Question

If \(S\) is module-finite over \(R\), then \(S\) is ring -finite over \(R\).

Short answer

Answer: No, module-finite over R and ring-finite over R are not the same, but they are closely related concepts. Module-finite over R means that the given structure can be generated by a finite set of elements over a ring as an R-module, while ring-finite over R means that the given structure can be generated by a finite set of elements over a ring as an R-algebra (which also involves the multiplicative structure). Although not always equivalent, in the given exercise, we showed that if S is module-finite over R, then it is also ring-finite over R.

Step-by-step solution

Module-finite over R

Given that \(S\) is module-finite over \(R\), there exists a finite set of generators, say \(\{s_1, s_2, \dots, s_n\}\), such that any element \(s \in S\) can be written as \(s = r_1s_1 + r_2s_2 + \dots + r_ns_n\) for some elements \(r_1, r_2, \dots, r_n \in R\).

Consider products of generators

Now, let's consider the products of the generators, \(s_i \cdot s_j\) for \(1 \leq i, j \leq n\). Each of these products can be written as an \(R\)-linear combination of the generators, i.e., \(s_i \cdot s_j = r_{ij}^1 s_1 + r_{ij}^2 s_2 + \dots + r_{ij}^n s_n\), where \(r_{ij}^k \in R\).

Generate the whole ring S

We will prove that the set \(\{s_1, s_2, \dots, s_n\}\) generates not only the \(R\)-module \(S\) but also the whole ring \(S\) as an algebra over \(R\). To do this, we need to show that the product of any two elements in \(S\) can be expressed as an \(R\)-linear combination of the generators. Consider \(s, t \in S\). We know from Step 1 that there exist \(r_1, r_2, \dots, r_n, r_1', r_2', \dots, r_n' \in R\) such that \(s = r_1s_1 + r_2s_2 + \dots + r_ns_n\) and \(t = r_1's_1 + r_2's_2 + \dots + r_n's_n\). Now, consider the product \(st\). Using the expressions for \(s\) and \(t\), the product can be written as: \(st = (r_1s_1 + r_2s_2 + \dots + r_ns_n)(r_1's_1 + r_2's_2 + \dots + r_n's_n)\) Expanding the product, every term will be of the form \((r_is_i)(r_j's_j)\) for some \(1 \leq i, j \leq n\). But from Step 2, each of these products can be written as an \(R\)-linear combination of the generators, which can be summed up to obtain a total \(R\)-linear combination of the generators for \(st\).

Conclusion

Therefore, we have shown that the set \(\{s_1, s_2, \dots, s_n\}\) generates the whole ring \(S\) as an algebra over \(R\), since the product of any two elements in \(S\) can be expressed as an \(R\)-linear combination of the generators. This means \(S\) is ring-finite over \(R\).

Key concepts

Understanding Algebraic Geometry

Algebraic geometry is a branch of mathematics that combines algebra, particularly commutative algebra, with geometry. It is fundamentally related to the study of shapes, geometric structures, and the properties of algebraic equations. In this realm, we explore varieties, which are solution sets to systems of polynomial equations. The ring of polynomials, typically denoted as R[x], is an essential object in algebraic geometry, representing the collection of all polynomials with coefficients from a ring R. Inside algebraic geometry, the concept of a module being finite over a ring is quite important as it often underpins the structure of these geometric objects.

For students delving into this field, considering modules as 'generalized vector spaces' over a ring can be a helpful analogy. Just as vector spaces are spanned by a finite number of vectors, when we say a module is finite over a ring, it signifies that the entire module can be generated from a finite set of elements within the ring, much like a finite dimensional vector space. This interplay between algebraic structures, such as rings and modules, and geometric notions is one of the key aspects that make algebraic geometry both challenging and fascinating.

What Does It Mean To Be Ring-Finite Over R?

When we say that a structure, specifically a ring S, is 'ring-finite over R' or 'algebra-finite over R', we're referring to the concept that S can be built from R using only a finite number of R's elements. This term implies that the elements of S are expressive enough that any element of S can be represented as a combination of these finitely many elements. More formally, this means that S is finitely generated as an R-algebra.

To understand this in the context of our exercise, we see that if S is module-finite over R, then every element of S can be expressed as an R-linear combination of some finite generating set. This principle is vital in understanding the foundational makeup of algebra structures and paves the way for more complex algebraic manipulations. Being able to express a ring as ring-finite over another ring indicates a form of algebraic compactness or completeness that is a convenient and powerful property to work with in various contexts of algebra.

The Role of R-linear Combinations

In both algebra and algebraic geometry, the notion of an R-linear combination is central. It is a sum of elements each multiplied by a corresponding element from the ring R. This is analogous to linear combinations in vector spaces, where scalars from a field are used. In the case of modules or rings, these scalars come from a ring, which is a more general algebraic structure than a field as it may lack properties like multiplicative inverse for each element.

Through R-linear combinations, we are able to construct new elements from existing ones within the structure. In the given exercise, the ability to express any product of elements in S as an R-linear combination of a finite set proves that we can encapsulate the whole ring with just a finite part of it. This concept is critical because it allows for the reduction of infinite possibilities into a finite, more manageable framework. Understanding R-linear combinations is thus a gateway to grasping more complex algebraic concepts and revealing the underlying simplicity in seemingly complex algebraic structures.