Principle Ideal in F[x] & Primitive nth root of unity

[mathjam0990]

Let ζ₅ be e^2πi/5. Find a monic polynomial of degree two in K(ζ + ζ⁻¹)

So, if E/F is a field extension, with α∈E then K(α) = {f(x)∈F[x] | f(α)=0} and m(x) is the minimal polynomial of α over F such that K(α) = [m(x)] where [m(x)] is the ideal generated by m(x).

I was thinking maybe (x- ζ - ζ⁻¹)(x + ζ - ζ⁻¹). Sorry I am not sure how to approach this exactly but I know of course ζ + ζ⁻¹ should be a root of this degree 2 polynomial with the x² coefficient as 1.

If anyone could provide an explanation that would be very helpful. Thank you.

 

[Deveno]

I hope you meant "irreducible monic polynomial", as there are several monic polynomials that are possible (take any monic quadratic from $\Bbb Q[x]$, for example).

You also don't say what "$K$" is, which is important. There are lots and lots of fields, and reducibility depends in an essential way on "which field we're in".

I will pretend $K = \Bbb Q$, the rational numbers.

First, some preliminary facts about $\zeta = e^{\frac{2\pi i}{5}}$:

Note that $\zeta^5 = 1$ which implies $\zeta^{-1} = \zeta^4$.

From $\zeta^5 = 1$ it is evident that $(\zeta^k)^5 = \zeta^{5k} = (\zeta^5)^k = 1^k = 1$. It is easy to see that:

$\zeta,\zeta^2,\zeta^3,\zeta^4$ are all distinct (use trigonometry) and thus that:

$1,\zeta,\zeta^2,\zeta^3,\zeta^4$ are 5 distinct roots of $x^5 - 1$, and thus that $\Bbb Q(\zeta)$ is the splitting field of:

$\dfrac{x^5 - 1}{x - 1} = x^4 + x^3 + x^2 + x + 1$.

The RHS is the 5th cyclotomic polynomial, and these are all irreducible over $\Bbb Q$. This is a non-trivial theorem, but the proof for the $p$-th cyclotomic polynomial (for a prime $p$) is somewhat easier. I will not prove this here, we can discuss it later, if you like.

So, evidently:

$x^4 + x ^3 + x^2 + x + 1 = (x - \zeta)(x - \zeta^2)(x - \zeta^3)(x - \zeta^4)$.

Thus this polynomial ($\Phi_5(x)$) is separable, and since adjoining one root gives us the entire splitting field, it is a Galois extension of $\Bbb Q$ (it is separable, and normal). So the Galois group is the same as $\text{Aut}(\Bbb Q(\zeta)/\Bbb Q)$, and has order 4.

Since $\sigma:\Bbb Q(\zeta) \to \Bbb Q(\zeta)$ defined by $\sigma(\zeta) = \zeta^2$ is an element of this group, and has order 4:

$\sigma^2(\zeta) = \sigma(\sigma(\zeta)) = \sigma(\zeta^2) = (\sigma(\zeta))^2 = \zeta^4$
$\sigma^3(\zeta) = \sigma(\sigma^2(\zeta)) = \sigma(\zeta^4) = (\sigma(\zeta))^4 = \zeta^8 = \zeta^3$
$\sigma^4(\zeta) = \sigma(\sigma^3(\zeta)) = \sigma(\zeta^3) = (\sigma(\zeta))^3 = \zeta^6 = \zeta$

we see this group is cyclic. By the Galois correspondence, we see that since we have just ONE subgroup of order 2 (namely: $\{\text{id},\sigma^2\}$), we have just one subfield of $\Bbb Q(\zeta)$ of dimension 2 over $\Bbb Q$ as a vector space. We will see shortly, that that field is $\Bbb Q(\zeta+\zeta^{-1})$.

Our automorphism group can be regarded as a permutation group acting on the roots of $x^4 + x^3 + x^2 + x + 1$, in this case, we have that:

$\sigma \mapsto (1\ 2\ 4\ 3)$
$\sigma^2 \mapsto (1\ 4)(2\ 3)$
$\sigma^3 \mapsto (1\ 3\ 4\ 2)$
$\sigma^4 \mapsto e$

So the subfield we are looking for is the field fixed by $\{\text{id}, \sigma^2\}$, which suggests grouping the polynomial factors like so:

$[(x - \zeta)(x - \zeta^4)][(x - \zeta^2)(x - \zeta^3)]$

Since each quadratic of the factorization above is fixed by $\sigma^2$ (it just swaps roots in each quadratic pair), the coefficients must lie in the fixed field of $\{\text{id}, \sigma^2\}$. In particular:

$(x - \zeta)(x - \zeta^4) = (x - \zeta)(x - \zeta^{-1}) = x^2 - (\zeta + \zeta^{-1})x + 1$

has coefficients in this fixed field. So the fixed field CONTAINS $\Bbb Q(\zeta + \zeta^{-1})$.

