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Chapter 12: Problem 4

Let \(A\) be a principal entire ring, and let \(K\) be its quotient field. Let o be a valuation ring of \(K\) containing \(A\), and assume \(0 \neq K\). Show that o is the local ring \(A_{(D)}\) for some prime element \(p\). [This applies both to the ring \(\mathbf{Z}\) and to a polynomial ring \(k[X]\) over a field \(k\).]

Short Answer

Expert verified
In summary, we have shown that the valuation ring o is the local ring \(A_{(D)}\) for some prime element p. We found a prime element p 鈭� A that is in the maximal ideal m of o and showed that every element in o can be written in the form a/b with a and b 鈭� A, and b not divisible by p. Conversely, we demonstrated that every element in \(A_{(D)}\) belongs to o. Therefore, the valuation ring o is indeed the local ring \(A_{(D)}\) for the prime element p.

Step by step solution

01

1. Definition of prime element, principal entire ring, and quotient field#

A prime element in a ring is an element p such that whenever p divides a product of two elements in the ring, p must divide at least one of the two factors. Formally, if p | xy, where x and y are elements in the ring, then p | x or p | y. A principal entire ring, denoted by A, is an integral domain in which every nonzero, non-unit element can be written as a product of prime elements, and the factorization is unique up to the order and unit multiples of the prime factors. The quotient field of an integral domain A, denoted by K, is the smallest field containing A, which can be obtained by taking the field of fractions of A. In other words, it is the set of rational expressions of the form a/b, where a and b are elements of A and b 鈮� 0.
02

2. Introduction of valuation ring#

A valuation ring is a subring o of a field K (in this case, the quotient field of A) that has the following properties: 1. For any element x in K, either x or its inverse (1/x) belongs to o. 2. The intersection of o with the set of nonzero elements in K is a multiplicative group. An important property of valuation rings is that they are local rings, that is, they have exactly one maximal ideal.
03

3. Local ring and its properties#

A local ring is a commutative ring with a unique maximal ideal. This maximal ideal, denoted by m, consists of all non-invertible elements in the ring. For a principal entire ring A and a prime element p, we can associate a local ring \(A_{(D)}\), which is obtained by inverting all elements of A that are not divisible by p. In other words, the local ring \(A_{(D)}\) is formed by taking fractions of the form a/b, where a and b are elements of A, and b is not divisible by p.
04

4. Show that the valuation ring o is the local ring \(A_{(D)}\) for some prime element p#

To show that o is the local ring \(A_{(D)}\) for some prime element p, we need to find a prime element p such that o consists of elements of the form a/b, where a and b are elements of A, and b is not divisible by p. Since o is a valuation ring, it is a local ring with a unique maximal ideal m. Note that A is a subring of o. Let p be an element of A that is in m. Since m is a unique maximal ideal, p must be prime in A (otherwise, there would be another ideal that contains p, which contradicts the uniqueness of m). Now let's prove that o is the local ring \(A_{(D)}\) corresponding to this prime element p. Let x be an element in o. Then x can be written as a/b for some elements a and b in A, since o is a subring of K, the quotient field of A. Because o is a valuation ring, either b or 1/b belongs to o. If b 鈭� o, then x 鈭� \(A_{(D)}\) by definition. If 1/b 鈭� o, then p does not divide b, because p 鈭� m, and m is the maximal ideal of o, which means that b should be invertible in o. Therefore, x 鈭� \(A_{(D)}\) as well. Conversely, let y be an element in \(A_{(D)}\). Then y can be written as a/b for some elements a and b in A, where b is not divisible by p. Since p does not divide b, b 鈭� o. Therefore, y = a/b 鈭� o, as o is closed under multiplication and inverses. Hence, the valuation ring o is indeed the local ring \(A_{(D)}\) for the prime element p we found.

