# Is a generated field independent of the extension over which it is generated?

Mathematics Asked by jam on December 16, 2020

This sounds like a very basic question of field theory. It’s probably a small detail, most authors and introductory sources to field theory seem to ignore it.

Given a field extension $$L/K$$ and a subset $$S$$ of $$L$$, there is a smallest subfield of $$L$$ that contains $$K$$ and $$S$$. It is the intersection of all subfields of $$L$$ that contain $$K$$ and $$S$$, and is denoted by $$K(S)$$. My confusion is about this notation $$K(S)$$ which does not denote explictly $$L$$. The first time I read this definition, I wanted to write it $$K_L(S)$$ because I don’t know whether it depends on $$L$$. But most sources just write $$K(S)$$ and that’s where my confusion is, because they seem to implicitly assume that it somehow does not depend on $$L$$ which I don’t find obvious.

My question is thus: If I have two extensions $$L/K$$ and $$L’/K$$ over the same field $$K$$, and $$Ssubseteq Lcap L’$$, is it true that $$K_L(S) = K_{L’}(S)$$ or at least that they are field isomorphic ?

I am not assuming any relationship between $$L$$ and $$L’$$, I am assuming the minimum for $$K_L(S)$$ and $$K_{L’}(S)$$ to be defined, $$Lcap L’$$ might not be a field at all.

If the two extensions form a tower, for example $$Lsubseteq L’$$ then I think it is true because $$K_L(S)$$ represents the smallest set closed under a finite number of field operations. However if $$L$$ and $$L’$$ are not a tower, i do not have an answer. I don’t even know if the sum/product in $$L$$ would correspond to the sum/product in $$L’$$ for two elements coming from $$S$$.
For that reason, i find it confusing to use the notation $$K(S)$$ without any explanation, which most sources do.

It seems like an important thing to understand, because I have noticed that most people then usually, don’t think of the extension $$L$$ at all, they seem to just think of $$K(V)$$ as "adjoining" some elements $$V$$ into a field $$K$$, and they really seem to think that $$K(V)$$ as if the concept could be defined just from $$K$$ and $$V$$. However my attempt to define "an extension generated from a set $$V$$ and a field $$K$$" just fails, without $$L$$ because i need some underlying operations to be defined.

EDIT: Thanks to Lee Mosher for his answer: Yes $$K_L(S)$$ depends on $$L$$.

I think a motivation for writing only $$K(S)$$ is the following:

So far what I have defined is, a generated subextension $$K_L(S)$$.
We can actually extend the concept to a generated extension.
An extension $$L/K$$ is said to be generated by $$Ssubseteq L$$ if $$L=K_L(S)$$.
According to that definition we can notice that $$K_L(S)$$, as an extension, is generated by $$S$$, because $$K_L(S) = K_{K_L(S)}(S)$$, which works because $$K rightarrow K_L(S) rightarrow L$$ is a tower of extensions.
Therefore it makes sense to just write $$K(S)$$ because it is constant in a given tower, and it avoids writing those packed subscripts.
Also that way, concepts like a simple extension can be elegantly defined as an extension generated by one of its elements and written $$K(alpha)$$ (but truly it still depends on L).

Where you wrote "they seem to implicity assume that it somehow does not depend on $$L$$", in fact the opposite is true, they implicitly assume that it does depend on $$L$$.

Without such an assumption, there is a kind of pedestrian construction of counterexamples, using the concept of "transport of structure"; it is to avoid such counterexamples that one implicitly assumes dependence on $$L$$. And transport of structure is in some sense a very general set-theoretic concept which can be used for any kind of mathematical structure on a set whatsoever. That, perhaps, explains why authors in field theory ignore this issue.

Here is such a counterexample.

Consider the following two field extensions of the rational numbers: $$L_1 = mathbb Q(sqrt{2})$$ $$L_2 = mathbb Q(sqrt{3})$$

Now let me define two new fields. Their underlying sets are $$L'_1 = bigl(mathbb Q(sqrt{2}) - {sqrt{2}}bigr) cup {s}$$ $$L'_2 = bigl(mathbb Q(sqrt{3}) - {sqrt{3}}bigr) cup {s}$$ Here $$s$$ is any object whatsoever which is neither an element of $$L_1$$ nor of $$L_2$$.

Let me define two bijections $$f_1 : L'_1 to L_1, quad f(x) = begin{cases} x & text{if x ne sqrt{2}} \ s &text{if x = sqrt{2}} end{cases}$$ and $$f_2 : L'_2 to L_2$$ is defined similarly using $$sqrt{3}$$ in place of $$sqrt{2}$$.

Finally, the field structures on $$L'_1$$ and on $$L'_2$$ are defined by transport of structure, for example given $$x,y in L'_1$$ we define $$x + y = f_1^{-1}(f_1(x)+f_1(y))$$. Using those field structures, it follows that $$f_i$$ is a field isomorphism between $$L'_i$$ and $$L_i$$ for each $$i=1,2$$.

However, in $$L'_1$$ the field $$mathbb Q(s)$$ is equal to $$L'_1$$ which is $$mathbb Q(sqrt{2})$$, whereas in $$L'_2$$ the field $$mathbb Q(s)$$ is equal to $$L'_2$$ which is $$mathbb Q(sqrt{3})$$. So the two versions of $$mathbb Q(s)$$ are not even isomorphic to each other, let alone equal to each other.

Correct answer by Lee Mosher on December 16, 2020