Subfield

**Carl** · March 15th 2009, 06:03 AM

Hi everybody. I have this exercise where I am to describe all the subfields $K_1$ and $K_2$ of $L_1$ and $L_2$ with $[K_1:\mathbb{Q}]=2$ and $[K_2:\mathbb{Q}]=2$ , where $L_1$ and $L_2$ are the splitting fields for the polynomials $g_1(x)=(x^3-2)$ and $g_2=(x^4+1)$ over $\mathbb{Q}$ .
(Remark: $L_2$ is the cyclotomic field $\mathbb{Q}_8$ )

Now, I would suppose that you have to use the transitivity theorem to say that; $[K_1:\mathbb{Q}]=[K_1:\mathbb{Q}(\epsilon_3)][\mathbb{Q}(\epsilon_3):\mathbb{Q}]$

$[K_2:\mathbb{Q}]=[K_2:\mathbb{Q}(\epsilon_8)][\mathbb{Q}(\epsilon_8):\mathbb{Q}]$

and then argue that $[K_1:\mathbb{Q}(\epsilon_3)]=[K_2:\mathbb{Q}(\epsilon_8)]=1$ or did I get that wrong? I'm just not sure, then, how to argue that I then have found ALL the subfields with this dimension?

**ThePerfectHacker** · March 15th 2009, 10:35 AM

Originally Posted by Carl

Hi everybody. I have this exercise where I am to describe all the subfields $K_1$ and $K_2$ of $L_1$ and $L_2$ with $[K_1:\mathbb{Q}]=2$ and $[K_2:\mathbb{Q}]=2$ , where $L_1$ and $L_2$ are the splitting fields for the polynomials $g_1(x)=(x^3-2)$ and $g_2=(x^4+1)$ over $\mathbb{Q}$ .
(Remark: $L_2$ is the cyclotomic field $\mathbb{Q}_8$ )

Now, I would suppose that you have to use the transitivity theorem to say that; $[K_1:\mathbb{Q}]=[K_1:\mathbb{Q}(\epsilon_3)][\mathbb{Q}(\epsilon_3):\mathbb{Q}]$

$[K_2:\mathbb{Q}]=[K_2:\mathbb{Q}(\epsilon_8)][\mathbb{Q}(\epsilon_8):\mathbb{Q}]$

and then argue that $[K_1:\mathbb{Q}(\epsilon_3)]=[K_2:\mathbb{Q}(\epsilon_8)]=1$ or did I get that wrong? I'm just not sure, then, how to argue that I then have found ALL the subfields with this dimension?

I start you out with $K_1$ and you think about $K_2$ . Since $L_1$ is the splitting of $x^3 - 2$ over $\mathbb{Q}$ it means $L_1 = \mathbb{Q}(\zeta,\sqrt[3]{2})$ . We will prove that $\text{Gal}(L_1/\mathbb{Q}) \simeq S_3$ . If $\sigma$ is an automorphism of $L_1$ then it is completely determined by $\sigma (\zeta)$ and by $\sigma (\sqrt[3]{2})$ . Now $\zeta$ is a zero of the irreducible polynomial $x^2+x+1$ , this polynomial has zeros $\zeta,\zeta^2$ , thus $\sigma (\zeta) = \zeta \text{ or }\zeta^2$ . By similar reasoning $\sigma (\sqrt[3]{2}) = \sqrt[3]{2},\zeta\sqrt[3]{2},\text{ or }\zeta^2\sqrt[3]{2}$ since $\sqrt[3]{2}$ is a zero of the irreducible polynomial $x^3-2$ . By isomorphism extension theorem there is an automorphism $\sigma$ so that $\sigma (\zeta) = \zeta^2$ and $\sqrt(\sqrt[3]{2}) = \sqrt[3]{2}$ . Also, there is an automorphism $\tau$ so that $\tau (\zeta) = \zeta$ and $\tau (\sqrt[3]{2}) = \zeta\sqrt[3]{2}$ . Notice that $\sigma (\zeta) = \zeta^2 = \overline{\zeta}$ so $\sigma$ is in fact complex-conjugation. We see that the order of $\sigma$ in the Galois group is $2$ and order of $\tau$ in the Galois group is $3$ , furthermore $\text{Gal}(L_1/\mathbb{Q}) = \left< \sigma,\tau\right>\simeq S_3$ . You want, as the problem says, to find all $K_1$ so that $[K_1:\mathbb{Q}] = 2$ . Remember by the fundamental theorem of Galois theory $[K_1:\mathbb{Q}] = [S_3:H]$ where $H$ is the corresponding subgroup i.e. $H = \text{Gal}(L_1/K_1)$ . Thus, $K_1$ corresponds to the subgroup of $S_3$ which has index $2$ i.e. that is $H = A_3$ and there is only one such subgroup. Therefore, there is only one subfield of $L_1$ which has $[K_1:\mathbb{Q}]=2$ . It is clear to see that $[\mathbb{Q}(\zeta):\mathbb{Q}]=2$ and so $K_1 = \mathbb{Q}(\zeta)$ .

