Question

Let \(K\) be a subfield of a field \(L\). (a) Show that the set of elements of \(L\) that are algebraic over \(K\) is a subfield of \(L\) containing \(K\). (Hint: If \(v^{n}+a_{1} v^{n-1}+\cdots+a_{n}=0\), and \(a_{n} \neq 0\), then \(v\left(v^{n-1}+\cdots\right)=-a_{n}\) ) (b) Suppose \(L\) is module-finite over \(K\), and \(K \subset R \subset L\). Show that \(R\) is a field.

Short answer

Question: Prove the following statements regarding field extensions: (a) Show that the set of elements in field \(L\) that are algebraic over \(K\) is a subfield of \(L\) containing \(K\). (b) Given that \(L\) is module-finite over \(K\) and \(K \subset R \subset L\), prove that \(R\) is a field. Answer: (a) To prove that the set of elements in \(L\) that are algebraic over \(K\) is a subfield of \(L\) containing \(K\), we showed that this set, denoted as \(A\), is non-empty, closed under addition, subtraction, multiplication, and division. (b) To prove that \(R\) is a field, we showed that every non-zero element in \(R\) has an inverse in \(R\), taking advantage of the module-finite property of \(L\) over \(K\).

Step-by-step solution

Define the set of algebraic elements

Let \(A\) be the set of all elements of \(L\) that are algebraic over \(K\). Our goal is to show that \(A\) is a subfield of \(L\).

Show A is non-empty

Since \(K\) is a subfield of \(L\), all elements of \(K\) are elements of \(L\) and trivially algebraic over \(K\). Therefore, \(A\) is non-empty.

Prove that A is closed under addition and subtraction

Let \(v, w \in A\). This means that both \(v\) and \(w\) are algebraic over \(K\). Now, let's consider the sum \(v+w\) and difference \(v-w\). There exist nonzero polynomials \(p(x), q(x) \in K[x]\) such that \(p(v) = 0\) and \(q(w) = 0\). Consider the polynomial \(r(x) = p(x)q(x-1) + q(x)p(x+1)\). It can be seen that this polynomial has coefficients in \(K\) and \(r(v+w) = 0\). Similarly, for the difference case, consider the polynomial \(s(x) = p(x)q(1-x) + q(x)p(1-x)\). We have \(s(v-w)=0\). Hence, \(A\) is closed under addition and subtraction.

Prove that A is closed under multiplication

Let \(v, w \in A\). This means that both \(v\) and \(w\) are algebraic over \(K\). Now let's consider the product \(vw\). There exist nonzero polynomials \(p(x), q(x) \in K[x]\) such that \(p(v) = 0\) and \(q(w) = 0\). Consider the polynomial \(t(x) = p(x)q(x)\). It can be seen that this polynomial has coefficients in \(K\) and \(t(vw) = p(v)q(w) = 0\). Hence, \(A\) is closed under multiplication.

Prove that A is closed under division

Let \(v, w \in A\) with \(w\neq0\). This means that both \(v\) and \(w\) are algebraic over \(K\). Now let's consider the quotient \(\frac{v}{w}\). There exist nonzero polynomials \(p(x), q(x) \in K[x]\) such that \(p(v) = 0\) and \(q(w) = 0\). We have a hint that if \(v^n + a_1v^{n-1}+\cdots+a_n = 0\) and \(a_n \neq 0\), then \(v(v^{n-1}+\cdots)= - a_n\). Using the hint, we have \(w(w^{m-1}q(w)+\ldots + q(w)) = -a_m\) for some \(m\), where \(a_m \neq 0\). Now take the polynomial \(r(x)= p(wx+a_m)\). It can be seen that this polynomial has coefficients in \(K\) and \(r\left(\frac{v}{w}\right)=0\). Hence, \(A\) is closed under division.

Prove R is a field

We now proceed to part (b). Since \(L\) is module-finite over \(K\), the module \(K + Kx_1 + \cdots + Kx_n\) is dense in \(L\), where \(x_1, \ldots, x_n \in L\). Therefore, for any non-zero \(y \in R\), there exists a linear combination \(z = k_0 + k_1x_1+ \cdots + k_nx_n\), where \(k_i \in K\), such that \(yz = 1\). It can be seen that \(z \in R\), since \(R\) is closed under addition and multiplication by elements of \(K\). Also, \(1 \in R\) as a consequence of \(K \subset R\). Therefore, every non-zero element \(y\) in \(R\) has an inverse, \(z\), in \(R\). Hence, \(R\) is a field.

Key concepts

Subfield of a Field

In the realm of algebra, the concept of a subfield is fundamental. Imagine a field, denoted as L, as a playground where certain numbers can play, interacting through addition, subtraction, multiplication, and division without breaking any rules of arithmetic (except dividing by zero). A subfield, denoted as K, can be seen as a smaller playground inside L that holds onto the same rules.

For a formal definition, a subfield K of a field L must satisfy the following three criteria: it should contain the additive and multiplicative identities of L, it must be closed under the operations of addition, subtraction, multiplication, and the inverse of multiplication (except by zero), and all these operations should comport to the field's axioms on both K and L. Thus, every element you play with in K behaves exactly as it would in the larger field L, and every 'game' you can play in L, you can also play in K, albeit with fewer players.

Algebraic Over a Field

Diving into the term algebraic over a field, we enter a scenario where elements of a larger field L have a special relationship with a subfield K. An element is considered algebraic over K if it can solve a polynomial equation where the coefficients are from K—essentially, the element can 'resolve drama' in K. Not just any drama, though: we're talking about zero-drama, where the polynomial equation equals zero.

For example, if we can find a polynomial with coefficients in K such that when we plug our element from L into the polynomial, the result is zero, then we've found an algebraic over K. The simplest instance is when L itself is an element of K; every element in K is trivially algebraic over K. In this case, we could say the elements of K are 'self-solving' actors; they always know their cue to zero in a polynomial script.

Closure Properties

Last but not least, the term closure properties is key in this algebraic playhouse. In this context, closure is like saying a playground is 'complete' meaning that if kids (elements) engage in any of the playground activities (algebraic operations), they won't fall out of the playground (field). To be closed under an operation, say addition, means if you take any two kids (elements) from the playground (field) and add them together, their sum is also a kid who belongs to the same playground (element of the field).

The same logic applies to subtraction, multiplication, and inverses (but we still do not divide by 'the kid who is not there,' zero!) When we consider the set of all algebraic elements over K, we see that this set A does keep all kids safely within the playground. Through clever algebraic manipulation with polynomials, as illustrated in the step-by-step solution, we can show that all operations keep us in A. That's what we mean when we say A is a subfield of L: no matter how the algebraic elements play with each other, they always stay within the world of algebraic elements.