# Determine the degree of the splitting field for $f(x)=x^{15}-1$.

Let $f(x)= x^{15} - 1$. Let $L$ be the splitting field of $f(x)$ over the field $K$. Determine the extension degree $[L:K]$ in each case.

a) $K= \Bbb{R}$

b) $K= \Bbb{Q}$

c) $K = \Bbb{F}_2$

d) $K = \Bbb{F}_{31}$

a) If we let $\Psi_d(x)$ denote the dth cyclotomic polynomial, then we have,

$$x^{15}-1= \Psi_{d|15}(x) = \Psi(x) \Psi_3(x) \Psi_5(x) = (x-1)(x^2+x+1)(x^4+x^3+x^2+x+1)$$

So the roots of $f(x)$ are $1, \zeta_3, \zeta_3^2, \zeta_5, \zeta_5^2, \zeta_5^3, \zeta_5^4$ where $\zeta_k$ denotes the kth primitive root of unity.

So the splitting field is $L=\Bbb{R}(\zeta_3, \zeta_5)$. We know that $$[L:\Bbb{R}] = [L: \Bbb{R}(\zeta_5)][\Bbb{R}(\zeta_5): \Bbb{R}]$$

Since $x^4+x^3+x^2+x+1 \in \Bbb{R}[x]$, we just need to check if we can factorize it into polynomials of lower degree. Of course, we cannot reduce it into polynomials of degrees 1 and 3, since we would then have a complex number in the reals. So we just need to check if we can reduce it into quadratic polynomials. Since $x^4+x^3+x^2+1=(x-\zeta_5)(x-\zeta_5^2)(x-\zeta_5^3)(x-\zeta_5^4)$. By multiplying $(x-\zeta_5^t)$ and $(x-\zeta_5^w)$ for $w,t=0,1,2,3,4$ and $w \not= t$, we find that we can reduce it into quadratics $(x-\zeta_5)(x-\zeta_5^4) = x^2 - (\zeta_5 + \zeta_5^4)x + 1$ and $(x-\zeta_5^2)(x - \zeta_5^3) = x^2 - (\zeta_5^3 + \zeta_5^2) + 1$. We know that $\zeta_5 + \zeta_5^4$ and $\zeta_5^3 + \zeta_5^2$ are real but looking at the unit circle, because it is symmetric with respect to the horizontal axis, and so we end up on that axis.

So $[L: \Bbb{R}(\zeta_5):\Bbb{R}]=2$ since the minimal polynomial over $\Bbb{R}$ is $x^2-(\zeta_5+\zeta_5^4)x+1$.

Now we need to compute $[L:\Bbb{R}(\zeta_5)]$.

Since $x^2+x+1$ is the minimal polynomial for $\zeta_3$ over $\Bbb{R}$ and since $\Bbb{R} \subseteq \Bbb{R}(\zeta_5)$, we just need to check if $x^2+x+1$ has any linear factors in $\Bbb{R}(\zeta_5)$. However, that would imply $\zeta_3 \in \Bbb{R}(\zeta_5)$, which is not true.

So $[L:\Bbb{R}]=4$

b) We just do the same thing, except that we cannot reduce the polynomial $x^4+x^3+x^2+x+1$, because $x^2-(\zeta_5 + \zeta_5^4) + 1 \not\in \Bbb{Q}[x]$, since $$\zeta_5 + \zeta_5^4 = \cos(2\pi/5) + i\sin(2\pi/5) + \cos(8\pi/5) + \sin(8\pi/5) = 2\cos(2\pi/5).$$

So $[L:\Bbb{R}]=8$.

c) For $x^4+x^3+x^2+x+1$, we check if it has linear factors in $\Bbb{F}_2$ by plugging in 0 and 1. Since there aren't any, we just need to check if it has a quadratic as a factor. Since $x^2+1=(x+1)^2$, we just need to check if $x^2+x+1$ is a factor, and we find that it isn't.

