Question

Let \(R\) be a subring of \(S, S\) a subring of (a domain) \(T .\) If \(S\) is integral over \(R\), and \(T\) is integral over \(S\), show that \(T\) is integral over \(R\). (Hint: Let \(z \in T\), so we have \(z^{n}+a_{1} z^{n-1}+\cdots+a_{n}=0, a_{i} \in S\). Then \(R\left[a_{1}, \ldots, a_{n}, z\right]\) is module-finite over \(\left.R .\right)\)

Short answer

Question: Show that if S is integral over R and T is integral over S, then T is also integral over R. Assume that R is a subring of S and S is a subring of T which is a domain (integral domain). Answer: The main concept behind the proof is to show that an arbitrary element in T, z, is integral over R, meaning that there exists a monic polynomial in R[x] that z satisfies. By considering subring R[a_1,...,a_n,z] and using the fact that R is a subring of S, we can demonstrate that z is integral over R, proving the original claim. Thus, if S is integral over R and T is integral over S, then T is integral over R.

Step-by-step solution

Start with an arbitrary element from T

Let us begin the proof by considering an arbitrary element \(z \in T\).

Construct the polynomial of z with coefficients in S

Since \(T\) is integral over \(S\), there exists a monic polynomial with coefficients in \(S\) such that \(z\) satisfies this polynomial. Let this polynomial be \(p(x) = x^n + a_1x^{n-1} + \dots + a_n\), where \(a_i \in S\) for \(1 \leq i \leq n\).

Consider the subring R[a_1, ..., a_n, z]

Create the subring \(R[a_1,\dots, a_n, z]\), which contains \(R\) and all the elements \(a_1, \dots, a_n\), and \(z\). This subring is module-finite over \(R\) as \(S\) is integral over \(R\) and therefore module-finite.

Use the fact that z is integral over R[a_1, ..., a_n, z]

By definition of integral, the element \(z\) is integral over the subring \(R[a_1,\dots, a_n, z]\). Therefore, there exists a monic polynomial with coefficients in \(R[a_1,\dots, a_n, z]\) that \(z\) satisfies. Let this polynomial be \(q(x) = x^m + b_1x^{m-1} + \dots + b_m\), where \(b_i \in R[a_1,\dots, a_n, z]\) for \(1 \leq i \leq m\).

Express the coefficients b_i in R[a_1, ..., a_n, z]

Each of the coefficients \(b_i \in R[a_1,\dots, a_n, z]\) can be written as a finite linear combination of elements from \(R\) and the elements \(a_1, \dots, a_n, z\). Since the subring \(R[a_1,\dots, a_n, z]\) is module-finite over \(R\), we know that these linear combinations are finite.

Substitute z into the polynomial q(x)

Substitute \(z\) into the polynomial \(q(x)\): $$q(z) = z^m + b_1z^{m-1} + \dots + b_m$$As each \(b_i\) is a finite linear combination of elements of \(R\), and the subring \(R[a_1,\dots, a_n, z]\) is module-finite over \(R\), the expression $$q(z) = z^m + b_1z^{m-1} + \dots + b_m$$will be an element of the subring \(R[a_1,\dots, a_n, z]\).

Show that z is integral over R

As \(q(z)\) is an element of the subring \(R[a_1,\dots, a_n, z]\) and \(R\) is a subring of \(S\), we can show that \(z\) is integral over \(R\), thus proving the original claim. The polynomial \(q(x)\) has coefficients in \(R\) and is monic, meaning that there exists a monic polynomial in \(R[x]\) such that \(z\) is a root. Therefore, \(T\) is integral over \(R\). Hence, if \(S\) is integral over \(R\) and \(T\) is integral over \(S\), then \(T\) is integral over \(R\).

Key concepts

Understanding a Subring

The concept of a subring serves as a cornerstone in the study of algebraic structures. A subring is essentially a subset of a ring that is a ring in its own right, equipped with the same operations of addition and multiplication as the larger ring. For a subring to be qualified as such, it must be non-empty and closed under subtraction and multiplication. Moreover, it must contain the multiplicative identity of the larger ring if the larger ring is unital.

In the context of our exercise, we are examining rings that are subrings of other rings. This forms a chain of subrings, in our case, with the ring R being a subring of S, and S being a subring of T. Each subring inherits the ring properties from its parent ring but may have additional properties or restrictions based on its own elements.

The Role of Monic Polynomials

A monic polynomial is a single-variable polynomial where the leading coefficient—the coefficient of the highest power—is one. This small but important detail has significant consequences. For instance, in the context of our exercise, the fact that a monic polynomial with coefficients in S satisfies an element z in T is a manifestation that z is integral over S.

To put it simply, being integral means that an element is a root of a monic polynomial with coefficients in a certain subring. Thus, monic polynomials bridge the relationship between elements and the subrings they are integral over. The existence of such polynomials is integral (pun intended) to demonstrating the concept of integral extensions within algebraic domains.

Module-Finite Extensions

Module-finite extensions deal with the notion of an algebra being finitely generated as a module over a subring. When we say that a subring A is module-finite over a subring B, we imply that A has a finite basis as a module, meaning it can be spanned by a finite number of elements from B. In the step-by-step solution provided, it is essential to understand that the subring formed by R along with the coefficients of our monic polynomial and the element z is module-finite over R. This suggests that there's a finite R-module including all combinations of these elements, thus providing a concrete structure through which we can prove the integral nature of T over R.

Algebraic Domain Fundamentals

The term 'algebraic domain,' within the scope of our exercise, refers to rings that are integral domains. An integral domain is a type of ring with no zero divisors; two non-zero elements multiplied together will never give zero. This property ensures that we can carry on the operations needed to consider integral elements and monic polynomials without running into complications caused by zero divisors.

In short, an algebraic domain provides a clean and predictable framework for discussing integral extensions. Since integral elements are roots of monic polynomials, these polynomials have more predictable behavior in the algebraic domain, meaning their roots cannot be trivialized by zero divisors. Thus, being able to work within an algebraic domain greatly simplifies the proof of integral extensions like the one at hand.