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I am wondering if anyone can offer some "intuition" on what is going on concerning decomposition of prime numbers in relation to galois extensions of a cyclotomic polynomial?

I was asked to:

1) find intermediate fields corresponding to the various subgroups of $\mathbb Q(\zeta_{27})/\mathbb Q$. Then:

2) how primes decompose in each of those extensions

I have found what may be similar questions The most closely related seems to be here. There is also a paper here. However I am not gaining any understanding of what is really happening. I figured out 1), and at least have some understanding of what is happening:

For 1) We know by Euler's Phi function that $G=\operatorname{Gal}\left(\Bbb Q(\zeta_{27})/\Bbb Q\right)\cong \left(\Bbb Z/27\Bbb Z\right)^*$ is a cyclic group of order 27 $\phi(27) = 27-9=18$. Now as cyclic subgroups have unique subgroups of every order dividing the order of the group order. This means that there are 6 subgroups of $G = <x>$. These subgroups are $\{1\}, G, \langle x^2\rangle, \langle x^3\rangle, \langle x^6\rangle, \langle x^9\rangle$.

The degree of an extension and the index of a fixed field are exactly the same! Using this uniqueness of the subgroups and that the degrees of the extensions corresponds to the index of a fixed field we can classify the fields.

Also that for a subgroup $H$ of $Gal(E/F)$, the corresponding field $E^H$ is the set of those elements in $E$ which are fixed by every automorphism in $H$. For any intermediate field $K$ of $E/F$, the corresponding subgroup is $Aut(E/K)$ which is the set of those automorphisms in $Gal(E/F)$ which fix every element of $K$. This means that the for $H \subset Gal(\mathbb(\zeta_{27})/\mathbb{Q})$ topmost field $E$ corresponds to the trivial subgroup of $Gal(E/F)$ and the base field $F$ corresponds to the whole group $Gal(E/F)$.

But for 2), I just do not see how to proceed - could someone offer some intuition of what is going on? I know that I will have a relation to a congruence modulo 27, and as, in the end, as 3 is the only factor of 27 as $3^3$ I am guessing something "special happens" in relation to the prime 3 (and only 3?) as every other prime than 3 (or ideal of 3) have no common factors (by definition of prime).

Forgive my informality but am trying to understand what is going on before I write it formally.

Any advice?

Thanks,

Brian

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  • $\begingroup$ Given your follow up, I think you must have not meant what you wrote for 1) - "find intermediate fields corresponding to the various subgroups of the galois group of order 27" - Rather, you mean "find the intermediate fields of $\mathbb Q(\zeta_{27})/\mathbb Q$," - correct? $\endgroup$ – peter a g Sep 10 '15 at 16:19
  • $\begingroup$ @peterag, thanks for the catch, indeed the subgroups of the cyclotomic polynomial of order 27 over the rationals. $\endgroup$ – Relative0 Sep 10 '15 at 16:43
  • $\begingroup$ Let $\omega = \frac{1+\sqrt{3}}{2}$ be a 3rd root of unity, so $\mathbb{Q}(\omega)$ will be one of your subfields. Over $\mathbb{Z}$ you get $\mathbb{Z}[\omega]/(p) = \mathbb{Z}[X]/(p, 1+X+X^2) = \mathbb{F}_p[X]/(1+X+X^2)$, and therefore $p$ is a prime element of $\mathbb{Z}[\omega]$ iff $1+X+X^2$ is irreducible over $\mathbb{F}_p[X]$. Is this really the case for (let's say) $p=7$? $\endgroup$ – j.p. Sep 11 '15 at 15:39
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For reference, but without proof, general theory says:

Suppose $L/k$ is an extension of number fields, with $n = [L:k]$, and ring of integers $A_L$ and $A_k$ respectively. A prime ideal $\pi$ of $A_k$ factors $$ \pi A_L=P_1^{e_i}\cdots P_g^{e_g}, $$ where $P_i$ are primes (prime ideals) of $A_L$, and $ A_L/P_i$ is a field extension $f_i$ of the the finite field $\mathbb F= A_k/\pi $; one denotes the degree of the extension $f_i$. Then $$ n = e_1f_1 + \cdots + e_gf_g.\tag {1}$$ If $L/k$ is Galois, the Galois group permutes the $P_i$ transitively, with the result that the $f_i$ and $e_i$ do not depend on $i$. Therefore one drops the subscript $i$, and $(1)$ simplifies to $n= efg$.

Vocabulary: One says that $\pi$ ramifies if $e>1$; this happens only finitely often (iff $\pi$ divides the relative discriminant). If $g =n$, we say the prime $\pi$ splits completely; if $f =n$, one says that $\pi$ is inert - $\pi A_L$ is a prime ideal.

Returning to your case: set $L =E = \mathbb Q(\zeta_{27})$.

