# Field decreases. sequel

This is an adhere to - approximately the inquiry on field reductions.

(EDIT : Originally this inquiry made use of the symbols $\{\mathbb{R}(\setminus a)\}$, today makes use of $\mathbb{R}(\setminus a)$ rather.)

There isn't an one-of-a-kind biggest subfield of $\mathbb{R}$ not having $a$, so allow $\mathbb{R}(\setminus a)$ represent the set of topmost subfields of $\mathbb{R}$ not having $a$,

EDIT : The inquiry "What is the cardinality of $\mathbb{R}(\setminus a)$? " has actually currently been asked on Mathoverflow.

From the previous inquiry if $F\in \mathbb{R}(\setminus a)$ after that $F$ is vast, yet what is the dimension of $\mathbb{R}\setminus F$ the enhance of $F$ in $\mathbb{R}$

Each $F \in \mathbb{R}(\setminus a)$ should be gotten to by getting rid of components of $\mathbb{R}$, yet if you start with $\mathbb{R}$ and also remove $a$ and afterwards remove even more components till you get an area after that undoubtedly you get to a solitary area, so where do the numerous topmost areas originate from? - It must be due to the fact that you get various areas relying on the order in which components are gotten rid of from $\mathbb{R}$, in which instance what is one of the most all-natural order to remove components to get a definition of a solitary approved subfield $\mathbb{R}(\setminus a)$ of $\mathbb{R}$ that does not have $a$, that we could call 'the' area decrease of $\mathbb{R}$ by $a$?

EDIT : Given Arturo Magidin is remarks and also solution, if we desert the suggestion of a notable component in the abstract/in basic, after that given that the symbols $\{\mathbb{R}(\setminus a)\}$ resembles a set having a solitary component called $\mathbb{R}(\setminus a)$, it makes even more feeling to go down the brackets and also simply write $\mathbb{R}(\setminus a)$ on the understanding that this is a set of area decreases i.e. a set of subfields not a solitary subfield.

From Pete L. Clark is response to the previous question, if $F(a)=\mathbb{R}$ after that $F=\mathbb{R}$, so allow is specify the idea of an extremely - expansion of $F$ by $a$, $F(a..)=R$ being the series of area expansions to reach $\mathbb{R}$ from $F\in\mathbb{R}(\setminus a)$. Extra usually, allow [email protected](\setminus a)$be a component of$K(\setminus a)$after that [email protected](\setminus a)(a..)=K$.