**Carl** · March 15th 2009, 11:28 AM

Ahh, ok, thanks a lot. I think I got the idea.

I will try to figure out $K_2$ then.

**ThePerfectHacker** · March 15th 2009, 12:24 PM

Originally Posted by Carl

Ahh, ok, thanks a lot. I think I got the idea.

I will try to figure out $K_2$ then.

The hint that I can give you is that $\text{Gal}(L_2/\mathbb{Q}) = \mathbb{Z}_2\times \mathbb{Z}_2$ and so there are $3$ subgroups of index two, these will correspond to $[K_2:\mathbb{Q}]=2$ . Thus, you will have three such subfields. For this problem this post may help you.

**Carl** · March 16th 2009, 11:16 AM

Ok, I think I got it now:
So, I say that $Gal(L_2/ \mathbb{Q})=V$
And since V has 3 subgroups of order 2, which are $\langle \sigma\rangle, \langle \tau \rangle$ and $\langle \sigma\tau\rangle$ , then there are 3 $\alpha$ such that $[\mathbb{Q}(\alpha):\mathbb{Q}]=2$ and those are $\sqrt{2}i, i$ and $\sqrt{2}$ since the splitting field for $x^4+1$ is $\mathbb{Q}(\sqrt{2},i)$ .
And then I just have to check that those 3 are independent from each other and thereby I have found the 3 $K_2$ 's
Did I get that right?

**ThePerfectHacker** · March 16th 2009, 06:08 PM

Originally Posted by Carl

Ok, I think I got it now:
So, I say that $Gal(L_2/ \mathbb{Q})=V$
And since V has 3 subgroups of order 2, which are $\langle \sigma\rangle, \langle \tau \rangle$ and $\langle \sigma\tau\rangle$ , then there are 3 $\alpha$ such that $[\mathbb{Q}(\alpha):\mathbb{Q}]=2$ and those are $\sqrt{2}i, i$ and $\sqrt{2}$ since the splitting field for $x^4+1$ is $\mathbb{Q}(\sqrt{2},i)$ .
And then I just have to check that those 3 are independent from each other and thereby I have found the 3 $K_2$ 's
Did I get that right?

Yes,

, and it is easy to show that $\mathbb{Q}(\sqrt{2}),\mathbb{Q}(i),\mathbb{Q}(i\sq rt{2})$ are distinct. It is easy to see they are distinct because $\mathbb{Q}(i) = \{ a+bi|a,b\in \mathbb{Q}\}$ , $\mathbb{Q}(\sqrt{2}) = \{a+b\sqrt{2}|a,b\in \mathbb{Q}\}$ , and $\mathbb{Q}(i\sqrt{2}) = \{ a + bi\sqrt{2}|a,b\in \mathbb{Q}\}$ . Now it is clear that these are distinct because $\sqrt{2}$ is irrational.

**Carl** · March 16th 2009, 11:53 PM

Allright, thanks a lot, you have been most helpful

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