MHB Splitting Fields - Dummit and Foote - Exercise 1, page 545

[Math Amateur]

Dummit and Foote Exercise 1 on page 545 reads as follows:

Determine the splitting field and its degree over \mathbb{Q} for x^4 - 2.

------------------------------------------------------------------------------------------------------

I have started on the solution to this exercise as follows:

The two roots of x^4 - 2 are \alpha = \sqrt[4]{2} and \beta = \sqrt[4]{2}i

Thus the splitting field is \mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i )

We note that $$ \mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i ) = \mathbb{Q}(\sqrt[4]{2}, i ) $$ since the product of $$ \sqrt[4]{2} $$ and i must be in $$ \mathbb{Q}(\sqrt[4]{2}, i ) $$The element \alpha = \sqrt[4]{2} \in \mathbb{R} is algebraic over \mathbb{Q} and so m_{\alpha}(x) = m_{\sqrt[4]{2}}(x) = x^4 - 2 is the minimal polynomial for \alpha = \sqrt[4]{2} over \mathbb{Q}.

Thus the degree of \alpha = \sqrt[4]{2} over \mathbb{Q} is the degree of m_{\alpha}(x) = 4

Hence $$ [\mathbb{Q}(\sqrt[4]{2}) \ : \ \mathbb{Q}] $$ = 4
The element \beta = \sqrt[4]{2}i \in \mathbb{C} is also algebraic over \mathbb{Q} and so m_{\beta}(x) = m_{\sqrt[4]{2}i}(x) = x^4 - 2 is the minimal polynomial for \beta = \sqrt[4]{2}i over \mathbb{Q}.

Thus the degree of \beta = \sqrt[4]{2}i over \mathbb{Q} is the degree of m_{\beta}(x) = 4

Hence $$ [\mathbb{Q}(\sqrt[4]{2}i) \ : \ \mathbb{Q}] $$ = 4
But where to from here - need to find the degree of the splitting field.

Can someone please confirm that my reasoning above is valid and show me the way forward from here?

Peter[Note that this has also been posted on MHF]

 

[caffeinemachine]

Dummit and Foote Exercise 1 on page 545 reads as follows:

Determine the splitting field and its degree over \mathbb{Q} for x^4 - 2.

------------------------------------------------------------------------------------------------------

I have started on the solution to this exercise as follows:

The two roots of x^4 - 2 are \alpha = \sqrt[4]{2} and \beta = \sqrt[4]{2}i

Thus the splitting field is \mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i )

We note that $$ \mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i ) = \mathbb{Q}(\sqrt[4]{2}, i ) $$ since the product of $$ \sqrt[4]{2} $$ and i must be in $$ \mathbb{Q}(\sqrt[4]{2}, i ) $$The element \alpha = \sqrt[4]{2} \in \mathbb{R} is algebraic over \mathbb{Q} and so m_{\alpha}(x) = m_{\sqrt[4]{2}}(x) = x^4 - 2 is the minimal polynomial for \alpha = \sqrt[4]{2} over \mathbb{Q}.

Thus the degree of \alpha = \sqrt[4]{2} over \mathbb{Q} is the degree of m_{\alpha}(x) = 4

Hence $$ [\mathbb{Q}(\sqrt[4]{2}) \ : \ \mathbb{Q}] $$ = 4
The element \beta = \sqrt[4]{2}i \in \mathbb{C} is also algebraic over \mathbb{Q} and so m_{\beta}(x) = m_{\sqrt[4]{2}i}(x) = x^4 - 2 is the minimal polynomial for \beta = \sqrt[4]{2}i over \mathbb{Q}.

Thus the degree of \beta = \sqrt[4]{2}i over \mathbb{Q} is the degree of m_{\beta}(x) = 4

Hence $$ [\mathbb{Q}(\sqrt[4]{2}i) \ : \ \mathbb{Q}] $$ = 4
But where to from here - need to find the degree of the splitting field.

