/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Let \(K\) be a field of a charac... [FREE SOLUTION] | 91影视

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Let \(K\) be a field of a characteristic zero, \(L\) a finite (algebraic) extension of \(K\). Then there is a \(z \in L\) such that \(L=K(z)\). Outline of Proof. Step (i): Suppose \(L=K(x, y)\). Let \(F\) and \(G\) be monic irreducible polynomials in \(K[T]\) such that \(F(x)=0, G(y)=0\). Let \(L^{\prime}\) be a field in which \(F=\) \(\prod_{i=1}^{n}\left(T-x_{i}\right), G=\prod_{j=1}^{m}\left(T-y_{i}\right), x=x_{1}, y=y_{1}, L^{\prime} \supset L\) (see Problems 1.52, 1.53). Choose \(\lambda \neq 0\) in \(K\) so that \(\lambda x+y \neq \lambda x_{i}+y_{j}\) for all \(i \neq 1, j \neq 1\). Let \(z=\lambda x+y, K^{\prime}=K(z)\). Set \(H(T)=G(z-\lambda T) \in K^{\prime}[T] .\) Then \(H(x)=0, H\left(x_{i}\right) \neq 0\) if \(i>0 .\) Therefore \((H, F)=\) \((T-x) \in K^{\prime}[T]\). Then \(x \in K^{\prime}\), so \(y \in K^{\prime}\), so \(L=K^{\prime}\) Step (ii): If \(L=K\left(x_{1}, \ldots, x_{n}\right)\), use induction on \(n\) to find \(\lambda_{1}, \ldots, \lambda_{n} \in k\) such that \(L=K\left(\sum \lambda_{i} x_{i}\right)\)

Short Answer

Expert verified
Question: Prove that for a field of characteristic zero, K, and a finite, algebraic extension of K, denoted as L, there exists a z 鈭 L such that L = K(z). Solution: Choose scalars 位鈧, ..., 位鈧 鈭 K for L = K(x鈧, ..., x鈧) such that L = K(鈭 位_i x_i). Then, let z = 鈭 位_i x_i. Now, L = K(z).

Step by step solution

01

Consider the case L=K(x,y)

We first suppose that L=K(x,y) where x and y are algebraic over K. Let F and G be the monic irreducible polynomials in K[T] for which F(x)=0 and G(y)=0. Now consider an extension field L' that contains L and satisfies the conditions mentioned in the exercise. We will try to find a z in L' of the form 位x + y, where 位 is a non-zero scalar in K. Choose a 位 鈭 K, 位 鈮 0, so that 位x + y 鈮 位x_i + y_j for all i 鈮 1 and j 鈮 1. Define z = 位x + y. Then, L' contains K(z) since K(z) is a subfield of L'. We will now prove that L = K(z). Define H(T) = G(z - 位T) which is a polynomial in K'(T). It can be observed that H(x) = 0 and H(x_i) 鈮 0 for all i > 0. Hence, (H,F) = (T-x) in K'[T]. This implies that x 鈭 K(z) and since z - 位x = y, we find that y 鈭 K(z) as well. Therefore, L = K(x,y) = K(z).
02

Induction on n for L = K(x1,...,xn)

Now, let L = K(x鈧, ..., x鈧). Our goal is to find scalars 位鈧, ..., 位鈧 鈭 K such that L = K(鈭 位_i x_i). We will use induction on n: Base Case: n = 1. In this case, we just have L = K(x鈧), and we can choose 位鈧 = 1. So our base case is true. Inductive step: Assume the statement is true for n = k. Now consider the case n = k+1. Let L鈧 = K(x鈧, ..., x鈧) and L鈧栤倞鈧 = K(x鈧, ..., x鈧, x鈧栤倞鈧). We have L鈧栤倞鈧 = L鈧(x鈧栤倞鈧). Apply step 1 to L鈧 and L鈧栤倞鈧, where x=x鈧栤倞鈧 and y=L鈧. It's known from the induction hypothesis that there exist 位鈧, ..., 位鈧 鈭 K such that L鈧 = K(鈭 位_i x_i). Now, let z = 位x鈧栤倞鈧 + L鈧 with z = 鈭 伪_i x_i, where 伪_i are elements in K, and 位 鈮 0. By step 1, we know that L鈧栤倞鈧=K(z). Let 位鈧栤倞鈧 = 位, and L鈧栤倞鈧 = K(鈭 位_i x_i + 位鈧栤倞鈧亁鈧栤倞鈧), completing the inductive step. Therefore, by mathematical induction, the statement is true for all n. This concludes the proof that for the given field of characteristic zero, K, and its finite, algebraic extension field L, there exists a z 鈭 L such that L = K(z).

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Key Concepts

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

Field Theory
Field Theory is a branch of mathematics in which we study the properties and structures of fields. A field is an algebraic structure equipped with two operations: addition and multiplication. These operations must satisfy certain properties such as associativity, commutativity, distributivity, and having identity and inverse elements.

Fields are foundational in algebra, providing a setting for analyzing polynomial equations and creating algebraic extensions. An algebraic extension is formed by adding roots of polynomials to an existing field. For instance, if you start with the field of rational numbers \( \mathbb{Q} \) and add a root of the polynomial \( x^2 - 2 = 0 \), you get the field \( \mathbb{Q}(\sqrt{2}) \).

