such polynomials are the building blocks of field extensions. They serve as the minimum polynomial for elements \(\alpha\), which are added to \(F\) to create the field extension \(F(\alpha)\).

In our given exercise, \(D(t)\) is the irreducible polynomial for \(\alpha\) over \(F\), defining the smallest field, \(F(\alpha)\), which includes \(\alpha\).
Norm and Trace
In the context of field theory, the norm and trace are essential concepts that measure properties of elements within field extensions. They are especially significant when dealing with algebraic extensions.

The **norm** of an element \(\alpha\) in the extension \(F(\alpha)\) is defined as the product of all the roots of its irreducible polynomial. If \(\alpha_1, \alpha_2, \ldots, \alpha_n\) are the roots of \(D(t)\), then the Norm \(N(\alpha)\) can be given by the equation:
\[ N(\alpha) = \prod_{i=1}^n \alpha_i \]
For the specific case here, it relates directly to the constant term of the irreducible polynomial, making \(N(\alpha) = (-1)^n a_0\).

The **trace**, on the other hand, sums all the roots of \(D(t)\):
\[ \operatorname{Tr}(\alpha) = \sum_{i=1}^n \alpha_i \]
In our scenario, it directly connects to the coefficient of \(t^{n-1}\) of the polynomial, equating to \(\operatorname{Tr}(\alpha) = -a_{n-1}\).

These expressions help in understanding how the modification of fields affects the nature of solutions. Additionally, they serve as tools to bridge the field characteristics back to the polynomial equations that define them.

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Most popular questions from this chapter

Let \(A=A_{1} \times \cdots \times A\), be a product of cyclic groups of orders \(d_{1} \ldots ., d_{x}\) respectively, and assume that \(d_{i} \mid n\) for all \(i\). Prove that $$ \operatorname{Hom}(A, C) \approx \prod_{i=1}^{r} \operatorname{Hom}\left(A_{i}, C\right) $$ and hence that Hom \((A, C)\) is isomorphic to \(A\), by using Exercise \(19(b)\).

Show that the polynomial \(t^{3}-p\) is irreducible over the rational numbers for cach prime number \(p\).

Let \(f(t)\) be a polynomial of degree 3 , irreducible over the field \(F\). Prove that the splitting field \(K\) of \(f\) contains at most one subfield of degree 2 over \(F\) namely \(F(\sqrt{D})\), where \(D\) is the discriminant. If \(D\) is a square in \(F\), then \(K\) does not contain any subfield of degree 2 over \(F\).

(a) Let \(A\) be a cyclic group of order \(n\). Let \(C\) be a cyclic group of order \(n\). Show that the group of homomorphisms of \(A\) into \(C\) is cyclic of order \(n\). We let \(\operatorname{Hom}(A, C)\) denote the group of homomorphisms of \(A\) into \(C\). (b) Let \(A\) be a cyclic group of order \(d\), and assume \(d \mid n\). Let again \(C\) be al cyclic group of order \(n\). Show that \(\operatorname{Hom}(A, C)\) is cyclic of order \(d\).

Let \(A, B\) be two finite abelian groups such that \(x^{*}=1\) for all \(x \in A\) and all \(x \in B\). Suppose given a bimultiplicative mapping $$ A \times B \rightarrow C \quad \text { denoted by } \quad(a, b) \mapsto\langle a, b\rangle $$ into a cyclic group \(C\) of order \(n\). Bimultiplicative means that for all \(a, a^{\prime} \in A\) and \(b, b^{\prime} \in B\) we have $$ \left\langle a d^{\prime}, b\right\rangle=\langle a, b\rangle\langle a, b\rangle \quad \text { and } \quad\left\langle a, b b^{\prime}\right\rangle=\langle a, b\rangle\langle a, b\rangle . $$ Define a perpendicular to \(b\) to mean that \(\langle a, b\rangle=1\). Axsume that if \(a \in A\) is perpendicular to all elements of \(B\) then \(a=1\), and also if \(b \in B\) is perpendicular to all clements of \(A\), then \(b=1\). Proce that there is a natural isomorphism $$ A \approx \operatorname{Hom}(B, C) $$ and hence by Exercise 20, that there is some isomorphism \(A \approx B\).

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