Field Extensions - Dummit and Foote - Exercise 13, Section 13.2 .... ....

[Math Amateur]

I am reading Dummit and Foote, Chapter 13 - Field Theory.

I am currently studying Section 13.2 : Algebraic Extensions

I need some help with Exercise 13 of Section 13.2 ... ... indeed, I have not been able to make a meaningful start on the problem

Exercise 13 of Section 13.2 reads as follows:
View attachment 6611
Definitions that may be relevant to solving this exercise include the following:View attachment 6612A result which seems relevant is Lemma 16 plus the remarks that follow its proof ... Lemma 16 and the following remarks read as follows:
View attachment 6613
As indicated above I need help in order to make a meaningful or significant start on the solution to this exercise .. ...

One thought, though ... there must be some way to use \(\displaystyle \alpha_i^2 \in \mathbb{Q}\) ... perhaps in establishing the dimension of \(\displaystyle F( \alpha_1 , \alpha_2, \ ... \ ... \ \ \alpha_{k + 1} )\) over \(\displaystyle F( \alpha_1 , \alpha_2, \ ... \ ... \ \ \alpha_{k } )\) ... and hence getting some knowledge of \(\displaystyle F( \alpha_1 , \alpha_2, \ ... \ ... \ \ \alpha_n )\) over \(\displaystyle \mathbb{Q}\) ... but even if we do gain such knowledge, how do we use it to show \(\displaystyle \sqrt [3]{2} \notin F\) ... ...

... AND FURTHER ... anyway ... what is implied by \(\displaystyle \alpha_i^2 \in \mathbb{Q}\) ... ... ?
Help will be much appreciated ...Peter

 

[Euge]

Hi Peter,

Let $F_0 = \Bbb Q$, $F_1 = \Bbb Q(\alpha_1)$, and $F_i = F_{i-1}(\alpha_i)$ for $1 \le i \le n$. Show that $[F_i:F_{i-1}]$ is either $1$ or $2$. Deduce that $[F:\Bbb Q]$ is a power of $2$. By degree considerations show that $\sqrt[3]{2}\notin F$.

 

[Math Amateur]

Hi Peter,

Let $F_0 = \Bbb Q$, $F_1 = \Bbb Q(\alpha_1)$, and $F_i = F_{i-1}(\alpha_i)$ for $1 \le i \le n$. Show that $[F_i:F_{i-1}]$ is either $1$ or $2$. Deduce that $[F:\Bbb Q]$ is a power of $2$. By degree considerations show that $\sqrt[3]{2}\notin F$.

Hi Euge ... thanks for the help ...

You advise that I show that \(\displaystyle [ F_i \ : \ F_{ i - 1 }]\) is either 1 or 2 ...

Now we are given that \(\displaystyle \alpha_i^2 \in \mathbb{Q}\)

Now ... \(\displaystyle \alpha_i^2 \in \mathbb{Q} \Longrightarrow \alpha_i\) is a root of a polynomial \(\displaystyle x^2 - r\) where \(\displaystyle r \in \mathbb{Q}\) Now, \(\displaystyle \alpha_i \) being the root of \(\displaystyle x^2 - r\) seems to imply that \(\displaystyle x^2 - r\) is the minimum polynomial for \(\displaystyle \alpha_i\) and for the extension \(\displaystyle F_i / F_{i - 1}\) ... ... BUT ... ... that is only true if \(\displaystyle x^2 - r\) is irreducible ... and it will not be if \(\displaystyle \alpha_i \in \mathbb{Q}\) ... which it may be ... then \(\displaystyle x^2 - r = (x - \alpha_i ) (x + \alpha_i ) \) ... that is \(\displaystyle x^2 - r\) is reducible ... and the minimal polynomial is \(\displaystyle x - \alpha_i\) and is of degree 1 ... ... so that \(\displaystyle [ F_i \ : \ F_{ i - 1 }] = 1 \) ... ... Is that set of remarks basically correct ...?How do argue rigorously that \(\displaystyle [ F_i \ : \ F_{ i - 1 }]\) may be 2 ... I think it is 2 if \(\displaystyle \alpha_i \notin \mathbb{Q}\) ... is that right ... but what is the rigorous and precise argument that \(\displaystyle [ F_i \ : \ F_{ i - 1 }]\) may be equal to 2 ... ...Can you help ... I am a bit unsure on this matter ...

Peter
NOTE 1 ... Despite being unsure ... I may have (almost) made an argument for \(\displaystyle [ F_i \ : \ F_{ i - 1 }] \) being either 1 or 2 ...

NOTE 2 ... I acknowledge some help from the Physics Forums ...

 

[Euge]

You have the right idea, but you're overthinking this a bit. Since $\alpha_i$ is a root of a quadratic polynomial in $F_{i-1}[x]$, its minimal polynomial over $F_{i-1}$ has degree no greater than $2$, and thus $[F_{i-1}(\alpha_i) : F_{i-1}] \le 2$, i.e., $[F_i:F_{i-1}]$ is either $1$ or $2$. That's all we need.

 

[Math Amateur]

You have the right idea, but you're overthinking this a bit. Since $\alpha_i$ is a root of a quadratic polynomial in $F_{i-1}[x]$, its minimal polynomial over $F_{i-1}$ has degree no greater than $2$, and thus $[F_{i-1}(\alpha_i) : F_{i-1}] \le 2$, i.e., $[F_i:F_{i-1}]$ is either $1$ or $2$. That's all we need.

Thanks Euge ... ... think I get the idea ...

Essentially then, it follows that ...

\(\displaystyle [ F \ : \ \mathbb{Q}] = 2^k\) where \(\displaystyle 0 \le k \le n\)
Now to complete the exercise ... ... suppose \(\displaystyle \sqrt [3]{2} \in F\) ... ... ( ... and try for a contradiction ... )Then since \(\displaystyle \mathbb{Q} \subseteq \mathbb{Q} ( \sqrt [3]{2} ) \subseteq F \)... we have that ...\(\displaystyle [ F \ : \ \mathbb{Q} ] = [ F \ : \ \mathbb{Q} ( \sqrt [3]{2} ) ] [ \mathbb{Q} ( \sqrt [3]{2} ) \ : \ \mathbb{Q}] \)So that \(\displaystyle 2^k = [ F \ : \ \mathbb{Q} ( \sqrt [3]{2} ) ] 3\)... BUT ... \(\displaystyle 3\) does not divide \(\displaystyle 2^k\) ... ... Contradiction!So \(\displaystyle \sqrt [3]{2} \notin F \)
Can someone please confirm that the above is correct ... ...

Peter

 

[Euge]

Now you've got it! Great job!