Can someone please confirm that my reasoning above is valid and show me the way forward from here?

Peter[Note that this has also been posted on MHF]

You have correctly figured out that the splitting field for $x^4-2$ over $\mathbb Q$ is $\mathbb Q(\sqrt[4]{2},i)=E$ (say).

write $\mathbb Q(\sqrt[4]{2})=F$

Also, you know that $[F:\mathbb Q]=4$. Now if you know what is $[E:F]$ you can apply the tower law to get the answer. Now $E=F(i)$. So all you need to find is the minimal polynomial of $i$ over $F$. This polynomial can easily be seen to be $x^2+1$ and hence $[E:F]=2$.

 

[Deveno]

Dummit and Foote Exercise 1 on page 545 reads as follows:

Determine the splitting field and its degree over \mathbb{Q} for x^4 - 2.

------------------------------------------------------------------------------------------------------

I have started on the solution to this exercise as follows:

The two roots of x^4 - 2 are \alpha = \sqrt[4]{2} and \beta = \sqrt[4]{2}i

A degree 4 polynomial has 4 roots. In this case, the complete factorization of $x^4 - 2$ in the splitting field is:

$x^4 - 2 = (x + \sqrt[4]{2})(x - \sqrt[4]{2})(x + i\sqrt[4]{2})(x - i\sqrt[4]{2})$

Thus the splitting field is \mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i )

It is more usual to call the splitting field: $\Bbb Q(\sqrt[4]{2},i)$. One can arrive at this splitting field like so:

Note that in $\Bbb Q(\sqrt{2}), x^4 - 2$ factors as:

$x^4 - 2 = (x^2 - \sqrt{2})(x^2 + \sqrt{2})$. This gives us a degree 2 extension of the rationals. Now $x^2 - 2$ is irreducible over $\Bbb Q$ (by Eisenstein), and is of degree 4, so since $\sqrt[4]{2}$ is a root of $x^4 - 2$, we conclude that $\Bbb Q(\sqrt[4]{2})$ is a degree 4 extension of $\Bbb Q$, which contains $\Bbb Q(\sqrt{2})$ as a subfield. But $\Bbb Q(\sqrt[4]{2})$ is a subfield of $\Bbb R$, which contains NO roots of $x^2 + \sqrt{2}$, so we conclude that $\Bbb Q(\sqrt[4]{2})$ is not the splitting field of $x^4 - 2$.

By the above, we also see that $x^2 + 1$ is irreducible over $\Bbb Q(\sqrt[4]{2})$, since if $x^2 + 1$ has no REAL roots, it certainly has no root in any subfield of $\Bbb R$ (we are assuming here that it can be taken as given $\sqrt[4]{2} \in \Bbb R$, a proof of which would be too long a digression, but can be shown by defining $\sqrt[4]{2}$ as the Dedekind cut:

$\{q \in \Bbb Q:q < 0, \text{or } q^4 - 2 < 0\}$).

Thus $[\Bbb Q(\sqrt[4]{2},i):\Bbb Q] = [\Bbb Q(\sqrt[4]{2},i):\Bbb Q(\sqrt[4]{2})]\ast[\Bbb Q(\sqrt[4]{2}):\Bbb Q] = 2 \ast 4 = 8$

We note that $$ \mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i ) = \mathbb{Q}(\sqrt[4]{2}, i ) $$ since the product of $$ \sqrt[4]{2} $$ and i must be in $$ \mathbb{Q}(\sqrt[4]{2}, i ) $$

This only shows one inclusion, to show the other you must show that $i \in \Bbb Q(\sqrt[4]{2},i\sqrt[4]{2})$, which is easy:

$i = \frac{1}{2}(\sqrt[4]{2})^3(i\sqrt[4]{2})$

The element \alpha = \sqrt[4]{2} \in \mathbb{R} is algebraic over \mathbb{Q} and so m_{\alpha}(x) = m_{\sqrt[4]{2}}(x) = x^4 - 2 is the minimal polynomial for \alpha = \sqrt[4]{2} over \mathbb{Q}.

