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Chapter 6: Problem 28

Let \(E\) be an algebraic extension of \(k\) such that every non-constant polynomial \(f(X\) in \(k[X]\) has at least one root in \(E\). Prove that \(E\) is algebraically closed. [Hint: Discus the separable and purely inseparable cases separately, and use the primitive clemen theorem.]

Short Answer

Expert verified
To prove that E is algebraically closed, we consider two cases: 1. For separable polynomials in E[X], we use the primitive element theorem to show that all the roots of a separable polynomial can be expressed as a linear combination of an element e 鈭� E and elements from k. Hence, all roots of f(X) are in E, and E is algebraically closed for separable polynomials. 2. For purely inseparable polynomials in E[X], we show that any polynomial of the form f(X) = (X^(p^m) - c) will have a root in E. This means that E is algebraically closed for purely inseparable polynomials as well. Since E is algebraically closed for both separable and purely inseparable polynomials, we conclude that E is algebraically closed.

Step by step solution

01

Case 1: The separable case

In this case, let's consider a non-constant polynomial f(X) in E[X] and suppose it is separable. This means the polynomial has distinct roots in its splitting field. Let these roots be r1, r2, ..., rn 鈭� E. Since f(X) 鈭� E[X], f(X) is also a polynomial in k[X]. Now, let's take a look at each root in turn: 1. r1: Since r1 is a root of f(X) 鈭� k[X], we know that at least one of its roots must be in E (by the given condition). So, r1 鈭� E. 2. r2, r3, ..., rn: We apply the primitive element theorem here. Since f(X) is a separable polynomial, there exists a primitive element e in E such that E = k(e). Since r1 鈭� E, we can write r1 = ce + d for some c, d 鈭� k. Following the same logic, all the remaining roots can also be expressed in the form of ae + b for some a, b 鈭� k. Therefore, all roots of f(X) are in E which implies that f(X) splits completely in E. Since any arbitrary separable polynomial in E[X] splits completely in E, E must be algebraically closed for separable polynomials.
02

Case 2: The purely inseparable case

In this case, let's consider a non-constant polynomial f(X) in E[X] and now suppose it is purely inseparable. This means that all the roots of the polynomial are repeated in its splitting field. For an algebraically closed field F, the polynomial will be of the form f(X) = (X^(p^m) - c), where p is the characteristic of F and c 鈭� F. Now, let's consider f(X) 鈭� k[X]: Since f(X) 鈭� k[X], it must have at least one root in E according to the given condition. Let this root be r 鈭� E. Hence, r^(p^m) = c for some c 鈭� k. Also, since f(X) is a polynomial in k[X], it will have all its roots in k. Therefore, for any algebraically closed extension F of k, the polynomial will have a root in F. As a result, any polynomial of the form f(X) = (X^(p^m) - c) will have a root in E. This means that E is algebraically closed for purely inseparable polynomials as well.
03

Conclusion

Since we have shown that E is algebraically closed for both separable and purely inseparable polynomials, we can conclude that E is indeed algebraically closed.

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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 fundamental concept in field theory, a branch of abstract algebra. If we have two fields, say k and E, where E contains k, we can consider E as an extension of k. Now, this extension is termed algebraic if every element e in E is a root of some non-zero polynomial with coefficients in k. Essentially, an element is algebraic over k if it solves a polynomial equation whose coefficients are from k. This contrasts with transcendental extensions, where some elements do not satisfy such polynomial equations.

In the light of our exercise, we are given that not only is E an extension of k, but it's also stipulated that every non-constant polynomial from k[X] has at least one root in E, which draws us into deeper discussion about what being algebraically closed entails.
Separable Polynomial
A separable polynomial is a special type of polynomial that plays a key role in field theory. In simple terms, a polynomial with coefficients in a field is called separable if all of its roots are distinct, meaning no repeats or duplicates. This is closely related to the concept of derivative: a polynomial is separable if its derivative is not identically zero, and as a consequence, the polynomial and its derivative have no common roots.

Separable polynomials are important because they have nice properties when we extend fields. For instance, they can help us construct larger fields where the polynomial can be factored completely into linear factors (i.e., splitting fields). Also, the existence of a separable polynomial assures us that we can apply the Primitive Element Theorem, which consolidates the extension's complexity into a single element.
Purely Inseparable Polynomial
On the other end of the spectrum, a purely inseparable polynomial is quite the opposite of a separable polynomial. When we speak of a purely inseparable polynomial, we refer to one in which every root is repeated; it lacks any distinct roots. These polynomials are found in extensions of fields with a characteristic > 0, and usually, they take a form where the variable is raised to a power of the characteristic, like X^(p^m).