Understanding field extensions helps us explore more complex systems of numbers. It is vital in understanding solvability of equations and in the development of number theory and cryptography. Essential concepts in field theory include subfields, extensions, algebraic and transcendental elements, and the notion of characteristic, which defines how the arithmetic behaves in a field. For example, characteristic zero means no positive integer multiples of the identity element (1) add up to zero, as seen in fields like \( \mathbb{Q} \) and \( \mathbb{R} \).
Irreducible Polynomials
In field theory, a polynomial is called irreducible if it cannot be factored into the product of two non-constant polynomials with coefficients in the same field. This concept is analogous to prime numbers in integer arithmetic. Irreducible polynomials are the building blocks of more complex polynomials.

When a polynomial is irreducible over a field \( K \), it means it lacks roots in \( K \). Considerable practical importance is given to irreducible polynomials in constructing field extensions. Given a polynomial that is irreducible, we can form an extension field by including a root of that polynomial.

For example, the polynomial \( x^2 - 2 \) is irreducible over \( \mathbb{Q} \), because it cannot be split into simpler polynomials over the rationals. However, by constructing the extension field \( \mathbb{Q}(\sqrt{2}) \), we include its root \( \sqrt{2} \). In the exercise, we use irreducible polynomials to express elements like \( x \) and \( y \) via equations \( F(x) = 0 \) and \( G(y) = 0 \), leading to the creation and manipulation of the appropriate extension field \( L \).
Inductive Proof
Inductive proof is a mathematical technique used to establish that a given statement is true for all natural numbers. It's based on two main steps: the base case and the inductive step.

The base case verifies the statement for the initial value, usually \( n = 1 \). The inductive step assumes the statement holds for \( n = k \) and then proves it for \( n = k+1 \). By these steps, a domino effect occurs, proving the statement for all \( n \).

In the given exercise, induction is applied to show how an extension field containing multiple elements \( x_1, x_2, \ldots, x_n \) can be reduced to a simple field extension with a single element. We first verify that it holds for a single element (base case), and then show that if it's true for \( k \) elements, it's true for \( k+1 \) elements too (inductive step).

This method proves indispensable in field theory, allowing the demonstration of results such as the ability to express a finite algebraic extension \( L \) of \( K \) as \( L = K(z) \), where \( z \) is some single element in \( L \). The beauty of induction lies in its simplicity and wide applicability across mathematics.

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Most popular questions from this chapter

Let \(F\) be an irreducible projective plane curve of degree \(n>\) 1. Let \(\Gamma_{h}(F)=k[X, Y, Z] /(F)=k[x, y, z]\), and let \(u, v, w \in \Gamma_{h}(F)\) be the residues of \(F_{X}, F_{Y}, F_{Z}\), respectively. Define \(\alpha: k[U, V, W] \rightarrow \Gamma_{h}(F)\) by setting \(\alpha(U)=u, \alpha(V)=v\), \(\alpha(W)=w .\) Let \(I\) be the kernel of \(\alpha\). (a) Show that \(I\) is a homogeneous prime ideal in \(k[U, V, W]\), so \(V(I)\) is a closed subvariety of \(\mathbb{P}^{2}\). (b) Show that for any simple point \(P\) on \(F,\left[F_{X}(P): F_{Y}(P): F_{Z}(P)\right]\) is in \(V(I)\), so \(V(I)\) contains the points corresponding to tangent lines to \(F\) at simple points. (c) If \(V(I) \subset\\{[a: b: c]\\}\), use Euler's Theorem to show that \(F\) divides \(a X+b Y+c Z\), which is impossible. Conclude that \(V(I)\) is a curve. It is called the dual curve of \(F\). (d) Show that the dual curve is the only irreducible curve containing all the points of (b). (See Walker's "Algebraic Curves" for more about dual curves when \(\operatorname{char}(k)=0 .\) )

Show that \(\left[x_{1}: \ldots: x_{n}\right] \mapsto\left[x_{1}: \ldots: x_{n}: 0\right]\) gives an isomorphism of \(\mathbb{P}^{n-1}\) with \(H_{\infty} \subset \mathbb{P}^{n} .\) If a variety \(V\) in \(\mathbb{P}^{n}\) is contained in \(H_{\infty}, V\) is isomorphic to a variety in \(\mathbb{P}^{n-1}\). Any projective variety is isomorphic to a closed subvariety \(V \subset \mathbb{P}^{n}\) (for some \(n\) ) such that \(V\) is not contained in any hyperplane in \(\mathbb{P}^{n}\).

Let \(X\) be a variety, \(f \in \Gamma(X)\). Let \(\varphi: X \rightarrow \mathbb{A}^{1}\) be the mapping defined by \(\varphi(P)=\) \(f(P)\) for \(P \in X\). (a) Show that for \(\lambda \in k, \varphi^{-1}(\lambda)\) is the pole set of \(z=1 /(f-\lambda)\). (b) Show that \(\varphi\) is a morphism of varieties.

Let \(Z \subset Y \subset X, X\) a topological space. Give \(Y\) the induced topology. Show that the topology induced by \(Y\) on \(Z\) is the same as that induced by \(X\) on \(Z\).

If there is a dominating rational map from \(X\) to \(Y\), then \(\operatorname{dim}(Y) \leq \operatorname{dim}(X)\).

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