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

#### Peter

##### Well-known member
MHB Site Helper
Dummit and Foote Exercise 1 on page 545 reads as follows:

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

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

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

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

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

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

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

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

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

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

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

Last edited:

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Dummit and Foote Exercise 1 on page 545 reads as follows:

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

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

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

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

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

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

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

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

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

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

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

Hence $$\displaystyle [\mathbb{Q}(\sqrt{2}i) \ : \ \mathbb{Q}]$$ = 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 $\mathbb Q(\sqrt{2},i)=E$ (say).

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

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

##### Well-known member
MHB Math Scholar
Dummit and Foote Exercise 1 on page 545 reads as follows:

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

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

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

The two roots of [TEX] x^4 - 2 [/TEX] are [TEX] \alpha = \sqrt{2}[/TEX] and [TEX] \beta = \sqrt{2}i [/TEX]
A degree 4 polynomial has 4 roots. In this case, the complete factorization of $x^4 - 2$ in the splitting field is:

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

Thus the splitting field is [TEX] \mathbb{Q}(\sqrt{2}, \sqrt{2}i ) [/TEX]
It is more usual to call the splitting field: $\Bbb Q(\sqrt{2},i)$. 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 $\sqrt{2}$ is a root of $x^4 - 2$, we conclude that $\Bbb Q(\sqrt{2})$ is a degree 4 extension of $\Bbb Q$, which contains $\Bbb Q(\sqrt{2})$ as a subfield. But $\Bbb Q(\sqrt{2})$ is a subfield of $\Bbb R$, which contains NO roots of $x^2 + \sqrt{2}$, so we conclude that $\Bbb Q(\sqrt{2})$ is not the splitting field of $x^4 - 2$.

By the above, we also see that $x^2 + 1$ is irreducible over $\Bbb Q(\sqrt{2})$, 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 $\sqrt{2} \in \Bbb R$, a proof of which would be too long a digression, but can be shown by defining $\sqrt{2}$ as the Dedekind cut:

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

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

We note that $$\displaystyle \mathbb{Q}(\sqrt{2}, \sqrt{2}i ) = \mathbb{Q}(\sqrt{2}, i )$$ since the product of $$\displaystyle \sqrt{2}$$ and i must be in $$\displaystyle \mathbb{Q}(\sqrt{2}, i )$$
This only shows one inclusion, to show the other you must show that $i \in \Bbb Q(\sqrt{2},i\sqrt{2})$, which is easy:

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

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

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

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

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

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

Hence $$\displaystyle [\mathbb{Q}(\sqrt{2}i) \ : \ \mathbb{Q}]$$ = 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.

#### Peter

##### Well-known member
MHB Site Helper
A degree 4 polynomial has 4 roots. In this case, the complete factorization of $x^4 - 2$ in the splitting field is:

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

It is more usual to call the splitting field: $\Bbb Q(\sqrt{2},i)$. 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 $\sqrt{2}$ is a root of $x^4 - 2$, we conclude that $\Bbb Q(\sqrt{2})$ is a degree 4 extension of $\Bbb Q$, which contains $\Bbb Q(\sqrt{2})$ as a subfield. But $\Bbb Q(\sqrt{2})$ is a subfield of $\Bbb R$, which contains NO roots of $x^2 + \sqrt{2}$, so we conclude that $\Bbb Q(\sqrt{2})$ is not the splitting field of $x^4 - 2$.

By the above, we also see that $x^2 + 1$ is irreducible over $\Bbb Q(\sqrt{2})$, 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 $\sqrt{2} \in \Bbb R$, a proof of which would be too long a digression, but can be shown by defining $\sqrt{2}$ as the Dedekind cut:

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

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

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

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

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