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

[Math Amateur]

Homework Statement

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 ... ...

As indicated above I need help in order to make a meaningful or significant start on the solution to this exercise .. ...Exercise 13 of Section 13.2 reads as follows:

Homework Equations

Definitions that may be relevant to solving this exercise include the following:

A result which seems relevant is Lemma 16 plus the remarks that follow its proof ... Lemma 16 and the following remarks read as follows:

The Attempt at a Solution

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As indicated above I need help in order to make a meaningful or significant start on the solution to this exercise .. ...

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

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

 

[fresh_42]

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

Good idea. What can you say about $\dim_\mathbb{Q} F\;$? And what about $\dim_\mathbb{Q}\mathbb{Q}[\sqrt[3]{2}]\;$? Then what do you know about the fields in between, from $\mathbb{Q}$ to $F\;:\;\mathbb{Q} \subseteq \mathbb{Q}[\alpha_1]\subseteq \mathbb{Q}[\alpha_1,\alpha_2]\subseteq \ldots \subseteq F\;$?

... AND FURTHER ... anyway ... what is implied by $\alpha_i^2 \in \mathbb{Q}$ ... ... ?

It means that $\alpha_i$ is a root of $x^2-r$ for some rational $r \in \mathbb{Q}$, ergo the minimal polynomial of $\alpha_i$ because $\alpha_i\notin \mathbb{Q}$ which rules out that the degree could be less than two.

 

[Math Amateur]

Good idea. What can you say about $\dim_\mathbb{Q} F\;$? And what about $\dim_\mathbb{Q}\mathbb{Q}[\sqrt[3]{2}]\;$? Then what do you know about the fields in between, from $\mathbb{Q}$ to $F\;:\;\mathbb{Q} \subseteq \mathbb{Q}[\alpha_1]\subseteq \mathbb{Q}[\alpha_1,\alpha_2]\subseteq \ldots \subseteq F\;$?

It means that $\alpha_i$ is a root of $x^2-r$ for some rational $r \in \mathbb{Q}$, ergo the minimal polynomial of $\alpha_i$ because $\alpha_i\notin \mathbb{Q}$ which rules out that the degree could be less than two.

Thanks for the help, Andrew ... ...

Just a simple clarification though ... ...

You write, in the context of $\alpha_i^2 \in \mathbb{Q}$, the following ... ...

" ... ... It means that $\alpha_i$ is a root of $x^2-r$ for some rational $r \in \mathbb{Q}$, ergo the minimal polynomial of $\alpha_i$ because $\alpha_i\notin \mathbb{Q}$ which rules out that the degree could be less than two ... ... "... BUT ..

... how do we know that $\alpha_i \notin \mathbb{Q}$ ... indeed maybe $\alpha_i$ is in $\mathbb{Q}$ ... then $\alpha_i$ will satisfy a polynomial of degree 1 in $\mathbb{Q}$ ... namely $x - \alpha_i$ ... ...

... so doesn't $\alpha_i^2 \in \mathbb{Q}$ mean that

$[ F( \alpha_1, \alpha_2, \ ... \ ... \ \ , \alpha_i ) \ : \ F( \alpha_1, \alpha_2, \ ... \ ... \ \ , \alpha_{i - 1} ) ]$

is either 1 or 2 ... ...

can you clarify/comment ?{Note : I have to confess that although I think it may well be true that $[ F( \alpha_1, \alpha_2, \ ... \ ... \ \ , \alpha_i ) \ : \ F( \alpha_1, \alpha_2, \ ... \ ... \ \ , \alpha_{i - 1} ) ]$ is either 1 or 2 (i.e a positive integer less than or equal to 2) I do not know the precise argument for why it cannot be larger than 2 ... say 3 r 4 or 37 or something ... can you explain ...}Peter

 

[fresh_42]

Hi Michael!

All cases in which $\alpha_i \in \mathbb{Q}$ can be omitted. Say $\alpha_1 , \ldots \alpha_{k-1} \in \mathbb{Q}$ and $\alpha_k, \ldots , \alpha_n \notin \mathbb{Q}$ then $F=\mathbb{Q}[\alpha_1,\ldots , \alpha_n]=\mathbb{Q}[\alpha_k,\ldots , \alpha_n]$ and we're back in the case where all $\alpha_i \notin \mathbb{Q}$. (The case in which there isn't a proper extension at all is also trivial, since $\sqrt[3]{2} \notin \mathbb{Q}$ and nothing has to be shown.)

