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Chapter 5: Problem 9

Prove that if \(R\) is Noetherian, then so is \(R\left[x_{1}, \ldots, x_{n}\right]\).

Short Answer

Expert verified
To prove that if a commutative ring \(R\) is Noetherian then so is \(R[x_1, \ldots, x_n]\), we proceed by induction on \(n\). First, we show that \(R[x_1]\) is Noetherian by considering an ideal \(I\) and the leading coefficients ideals \(J_d\). We show that each \(J_d\) is finitely generated as \(R\) is Noetherian and that \(I\) is finitely generated by elements in the set \(I_d\). Next, we assume that \(R[x_1, \ldots, x_{n-1}]\) is Noetherian and consider the polynomial ring \(R[x_1, \ldots, x_n] \cong R[x_1, \ldots, x_{n-1}][x_n]\). By applying similar reasoning as in the first step, we show that an ideal \(I\) of \(R[x_1, \ldots, x_n]\) is finitely generated, proving that \(R[x_1, \ldots, x_n]\) is Noetherian. This completes the induction and proves the statement for any positive integer \(n\).

Step by step solution

01

Show that R[x_1] is Noetherian

We begin by showing that the polynomial ring in one variable, \(R[x_1]\), is Noetherian. Let \(I\) be an ideal of \(R[x_1]\). For each positive integer \(d\), let \(J_d\) denote the ideal of \(R\) generated by the leading coefficients of all polynomials in \(I\) of degree \(d\). Part 1: Each ideal \(J_d\) is finitely generated. Since \(R\) is Noetherian, it follows that each \(J_d\) is finitely generated, say \(J_d = (a_{d1}, \ldots, a_{dk_d})\). Part 2: The ideal \(I\) is finitely generated. Let \(I_d\) be the set of all \(k_d\) polynomials in \(I\) with the smallest possible degrees corresponding to each \(a_{dj} \in J_d\). Then, we claim that every polynomial \(f(x_1) \in I\) can be written as a finite sum of elements of \(I_d\). Indeed, we can use long division to repeatedly subtract elements of \(I_d\) to reduce the degree of \(f(x_1)\), until the degree of the remaining polynomial is smaller than any polynomial in \(I_d\). Since the coefficients belong to the ideal generated by the leading coefficients of polynomials in \(I_d\), the remainder must be zero. Hence, \(I\) is finitely generated by elements in \(I_d\) and \(R[x_1]\) is Noetherian.
02

Inductive step (Prove R[x_1, ..., x_n] is Noetherian)

Now assume that \(R[x_1, \ldots, x_{n-1}]\) is Noetherian, and consider the polynomial ring \(R[x_1, \ldots, x_n] \cong R[x_1, \ldots, x_{n-1}][x_n]\). We want to show that this polynomial ring is also Noetherian. Let \(I\) be an ideal of \(R[x_1, \ldots, x_n]\). By our assumption, each ideal in \(R[x_1, \ldots, x_{n-1}]\) is finitely generated. Thus, each ideal of leading coefficients of polynomials in \(I\) is finitely generated. Now, we can apply the same reasoning as in Step 1 to find a finite set \(I_d\) of polynomials in \(I\) with the smallest degrees corresponding to each generator of the leading coefficient ideal in each degree \(d\). This shows that I is finitely generated, and thus \(R[x_1, \ldots, x_n]\) is Noetherian. This completes the induction and proves the statement for any positive integer \(n\).

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Key Concepts

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

Polynomial Rings
A polynomial ring is formed when we take a ring, which is a foundational mathematical structure encapsulating concepts like addition and multiplication, and extend it to include polynomials. Polynomials are expressions involving a variable, usually denoted by 鈥榵鈥�, raised to various powers that are whole numbers, and coefficients that belong to the original ring. In essence, a polynomial ring is a set consisting of all possible polynomials with coefficients in a given ring.

For instance, consider a ring denoted by 鈥楻鈥�. The polynomial ring in one variable over 鈥楻鈥� is represented as 鈥楻[x]鈥�. If R is a Noetherian ring, meaning it satisfies the condition that every ideal of R is finitely generated, we can investigate how this property extends to polynomial rings. The extension from a Noetherian ring to its polynomial ring is crucial since it preserves the property that allows for the efficient handling of ideals鈥攆undamental sets within ring theory that follow certain operational rules.
Ideals in Rings
Ideals are special subsets of a ring that play a central role in the algebraic structure of the ring. They are defined such that they are closed under addition, and when any element of the ideal is multiplied by any element from the ring, the result still lies within the ideal. You can think of ideals as 鈥榬ing-compatible鈥� sets that provide a way to construct new rings, known as quotient rings, and to study the structure of the original ring efficiently.

Within the context of Noetherian rings, which include polynomial rings, the concept of ideals becomes incredibly important. To be Noetherian implies that every ideal in the ring is generated by a finite number of elements, a property that also contributes to the name 'finitely generated鈥�. Each ideal can be written as a combination of these generators, which simplifies many aspects of ring analysis and problem-solving.

The proof that a polynomial ring over a Noetherian ring is itself Noetherian hinges heavily on understanding how ideals are constructed and how their properties are preserved under the operation of forming a polynomial ring.
Inductive Proof
An inductive proof is a mathematical technique used to prove statements that are phrased in terms of natural numbers. It relies on the principle of mathematical induction: if a statement holds for the first natural number (usually 1) and if the truth of that statement for a number 'n' implies its truth for 'n + 1', then the statement holds for all natural numbers.

This form of proof is particularly suited to demonstrating properties that are thought to extend indefinitely, like the Noetherian property of polynomial rings in an increasing number of variables. The exercise tackled this by first proving the base case 鈥� that a polynomial ring in one variable over a Noetherian ring is Noetherian. Then, by assuming virtually that a polynomial ring with 'n-1' variables is Noetherian, the inductive step extends this assumption to prove that adding one more variable (making it 'n' variables in total) preserves the Noetherian property.

Using inductive proof, broad and powerful results can be validated by breaking them down into a base case and an inductive step, thereby affirming the statement's validity in a step-by-step progression.

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Most popular questions from this chapter

Show that an \(R\)-module \(M\) satisfies the ACC for submodules if and only if the following condition holds. Every nonempty collection \(\mathcal{S}\) of submodules of \(M\) has a maximal element. That is, for every nonempty collection \(\mathcal{S}\) of submodules of \(M\) there is an \(S \in \mathcal{S}\) with the property that \(T \in \mathcal{S} \Rightarrow T \subseteq S .\)

If \(R\) is Noetherian and \(\mathcal{I}\) is an ideal of \(R\) show that \(R / \mathcal{I}\) is also Noetherian.

Prove that the union of an ascending chain of submodules is a submodule.

Prove that any \(R\)-module \(M\) is isomorphic to the quotient of a free module \(F\). If \(M\) is finitely generated, then \(F\) can also be taken to be finitely generated

Let \(\mathcal{I}\) be an ideal of \(R\). Prove that if \(R / \mathcal{I}\) is a free \(R\)-module, then \(\mathcal{I}\) is the zero ideal.
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