A critical point about purely inseparable polynomials is that they do not necessarily behave nicely across field extensions. However, in an algebraically closed field, which contains all its algebraic elements, even purely inseparable polynomials split completely, albeit into repeated linear factors.
Primitive Element Theorem

Utility in Field Extensions

The Primitive Element Theorem is a cornerstone in the field theory, providing a powerful simplification for certain algebraic extensions. It states that if we have a finite separable extension of fields, then there exists a single element in the larger field such that the entire extension can be generated by this element over the smaller field. Symbolically, if E is a separable extension of k, there exists an element e in E for which E = k(e). This theorem is incredibly useful because it allows us to represent the entirety of a possibly complex field extension in terms of just one element, simplifying calculations and theory.

Role in the Exercise

In our exercise, the Primitive Element Theorem assures us that if we have a separable polynomial, then all the roots of our polynomial can be expressed using just one 'primitive element', which ties into showing that an extension field E is algebraically closed if it can handle both separable and purely inseparable cases.
Splitting Field
Finally, let's shed light on the concept of a splitting field, which is essentially the stage on which the dramas of separable and purely inseparable polynomials play out. A splitting field of a polynomial with coefficients in some field k is the smallest field extension E of k in which the polynomial can be broken down (split) into linear factors. This means if you have a polynomial f(X), its splitting field is the smallest field where f(X) equals the product of (X - r_i) terms, where r_i are the roots.

In our exercise context, establishing that E is a splitting field for every polynomial in k[X] hinges on proving that E contains every possible root of those polynomials. By addressing separable and purely inseparable polynomials separately, we can conclude that E is indeed algebraically closed鈥攕ince it ultimately acts as a splitting field for all polynomials in question.

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Most popular questions from this chapter

Let \(K / k\) be a finite Galois extension with group \(G\). Let \(\alpha \in K\) be such that \(\\{\sigma \alpha\\}_{\sigma \in G}\) is a normal basis. For each subset \(S\) of \(G\) let \(S(\alpha)=\sum_{\sigma \in s} \sigma \alpha\). Let \(H\) be a subgroup of \(G\) and let \(F\) be the fixed field of \(H\). Show that there exists a basis of \(F\) over \(k\) consisting of elements of the form \(S(\alpha)\).

(a) Let \(k\) be a field of characteristic \(\chi 2 n\), for some odd integer \(n \geq 1\), and let \(\zeta\) be a primitive \(n-\) th root of unity, in \(k .\) Show that \(k\) also contains a primitive \(2 n\) -th root of unity. (b) Let \(k\) be a finite extension of the rationals. Show that there is only a finite number of roots of unity in \(k\).

(a) Let \(K\) be a cyclic extension of a field \(F\), with Galois group \(G\) generated by \(\sigma\). Assume that the characteristic is \(p\), and that \([K: F]=p^{m-1}\) for some integer \(m \geqq 2\). Let \(\beta\) be an element of \(K\) such that \(\operatorname{Tr}_{F}^{K}(\beta)=1\). Show that there exists an element \(\alpha\) in \(K\) such that $$ \sigma \alpha-\alpha=\beta^{p}-\beta $$ (b) Prove that the polynomial \(X^{\prime}-X-\alpha\) is irreducible in \(K[X]\). (c) If \(\theta\) is a root of this polynomial, prove that \(F(\theta)\) is a Galois, cyclic extension of degree \(p^{m}\) of \(F\), and that its Galois group is generated by an extension \(\sigma^{*}\) of \(\sigma\) such that $$ \sigma^{*}(\theta)=\theta+\beta $$

Let \(k\) be a field, \(k^{a}\) an algebraic closure, and \(\sigma\) an automorphism of \(k^{a}\) leaving \(k\) fixed. Let \(F\) be the fixed field of \(\sigma .\) Show that every finite extension of \(F\) is cyclic. (The above two problems are examples of Artin. showing how to dig holes in an algebraically closed field.)

Let \(f(X) \in \mathrm{Q}[X]\) be a polynomial of degree \(n\), and let \(K\) be a splitting field of \(f\) over \(Q\). Suppose that \(\operatorname{Gal}(K / Q)\) is the symmetric group \(S_{n}\) with \(n>2\). (a) Show that \(f\) is irreducible over \(\mathbf{Q}\). (b) If \(\alpha\) is a root of \(f\), show that the only automorphism of \(\mathbf{Q}(\alpha)\) is the identity. (c) If \(n \geqq 4\), show that \(\alpha^{n} \notin Q\).
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