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).

One Answer

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

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