We can additionally lower $K$ by a set $A$, with $K(\setminus A)$ being the set of topmost subfields that do not have any kind of components from $A$, and also once more write the extremely - expansion [email protected](\setminus A)(A..)=K$. Allow$\mathbb{A}'$be the set of non - sensible algebraic numbers. What is the cardinality of$\mathbb{R}(\setminus \mathbb{A}')$? 0 2019-12-02 02:52:48 Source Share Answers: 1 I'll position something as a solution for not having inquiries without solutions. I do not recognize what the cardinality of$\{\mathbb{R}(\setminus a)\}$is, yet I would certainly be stunned if it is not vast. I'll see if I can think of it some in the next number of days. When it comes to your last paragraph, the various areas come not from the order in which you would certainly remove components, yet instead on the selections you make regarding what to remove at each (of definitely several) actions. As I kept in mind in the remarks, consider the comparable procedure whereby we seek subgroups of an offered team that are topmost amongst those that do not have an offered component. If we start with the Klein$4$- team,$C_2\times C_2$, and also remove the component$a=(1,1)$, after that we get 2 various selections of what various other component to remove so as to get a team : we can remove either$(1,0)$or$(0,1)$; each of the selections will certainly bring about a various subgroup. In a similar way, take into consideration as an example the instance where$a=\sqrt{2}$; after that as kept in mind in your previous inquiry, a topmost subfield of$\mathbb{R}$that does not have$a$might have either$\sqrt{3}$or$\sqrt{6}$, yet not both, and also you get one area that is topmost and also does have$\sqrt{3}$(straightforward application of Zorn is Lemma), and also one that is maxiimal and also does have$\sqrt{6}$(it's the same) ; so with your procedure you can either remove$\sqrt{3}$at some time, or remove$\sqrt{6}$at some time, and also these selections bring about various topmost subfields. The actual word "choices" recommends just how the Axiom of Choice is contributing in this "top - down" building and construction of components of$\{\mathbb{R}(\setminus a)\}$(in the "bottoms - up" strategy, it contributes using Zorn is Lemma). In order to have the ability to get something one can call the "field reduction", we would certainly require some procedure or algorithm that establishes which selection we make at any kind of phase. One means would certainly be by granting$\mathbb{R}$with a well - getting, and afterwards at each action we can take into consideration the set of all components we can remove ; if it is vacant, we are done ; if it is not vacant, we remove the least component of the set. Yet, certainly, well - getting$\mathbb{R}$is a little a trouble per se : there is no continually concurred - upon well - getting of$\mathbb{R}$(and also in many cases, not also a specific one). We could attempt to select a certain version in which there are specific well - purchasings (as stated by Adres Caicedo in the remarks, such versions exist), and also a certain well - getting therein, and also after that specify "the" area decrease, yet that appears to be as much of a trouble per se : which version, and also which order, and also why? Probably there are various other devices wherein one can attempt to develop a notable component in$\{\mathbb{R}(\setminus a)\}$(or probably, a notable component for details circumstances of$a$), yet it appears to me to be a helpless job in the abstract/in general. Included and also modified : Recall that any kind of area expansion$K/k$can be separated as$K/F/k$, where$K/F$is algebraic and also$F/k$is a totally transcendental expansion. Specifically, if$T$is a transcendence basis for$K$over$k$, after that$k(T)$will certainly be a topmost subfield of$K$which contains$k$and also does not have any kind of components of$K$that are algebraic over$k$yet not in$k$. So for any kind of area expansion$K/k$, if we allow \begin{equation*} \mathbb{A}_{K/k} =\bigl\{ a \in K\ \bigm|\ \mbox{$a$is algebraic over$k$,$a\notin k$}\bigr\}, \end{equation*} after that $$K(\setminus \mathbb{A}_{K/k}) = \{ k(T)\ |\ \mbox{T is a transcendence basis for K over k}\}.$$ If$K$is algebraic over$k$, after that you simply get$k$, certainly ; if$K$is totally transcendental over$k$, after that$\mathbb{A}_{K/k}$is vacant, so you get$K$. I had actually initially created that the set would certainly constantly be singleton, yet this is not the instance. If$T$is any kind of transcendence basis for$\mathbb{R}$over$\mathbb{Q}$, you can change a component of$T$by increasing it by a non - sensible algebraic number and also you get a somewhat various subfield. As an example, solution$t\in T$and also change$t$with$\sqrt{2}t$to get a new set$T'$. This is still a transcendence basis for$\mathbb{R}$over$\mathbb{Q}$; yet whereas$\sqrt{2}t\in \mathbb{Q}(T')$, it does not hinge on$\mathbb{Q}(T)$. If it did, after that we would quickly get$\sqrt{2}\in\mathbb{Q}(T)$, negating the reality that$\mathbb{Q}(T)$is totally transcendental over$\mathbb{Q}$. You can do the very same point taking approximate parts of$T$and also increasing them via by a nonrational algebraic number. Given that$|T|=\mathfrak{c}$, this offers at the very least$2^{\mathfrak{c}}$distinctive components in$\mathbb{R}(\setminus\mathbb{A}')$, and also because that is the variety of parts of$\mathbb{R}$, it adheres to that the cardinality of$\mathbb{R}(\setminus\mathbb{A}')$is$2^\mathfrak{c}\$.

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2019-12-03 02:12:28
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