Thus the degree of \alpha = \sqrt[4]{2} over \mathbb{Q} is the degree of m_{\alpha}(x) = 4

Hence $$ [\mathbb{Q}(\sqrt[4]{2}) \ : \ \mathbb{Q}] $$ = 4
The element \beta = \sqrt[4]{2}i \in \mathbb{C} is also algebraic over \mathbb{Q} and so m_{\beta}(x) = m_{\sqrt[4]{2}i}(x) = x^4 - 2 is the minimal polynomial for \beta = \sqrt[4]{2}i over \mathbb{Q}.

Thus the degree of \beta = \sqrt[4]{2}i over \mathbb{Q} is the degree of m_{\beta}(x) = 4

Hence $$ [\mathbb{Q}(\sqrt[4]{2}i) \ : \ \mathbb{Q}] $$ = 4
But where to from here - need to find the degree of the splitting field.

Can someone please confirm that my reasoning above is valid and show me the way forward from here?

Peter[Note that this has also been posted on MHF]

I no longer visit MHF, the malware and the bot infestation is too sad to endure.

 

[Math Amateur]

A degree 4 polynomial has 4 roots. In this case, the complete factorization of $x^4 - 2$ in the splitting field is:

$x^4 - 2 = (x + \sqrt[4]{2})(x - \sqrt[4]{2})(x + i\sqrt[4]{2})(x - i\sqrt[4]{2})$
It is more usual to call the splitting field: $\Bbb Q(\sqrt[4]{2},i)$. One can arrive at this splitting field like so:

Note that in $\Bbb Q(\sqrt{2}), x^4 - 2$ factors as:

$x^4 - 2 = (x^2 - \sqrt{2})(x^2 + \sqrt{2})$. This gives us a degree 2 extension of the rationals. Now $x^2 - 2$ is irreducible over $\Bbb Q$ (by Eisenstein), and is of degree 4, so since $\sqrt[4]{2}$ is a root of $x^4 - 2$, we conclude that $\Bbb Q(\sqrt[4]{2})$ is a degree 4 extension of $\Bbb Q$, which contains $\Bbb Q(\sqrt{2})$ as a subfield. But $\Bbb Q(\sqrt[4]{2})$ is a subfield of $\Bbb R$, which contains NO roots of $x^2 + \sqrt{2}$, so we conclude that $\Bbb Q(\sqrt[4]{2})$ is not the splitting field of $x^4 - 2$.

By the above, we also see that $x^2 + 1$ is irreducible over $\Bbb Q(\sqrt[4]{2})$, since if $x^2 + 1$ has no REAL roots, it certainly has no root in any subfield of $\Bbb R$ (we are assuming here that it can be taken as given $\sqrt[4]{2} \in \Bbb R$, a proof of which would be too long a digression, but can be shown by defining $\sqrt[4]{2}$ as the Dedekind cut:

$\{q \in \Bbb Q:q < 0, \text{or } q^4 - 2 < 0\}$).

Thus $[\Bbb Q(\sqrt[4]{2},i):\Bbb Q] = [\Bbb Q(\sqrt[4]{2},i):\Bbb Q(\sqrt[4]{2})]\ast[\Bbb Q(\sqrt[4]{2}):\Bbb Q] = 2 \ast 4 = 8$
This only shows one inclusion, to show the other you must show that $i \in \Bbb Q(\sqrt[4]{2},i\sqrt[4]{2})$, which is easy:

$i = \frac{1}{2}(\sqrt[4]{2})^3(i\sqrt[4]{2})$

I no longer visit MHF, the malware and the bot infestation is too sad to endure.

Thanks Deveno, you give details of a number of points that are extremely important to understanding the exercise and the theory in general

appreciate such detailed help.

Peter