For $x^2+x+1$ (polynomial for $\zeta_3$), we just need to check if it has linear factors, and it doesn't. So $[L:\Bbb{F}_2]=8$

d) I can do what I did with $\Bbb{F}_2$, but that seems like it would take a long time, since there are 31 elements. It will also take a very long time trying to figure out the irreducible quadratics to see if they would divide $x^4+x^3+x^2+x+1$. So I was wondering if there was a simpler way to do this or if I had to do it the long way...

Are the other three correct?

-
b) is definitely correct. It's worth noting that you can take a slight shortcut here by noting that $[\mathbb{Q}(\zeta_{n}):\mathbb{Q}] = \varphi(n)$ for all $n$, where $\varphi$ is the euler-totient function. (This is a standard result proved in Dummit and Foote, and many other places I'm sure.) – Alex Wertheim Aug 7 '13 at 21:34
For a), note that $\mathbb{R}[\alpha] = \mathbb{C}$ for every $\alpha \in \mathbb{C}\setminus\mathbb{R}$, and $\mathbb{C}$ is algebraically closed. – Daniel Fischer Aug 7 '13 at 21:35
You appear to have left out the factor $$\psi_{15}(x)=x^8-x^7+x^5-x^4+x^3-x+1.$$ This did not cause errors later on, as the roots of this ($\zeta_{15}^a,\gcd(a,15)=1$) can be written in terms of $\zeta_5$ and $\zeta_3$. – Jyrki Lahtonen Aug 7 '13 at 21:52
In part (a) you wrote $$x^{15}-1= \Psi_{d|15}(x) = \Psi(x) \Psi_3(x) \Psi_5(x) = (x-1)(x^2+x+1)(x^4+x^3+x^2+x+1)$$ How can this be so? The thing on the right is a degree 7 polynomial. – Fly by Night Aug 7 '13 at 21:52

a) Algebraic extensions of $\mathbb{R}$ have degree $1$ or $2$. Since not all zeros of $x^{15} - 1$ are in $\mathbb{R}$, we get $[L : \mathbb{R}] = 2$.

b) Let $\zeta$ be a primitive $15$th root of unity. We have $L = \mathbb Q(\zeta)$. The minimal polynomial of $\zeta$ over $\mathbb Q$ is the $15$th cyclotomic polynomial of degree $\varphi(15) = 8$. So $[L : \mathbb Q] = 8$.

c) Using the fact that the multiplicative group of $\mathbb F_{2^r}$ is cyclic of order $2^r - 1$, $L$ is the smallest extension of $\mathbb F_2$ containing elements of order $15$. So $L = \mathbb F_{2^r}$ where $r$ is the smallest number such that $15\mid 2^r - 1$. Trying $r = 1,2,\ldots$, we find $r = 4$. Hence $[L : \mathbb F_2] = 4$.

d) Since $\mathbb F_{31}^\times$ is cyclic of order $30$, $\mathbb F_{31}$ contains elements of order $15$. So $[L : \mathbb F_{31}] = 1$.

-
@Thanks a lot. For c), $\Bbb{F}_{k}$ is a field if and only if $k$ is prime, right? So how is $\Bbb{F}_{2^r}$ a field if $r > 1$? Because if $k$ is not prime, then it is no longer an integral domain, right? – user58289 Aug 8 '13 at 13:26
Right? No, wrong. There is a field of each finite cardinality $p^r$, $p$ being prime and $r$ being positive. In particular, there is a field of cardinality $16$, which we call $\mathbb F_{16}$. And up to isomorphism, there is only one field of each possible cardinality. – Lubin Aug 8 '13 at 13:42

A hint for parts C and D: Use the fact that the multiplicative groups of finite fields are cyclic. So if the number of non-zero elements in the field (or an extension field) is $\equiv 1\pmod{15}$ that field has a lot of fifteenth roots of unity.

Another hint for part D: By Little Fermat $$a^{30}\equiv1\pmod{31}$$ for all $a=1,2,\ldots,30$. So $(a^2)^{15}=?$

-