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 $$x^4 - 2$$.

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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 \(\displaystyle \mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i ) = \mathbb{Q}(\sqrt[4]{2}, i ) \) since the product of \(\displaystyle \sqrt[4]{2} \) and i must be in \(\displaystyle \mathbb{Q}(\sqrt[4]{2}, i ) \)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 \(\displaystyle [\mathbb{Q}(\sqrt[4]{2}) \ : \ \mathbb{Q}] \) = 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 \(\displaystyle [\mathbb{Q}(\sqrt[4]{2}i) \ : \ \mathbb{Q}] \) = 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 \(\displaystyle \mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i ) = \mathbb{Q}(\sqrt[4]{2}, i ) \) since the product of \(\displaystyle \sqrt[4]{2} \) and i must be in \(\displaystyle \mathbb{Q}(\sqrt[4]{2}, i ) \)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 \(\displaystyle [\mathbb{Q}(\sqrt[4]{2}) \ : \ \mathbb{Q}] \) = 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 \(\displaystyle [\mathbb{Q}(\sqrt[4]{2}i) \ : \ \mathbb{Q}] \) = 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 $$x^4 - 2$$.

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

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 \(\displaystyle [\mathbb{Q}(\sqrt[4]{2}) \ : \ \mathbb{Q}] \) = 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 \(\displaystyle [\mathbb{Q}(\sqrt[4]{2}i) \ : \ \mathbb{Q}] \) = 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

 

[blue_raver22]

Hi Peter,

Your reasoning is indeed correct so far. To find the degree of the splitting field, we can use the fact that the degree of the splitting field is equal to the degree of the minimal polynomial of the largest degree among all the elements in the splitting field.

In this case, the minimal polynomial for both \alpha = \sqrt[4]{2} and \beta = \sqrt[4]{2}i is x^4 - 2 , which has degree 4. Therefore, the degree of the splitting field is also 4.

To prove this more formally, we can use the Tower Law, which states that for fields K \subset F \subset L , we have [L:K] = [L:F][F:K] . In this case, we have \mathbb{Q} \subset \mathbb{Q}(\sqrt[4]{2}, i) \subset \mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i) . Using the Tower Law, we get [\mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i) : \mathbb{Q}] = [\mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i) : \mathbb{Q}(\sqrt[4]{2}, i)][\mathbb{Q}(\sqrt[4]{2}, i) : \mathbb{Q}] . Since both [\mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i) : \mathbb{Q}(\sqrt[4]{2}, i)] and [\mathbb{Q}(\sqrt[4]{2}, i) : \mathbb{Q}] are equal to 4, we get [\mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i) : \mathbb{Q}] = 4.

Hope this helps! Let me know if you have any further questions.