Note as well, that complex conjugation restricted to $\Bbb Q(\zeta)$ is an automorphism of $\Bbb Q(\zeta)$ of order 2 that fixes $\Bbb Q$, so $\sigma^2$ IS complex conjugation.

What that tells us, is that $\zeta^4 = \overline{\zeta}$, and thus that $\Bbb Q(\zeta+\zeta^{-1}) \subseteq \Bbb R$.

Since the roots of $x^2 - (\zeta + \zeta^{-1})x + 1 = (x - \zeta)(x - \overline{\zeta})$ are not real, they certainly do not lie in $\Bbb Q(\zeta+\zeta^{-1})$, and thus:

$x^2 - (\zeta + \zeta^{-1})x + 1$ is irreducible over $\Bbb Q(\zeta + \zeta^{-1})$, which is presumably what you were looking for.

As promised, we will now show the fixed field of the unique subgroup of the Galois group of order 2 IS $\Bbb Q(\zeta + \zeta^{-1})$ (we already know it contains it).

To this end, some preliminary algebra:

$(\zeta + \zeta^{-1})^2 = \zeta^2 + 2(\zeta\zeta^{-1}) + (\zeta^{-1})^2 = \zeta^2 + 2 + \zeta^{-2} = \zeta^2 + \zeta^3 + 2$.

Now the elements of $\Bbb Q(\zeta)$ are of the form:

$a_0 + a_1\zeta + a_2\zeta^2 + a_3\zeta^3: a_0,a_1,a_2,a_3 \in \Bbb Q$.

If such an element $u$ is fixed by $\sigma^2$ (that is $\sigma^2(u) = u$), then we have:

$a_0 + a_1\zeta + a_2\zeta^2 + a_3\zeta^3 = \sigma^2(a_0 + a_1\zeta + a_2\zeta^2 + a_3\zeta^3)$

$= a_0 + a_1\sigma^2(\zeta) + a_2\sigma^2(\zeta^2) + a_3\sigma^2(\zeta^3)$

$= a_0 + a_1\zeta^4 + a_2\zeta^3 + a_3\zeta^2$.

Recall that $\zeta^4 + \zeta^3 + \zeta^2 + \zeta + 1 = 0$, so:

$\zeta^4 = -\zeta^3 - \zeta^2 - \zeta - 1$, so that:

$a_0 + a_1\zeta^4 + a_2\zeta^3 + a_3\zeta^2 = (a_0 - a_1) - a_1\zeta + (a_3 - a_1)\zeta^2 + (a_2 - a_1)\zeta^3$.

Equating the coefficients (they must be equal, since by the linear independence of $\{1,\zeta,\zeta^2,\zeta^3\}$ over $\Bbb Q$ these coefficients are unique), we have:

$a_0 = a_0 - a_1$
$a_1 = -a_1$
$a_2 = a_3 - a_1$
$a_3 = a_2 - a_1$, so that $a_1 = 0$, and $a_2 = a_3$.

Thus the elements fixed by $\sigma^2$ are of the form:

$a_0 + a_2\zeta^2 + a_2\zeta^3 = a_0 + a_2(\zeta^2 + \zeta^3) = a_0 - 2a_2 + a_2(\zeta^2 + \zeta^3 + 2)$

$= a_0 - 2a_2 + a_2(\zeta + \zeta^{-1})^2 \in \Bbb Q(\zeta + \zeta^{-1})$

and we see the fixed field is contained in $\Bbb Q(\zeta + \zeta^{-1})$, and thus the two are equal.

**************

Perhaps you were only looking for the "grouping" of factors I did earlier-this is rather the scenic route, and there are shorter ways to get from point A to point B, but I believe it helps to see how the information we get from group theory "forces our hand" in discussing the field structure (which would otherwise be "harder to arrive at").

 

[mathjam0990]

Thank you very much. This was helpful!