The prime $3$ is a special case, and is the only prime to ramify, and ramifies totally, i.e, $$ (3) = ( 1 - \zeta_{27})^{18} $$ in $E$ (as you say, $\phi(27) =27-9 = 18$).

Again, as you say, $E$ is Galois, with group $G=\mathop { \rm Gal} (E/\def\Q {{\mathbb Q}} \Q) = (\def\R {\mathbb Z/27\mathbb Z}\R)^* $, where the identification is given by $$ \sigma_a \colon \zeta \mapsto \zeta ^a,$$ for $a\in(\R)^*$. Suppose $H < G$ is a subgroup, and $F = E^H$ the corresponding fixed field. Since $G$ is abelian, $H$ is normal, and $F/\mathbb Q$ is Galois and abelian.

For any prime $p$ ($\ne 3$), its behavior in $F=E^H$ is determined by its residue class in $G/H$, and we can read it off from the order of $\sigma_p$, which is $f$: for any subgroup $H<G$, if $f$ is the smallest integer such that $p^f \in H$, then for $P$ a prime of $F= E^H$ over $p$, one has $A_F/P$ is a field extension of degree $f$ over $\mathbb F_p$, and conversely; the number of $P$ above $p$ is $ g = n/f$.

For example, taking $H=1$: then $F=E=E^H$, and $p$ splits completely in $F=E$ iff the residue class $p\in H$, i.e., $p \equiv 1 \pmod {27}$. So, for instance, $p= 4\cdot 27 +1 = 109$ splits completely in $E$. On the other hand the prime $p=7$ (j.p.'s prime of example above) has order multiplicative order $9$ (i.e., $7^9\equiv 1 \pmod {27}$, and $9$ is smallest positive such...), so $ 7 A_E = P_1 P_2$, with $A_E/P_i$ a field of degree $9$ over $\mathbb F_{7}$. $5$ has order $18$, so is inert, $19$ has order $3$, so factors as a product of $6$ primes etc.

Likewise, there is a unique extension $F$ (also a cyclotomic extension, $\mathbb Q( \zeta_3)$) of degree $2$. It corresponds to the subgroup $H$ of $G$ of index $2$: $H= \{a\in \R\ | \ a \equiv 1 \pmod 3\}$. Then $p$ splits in $F$ if $ p \equiv 1 \pmod 3$, is inert if $p \equiv 2 \pmod 3$, and if $p =3$, it ramifies: $$ (3) = ( 1 - \zeta_3)^2.$$ To take j.p.'s example, $7\equiv 1 \pmod 3$, and it splits in $F$, but $5$ is inert - it remains prime.

(A stupid example is $H=G$. Then $F= \mathbb Q$, and the theory successfully manages not to contradict that a prime $p$ is a prime.)

The other cases (other $H$) are similar, says he, manifestly wimping out. But, they are the subfields $\mathbb Q(\zeta_9)$, $\mathbb Q(\zeta_{27} + \bar \zeta_{27})$, and $ \mathbb Q(\zeta_{9} + \bar \zeta_{9})$, and correspond (up to permutation) to your enumeration of the subgroups of $G$.

A few general remarks... The above certainly doesn't prove much - sorry! On the other hand, write $ \mu_m(A) = \{ a \in A \ | \ a^m = 1\}$ for any ring $A$. Then if $(m,p)=1$, the natural reduction map (after choosing a prime above $p$) $$ \mu_m(\bar{\mathbb Q}) \rightarrow \mu_m(\bar {\mathbb F_p}) $$ is an isomorphism. This is because the size of the two sets are the same, and the map is injective (because if $g(x) = x^m-1$, then the reduction of $g'(\zeta)\ne 0$), and therefore onto. On the other hand, if the $m$th roots of unity are a subgroup of $\mathbb F_{p^f}^*$, we must have $ m | (p^f-1)$, and conversely. This is the main point in the above.

The assertions about which primes ramify depend on the explicit calculation of the ring of integers of $\mathbb Q(\zeta_{p^k})$, which is $ \mathbb Z[ \zeta_{p^k}] = \mathbb Z[ \zeta_{p^k} -1] $, and of the calculation of the discriminant, which is the norm of $ g'(\zeta_{p^k})$ (up to sign, perhaps), where $ g(x) = x^{p^k} -1$. Both are probably due to Kummer, I would think.

References: Lang's ANT, esp. Chapter IV, or Milne's notes http://www.jmilne.org/math/CourseNotes/ANTe6.pdf, especially around p 180, for the discriminant and ring of integers calculation.

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  • $\begingroup$ @Relativeo I just looked at how bounties worked - I see I owe you a thank you: thank you! Sincerely if belatedly... $\endgroup$ – peter a g Oct 14 '15 at 17:59

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