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Key Concepts

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

Principal Entire Ring
A principal entire ring, often referred to simply as a principal ring, is a specific type of integral domain. An integral domain itself is a ring where the product of any two non-zero elements is non-zero. What distinguishes a principal entire ring is that every ideal in the ring can be generated by a single element. This means that you can express any ideal of the ring as multiples of just one element, which is a big simplification when dealing with calculations in ring theory.
This unique factorization property has practical implications, particularly when solving polynomial equations, as it helps to simplify the expressions considerably. All elements in a principal entire ring can be expressed as multiples of primes (just like integers can be expressed in terms of prime numbers). This characteristic makes principal rings highly relevant in algebra and simplifies understanding structure and divisibility within the ring.
Quotient Field
The quotient field of an integral domain is a concept that extends the idea of fractions familiar from numbers to more abstract algebraic structures. If you have an integral domain, say A, its quotient field K is the smallest field that contains A. It is somewhat similar to taking the integers and constructing the rational numbers as fractions of integers.
To form the quotient field, you consider expressions of the form \(\frac{a}{b}\), where \(a\) and \(b\) are elements of the integral domain A, and \(b eq 0\). This operation allows us to construct a field in which every non-zero element has an inverse, and operations like addition, subtraction, multiplication, and division (except by zero) are possible. The quotient field is important as it allows us to work with more flexible mathematical operations and maintain properties that may not hold in the ring itself.
Local Ring
A local ring is a commutative ring with a special property: it has a unique maximal ideal. In simpler terms, it means there is only one biggest ideal beyond which you can't get any other ideals in a meaningful way. This property makes local rings particularly easy to work with in algebraic contexts since they simplify a lot of otherwise complex structures.
In practical terms, the maximal ideal of a local ring consists of all the non-invertible elements of the ring. If you have a principal entire ring and you localize it around a prime element, you get a local ring \(A_{(D)}\). Here, the elements are fractions \(\frac{a}{b}\) where \(a\) and \(b\) are in A, and \(b\) is not divisible by some prime element. Local rings feature prominently in algebraic geometry and number theory because of their specialized yet straightforward structures.
Prime Element
A prime element within a ring is something you'll encounter when breaking down elements into factors. Specifically, a prime element \(p\) is one where if \(p\) divides the product of two elements in the ring, \(p\) must divide at least one of those elements. It's a concept similar to prime numbers, but extended to polynomial rings and other algebraic structures.
Prime elements help in maintaining the uniqueness of factorization in a ring. They act as the building blocks for constructing other elements, in the same way, prime numbers are the building blocks of whole numbers. Understanding prime elements is crucial when solving polynomial equations and in the study of mathematical structures like principal entire rings or valuation rings. In these contexts, prime elements help determine the structure of the ring and facilitate calculating local properties of the ring like local rings.

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Most popular questions from this chapter

Show that the \(p\) -adic numbers \(Q_{p}\) have no automorphisms other than the identity. [Hint: Show that such automorphisms are continuous for the \(p\) -adic topology. Use Corollary \(7.5\) as an algebraic characterization of elements close to \(1 .]\)

(a) Let \(f(X)\) be a polynomial of degree 1 in \(\mathbf{Z}[X]\). Show that the values \(f(a)\) for \(a \in \mathbf{Z}\) are divisible by infinitely many primes. (b) Let \(F\) be a finite extension of \(\mathbf{Q}\). Show that there are infinitely many primes \(p\) such that all conjugates of \(F\) (in an algebraic closure of \(\mathbf{Q}_{p}\) ) actually are contained in \(\mathbf{Q}_{p}\). [Hint: Use the irreducible polynomial of a generator for a Galois extension of \(\mathbf{Q}\) containing \(F .]\)

Prove that the field \(\mathbf{Q}_{p}\) contains all \((p-1)\) -th roots of unity. [Hint : Use Proposition 7.6, applied to the polynomial \(X^{p-1}-1\) which splits into factors of degree 1 in the residue class field. ] Show that two distinct \((p-1)\) -th roots of unity cannot be congruent mod \(p\).

Let \(K\) be a field complete with respect to a discrete valuation, let o be the ring of integers of \(K\), and assume that o is compact. Let \(f_{1}, f_{2}, \ldots\) be a sequence of polynomials in \(n\) variables, with coefficients in o. Assume that all these polynomials have degree \(\leqq d\), and that they converge to a polynomial \(f\) (i.e. that \(\left|f-f_{i}\right| \rightarrow 0\) as \(i \rightarrow \infty\) ). If each \(f_{i}\) has a zero in 0 , show that \(f\) has a zero in o. If the polynomials \(f_{i}\) are homogeneous of degree \(d\), and if each \(f_{1}\) has a non-trivial zero in o, show that \(f\) has a non-trivial zero in o. [Hint: Use the compactness of o and of the units of o for the homogeneous case.]

Let \(F\) be a complete field with respect to a discrete valuation, let o be the valuation ring. \(\pi\) a prime element, and assume that \(0 /(\pi)=k\). Prove that if \(a, b \in 0\) and \(a \equiv b\left(\bmod \pi^{\prime}\right)\) with \(r>0\) then \(a^{r^{n}} \equiv b^{p^{n}}\left(\bmod \pi^{r+\pi}\right)\) for all integers \(n \geqq 0\).
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