Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a question requires to compute the Galois group of $X^4+1$ over $\mathbb{Q}$, $\mathbb{Q}(i)$, $\mathbb{F}_3$ and $\mathbb{F}_5$.

Here is a brief of what I can think of.

For the first two, the splitting field is $\mathbb{Q}(\sqrt{2},i)$. Since $X^2+1$ and $X^2-2$ are irreducible, the Galois group should be $C_2 \times C_2$ and $C_2$.

I am not sure what happens when the field is finite.

=================================

I thought about the first case again last night but I messed up myself a little bit. Take $\zeta_8 = \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ to be the eighth partition of unity. The splitting field of $X^4+1$ over $\mathbb{Q}$ should be $\mathbb{Q}(\zeta_8)$, since $\zeta_8$ generates other roots which are $\zeta_8^3,\zeta_8^5, \zeta_8^7$. It follows $Gal(\mathbb{Q}(\zeta_8)/\mathbb{Q})=Aut(C_8)=C_4$.

On the other hand, $\zeta_8$ and $\zeta_8^3$ generates $\sqrt{2}$ and $i$, which leads to the answer I tried to gave last time. In this case, the Galois group seems to be $C_2 \times C_2$.

For sure the degree of extension is $4$, but I cannot figure out which approach is correct.

share|improve this question
2  
When the field is finite, the Galois group is generated by the Frobenius map (exercise), and the only thing to calculate is its order. –  Qiaochu Yuan Aug 21 '12 at 17:36
    
In the case of finite fields the factorization (and hence also the splitting field) of this polynomial has been studied in this question. The order of the Galois group then follows as in Qiaochu's comment. –  Jyrki Lahtonen Aug 21 '12 at 17:51
1  
I don't see how "It follows ... $C_4$." Each automorphism $\sigma$ taking $\zeta$ to $\zeta^j$ has $\sigma^2=1$. –  Gerry Myerson Aug 22 '12 at 6:18
    
@GerryMyerson, thanks for the reminder, I got the correct computation. –  Honghao Aug 22 '12 at 6:28
add comment

2 Answers 2

up vote 3 down vote accepted

Over $F_3$ you have $\sqrt{2} = \sqrt{-1} = i$, so the extension is quadratic. Over $F_5$, you have $i = \sqrt{-1} = \sqrt{4} = \pm 2$, so again the extension is quadratic. Since extensions of finite fields are always cyclic, this was to be expected.

share|improve this answer
add comment

Another way to see what's going on:

The quadratic extension of ${\bf F}_3$ has 9 elements, so its multiplicative group has 8 elements, but the multiplicative group of a finite field is always cyclic, so there's an element $a$ with order 8; then $a$ is a zero of $x^8-1=(x^4-1)(x^4+1)$ but not of $x^4-1$, hence of $x^4+1$.

Similarly for the quadratic extension of ${\bf F}_5$ and its multiplicative group, cyclic of order 24, which number is a multiple of 8.

share|improve this answer
    
I understood the part in this approach which gives a root of $x^4+1$ in the quadratic extension of $F_3$. How can we see the polynomial split in the extension? –  Honghao Aug 22 '12 at 6:34
1  
I should have noted that there are 4 elements of order 8 in the cyclic group of order 8, so 4 roots of $x^4+1$, so it splits. –  Gerry Myerson Aug 22 '12 at 6:38
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.