Question

Let \(R\) be a domain with quotient field \(K\), and let \(L\) be a finite algebraic extension of \(K\). (a) For any \(v \in L\), show that there is a nonzero \(a \in R\) such that \(a v\) is integral over \(R\). (b) Show that there is a basis \(v_{1}, \ldots, v_{n}\) for \(L\) over \(K\) (as a vector space) such that each \(v_{i}\) is integral over \(R\).

Short answer

In this exercise, we have proved two statements related to the integral closure of a domain \(R\) in a finite algebraic extension \(L\) of its quotient field \(K\). For part (a), we have shown that for any element \(v \in L\), there is a nonzero element \(a \in R\) such that its product \(av\) is integral over \(R\). We used the property of \(L\) being a finite algebraic extension of \(K\) and the definition of integral elements in our proof. For part (b), we demonstrated that there is a basis for the vector space of \(L\) over \(K\) such that each of the basis elements is integral over \(R\). We used the result of part (a) and the properties of finite-dimensional vector spaces in our proof.

Step-by-step solution

Part (a): Proof that there exists a nonzero \(a \in R\) such that \(av\) is integral over \(R\)

Since \(L\) is a finite algebraic extension of \(K\), it means that for any \(v \in L\), there is a monic polynomial \(f(x) \in K[x]\) such that \(f(v) = 0\). We can write \(f(x) = x^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0\) with \(a_i \in K\) for \(i = 0, \dots, n-1\). Now we apply the monic polynomial to \(v\): $$f(v) = v^n + a_{n-1}v^{n-1} + \dots + a_1 v + a_0 = 0.$$ As the coefficients \(a_i\) are in \(K\), each \(a_i\) can be written as a fraction \(a_i = \frac{r_i}{s_i}\) where \(r_i, s_i \in R\) and \(s_i \neq 0\). Now define \(b = s_0 s_1 \dots s_{n-1}\). Then, $$b^n f(v) = b^n(v^n + \frac{r_{n-1}}{s_{n-1}}v^{n-1} + \dots + \frac{r_1}{s_1}v + \frac{r_0}{s_0}) = 0.$$ By clearing the denominators, we obtain a new polynomial \(g(v) = b^n v^n + c_{n-1} b^{n-1} v^{n-1}+ \dots + c_1 b v + c_0 = 0\), where \(c_i = b^i r_i\) for \(i = 0, \dots, n-1\). Note that this new polynomial has coefficients in \(R\) and \(g(bv) = 0\), which makes \(bv\) integral over \(R\). Since \(b\) is a nonzero element in \(R\), we have proved part (a).

Part (b): Proof that there exists a basis for \(L\) over \(K\) whose elements are integral over \(R\)

First, observe that \(L\) is a finite-dimensional \(K\)-vector space, since it is a finite algebraic extension of \(K\). Let \(\{w_1, w_2, \dots, w_n\}\) be a basis for \(L\) over \(K\). For each \(i = 1, \dots, n\), by part (a), there exists a nonzero \(a_i \in R\) such that \(a_i w_i\) is integral over \(R\). Let \(b = a_1 \dots a_n\). Then, \(b w_i \in R[a_i w_i]\) for each \(i\), and \(R[a_i w_i]\) is finitely generated as an \(R\)-module. Thus, \(b w_i\) is integral over \(R\) for each \(i = 1, \dots, n\). So, define the new basis of \(L\) over \(K\) as \(\left\{\frac{1}{b} w_i\right\}_{i=1}^n\). We have that \(b\left(\frac{1}{b} w_i\right) = w_i\), which means each basis element is integral over \(R\), completing the proof for part (b).

Key concepts

Integral Elements

When discussing algebraic extensions, the term "integral element" comes into play. But what does it exactly mean? Well, an element is said to be integral over a ring if it satisfies a polynomial equation with coefficients from the ring, whose leading term is 1. This polynomial is also known as a monic polynomial.

In this context, consider a ring, say domain \(R\), with a quotient field \(K\), and a larger field \(L\), which is a finite algebraic extension of \(K\). The exercise proposes that for any element \(v\) from \(L\), we can find a nonzero element \(a\) from \(R\) such that \(a\cdot v\) becomes integral over \(R\).

Here's the process: since \(L\) is a finite extension of \(K\), \(v\) already satisfies some monic polynomial over \(K\). Given this polynomial, we aim to clear any denominators in the coefficients by multiplying through by an appropriate element from \(R\). By doing so, we demonstrate that \(av\) is integral over \(R\), thus confirming the integral property.

Vector Space Basis

A vector space basis is a fundamental concept in linear algebra that provides a framework for describing every element within a vector space using linear combinations of the basis elements. When talking about field extensions like \(L/K\), we're treating \(L\) as a vector space over \(K\).

To find a basis for \(L\) over \(K\), we look for a minimal set of elements in \(L\) such that any other element of \(L\) can be obtained by linearly combining these basis elements with coefficients from \(K\). The exercise includes a basis \(\{w_1, w_2, \ldots, w_n\}\) of \(L\) over \(K\). This means every element in \(L\) can be expressed as a combination of these basis elements.

Further, the exercise illustrates that each of these basis elements can be made integral over \(R\). By using the established integral element \(a_i\) for each basis vector \(w_i\) (from the part about integral elements), we build a new basis with elements \(b w_i\) that are integral over \(R\). Such transformation keeps the algebraic structure but adds the property of integrality, which is beneficial for certain computations or theoretical investigations in algebra.

Polynomial Roots

In algebra, roots of polynomials are not only key in solving equations but also crucial in understanding algebraic extensions. When we state that \(v\) in \(L\) is a root of a polynomial in \(K[x]\), we mean that plugging \(v\) into the polynomial yields zero. This result solidifies \(v\) as an algebraic element over \(K\).

For a given polynomial \(f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0\), if \(f(v) = 0\), it indicates that \(v\) lies within some extension of \(K\) — here, it's \(L\). Solving such polynomials often gives insight into how \(L\) is structured above \(K\) and helps identify important elements like integral elements and vector space bases as described earlier.

Also, cleaning denominators in the coefficients helps define new polynomial expressions whose roots maintain the required algebraic properties while simplifying integration over \(R\). Such methods are vital in establishing the concept of integrality and ensuring elements fit neatly within the chosen algebraic structure.