So key to this exercise lies in the formula for finite field extensions:
Let $\mathbb{F} \subseteq \mathbb{K} \subseteq \mathbb{L}$ be finite field extensions, then $[\mathbb{L}:\mathbb{F}] = [\mathbb{L} : \mathbb{K}]\,\cdot \,[\mathbb{K}:\mathbb{F}]\;$.

In case you cannot use this equation, you might investigate how the roots of the minimal polynomial $m_{\mathbb{Q}}(\sqrt[3]{2})(x)$ can be or cannot be expressed by rationals and roots of the form $x = \alpha_i = \sqrt{r_i}$ with $r_i \in \mathbb{Q}\;$.

 

[Math Amateur]

Hi Michael!

All cases in which $\alpha_i \in \mathbb{Q}$ can be omitted. Say $\alpha_1 , \ldots \alpha_{k-1} \in \mathbb{Q}$ and $\alpha_k, \ldots , \alpha_n \notin \mathbb{Q}$ then $F=\mathbb{Q}[\alpha_1,\ldots , \alpha_n]=\mathbb{Q}[\alpha_k,\ldots , \alpha_n]$ and we're back in the case where all $\alpha_i \notin \mathbb{Q}$. (The case in which there isn't a proper extension at all is also trivial, since $\sqrt[3]{2} \notin \mathbb{Q}$ and nothing has to be shown.)

So key to this exercise lies in the formula for finite field extensions:
Let $\mathbb{F} \subseteq \mathbb{K} \subseteq \mathbb{L}$ be finite field extensions, then $[\mathbb{L}:\mathbb{F}] = [\mathbb{L} : \mathbb{K}]\,\cdot \,[\mathbb{K}:\mathbb{F}]\;$.

In case you cannot use this equation, you might investigate how the roots of the minimal polynomial $m_{\mathbb{Q}}(\sqrt[3]{2})(x)$ can be or cannot be expressed by rationals and roots of the form $x = \alpha_i = \sqrt{r_i}$ with $r_i \in \mathbb{Q}\;$.

Thanks fresh_42 ... appreciate your help ...

Just reflecting on what you have said ... but obviously most helpful ...

Thanks again,

Peter

 

[Math Amateur]

Hi Michael!

All cases in which $\alpha_i \in \mathbb{Q}$ can be omitted. Say $\alpha_1 , \ldots \alpha_{k-1} \in \mathbb{Q}$ and $\alpha_k, \ldots , \alpha_n \notin \mathbb{Q}$ then $F=\mathbb{Q}[\alpha_1,\ldots , \alpha_n]=\mathbb{Q}[\alpha_k,\ldots , \alpha_n]$ and we're back in the case where all $\alpha_i \notin \mathbb{Q}$. (The case in which there isn't a proper extension at all is also trivial, since $\sqrt[3]{2} \notin \mathbb{Q}$ and nothing has to be shown.)

So key to this exercise lies in the formula for finite field extensions:
Let $\mathbb{F} \subseteq \mathbb{K} \subseteq \mathbb{L}$ be finite field extensions, then $[\mathbb{L}:\mathbb{F}] = [\mathbb{L} : \mathbb{K}]\,\cdot \,[\mathbb{K}:\mathbb{F}]\;$.

In case you cannot use this equation, you might investigate how the roots of the minimal polynomial $m_{\mathbb{Q}}(\sqrt[3]{2})(x)$ can be or cannot be expressed by rationals and roots of the form $x = \alpha_i = \sqrt{r_i}$ with $r_i \in \mathbb{Q}\;$.

Thanks to Andrew and fresh_42 for all the help ...

Now to finish the exercise based on the help you gave ...Essentially then, it follows that ...

$[ F \ : \ \mathbb{Q} ] = 2^k$ where $0 \le k \le n$Now ... ... suppose $\sqrt [3]{2} \in F$ ... (and try for a contradiction)Then since $\mathbb{Q} \subseteq \mathbb{Q} ( \sqrt [3]{2} ) \subseteq F$

... we have that ...

$[ F \ : \ \mathbb{Q} ] = [ F \ : \ \mathbb{Q} ( \sqrt [3]{2} ) ] [ \mathbb{Q} ( \sqrt [3]{2} ) \ : \ \mathbb{Q} ]$

So that$2^k = [ F \ : \ \mathbb{Q} ( \sqrt [3]{2} ) ] . 3$... BUT ...$3$ does not divide $2^k$ ... ... Contradiction!So $\sqrt [3]{2} \notin F$
Can someone please confirm that the above is correct ... ...

Peter

 

[fresh_42]

Yes, this is correct. At some place, an extension of degree three (or at least $3m$) would be needed.