5
$\begingroup$

Dedekind's theorem states that if a polynomial in $\mathbb Z[x]$ is factored into irreducibles modulo a prime not dividing the discriminant, then the Galois group of the polynomial, considered as a subgroup of $S_n$, contains a permutation whose cycle type corresponds with the degrees of the irreducible factors.

I've looked around a bit for a proof of this but couldn't find one. Can someone please point me to a proof?

$\endgroup$
  • $\begingroup$ That paper doesn't seem to make the connection to Galois groups. is there one that does? $\endgroup$ – Nishant Jun 17 '14 at 5:15
  • $\begingroup$ Oh, sorry, wrong Dedekind's theorem. I thought Keith Conrad had written something about how to prove this without using decomposition groups but now I can't find it. $\endgroup$ – Qiaochu Yuan Jun 17 '14 at 5:23
  • 1
    $\begingroup$ @QiaochuYuan you are thinking of what I wrote about proving the existence of Frobenius elements without using decomposition groups (which was really just the original proof by Frobenius). There is no simple proof of Dedekind's theorem that avoids algebraic number theory (residue fields at prime ideals). Jacobson's Basic Algebra I has a proof trying to avoid such machinery, due I believe to Tate, but it makes the proof look more complicated than it would be if the reader knew a bit of algebraic number theory first (as Dedekind did). In college I tried reading the proof in Jacobson and (contd.) $\endgroup$ – KCd Sep 5 '16 at 16:30
  • $\begingroup$ did not understand it at all. After learning alg. number theory I read a proof from that perspective instead and it was very easy to follow what was going on. $\endgroup$ – KCd Sep 5 '16 at 16:35
6
$\begingroup$

A quick proof depending on the basic theory of decomposition groups can be found on page 15 of these notes (see Corollary 2.7).

Edit: I decided to elaborate on what's included in the notes linked above. As stated there, this result essentially follows from the surjectivity of the natural map $$ D_{\mathfrak{P}|\mathfrak{p}} = \{\sigma \in \text{Gal}(L/K) : \sigma(\mathfrak{P}) = \mathfrak{P}\} \to \text{Gal}((\mathcal{O}_L/\mathfrak{P})/(\mathcal{O}_K/\mathfrak{p})) $$ for $L/K$ finite Galois extension of number fields, $\mathfrak{p}$ a prime of $K$, and $\mathfrak{P}$ a prime of $L$ above $\mathfrak{p}$.

To apply this here, one assumes $L$ is the splitting field over $K$ of a monic irreducible $f \in \mathcal{O}_K[X]$ and $\mathfrak{p}$ is a prime of $K$ modulo which $f(X)$ factors into a product of distinct irreducible polynomials. (So in particular, one can take $K = \mathbb{Z}$ and a prime $p$ of the desired form.) For such a prime $\mathfrak{p}$, choose any prime $\mathfrak{P}$ of $L$ over $\mathfrak{p}$ and let $S = \{\alpha \in L : f(\alpha) = 0\}$ (a set of $n = \deg(f)$ distinct elements of $L$). Then $D_{\mathfrak{P}|\mathfrak{p}}$ acts on $S$, and this action gives us a homomorphism $D_{\mathfrak{P}|\mathfrak{p}} \to H \leq S_n$. By the result quoted above, this $H \cong \text{Gal}((\mathcal{O}_L/\mathfrak{P})/(\mathcal{O}_K/\mathfrak{p}))$, and (as the Galois group of a finite extension of finite fields) the RHS is a cyclic group, say generated by some $\sigma$. More explicitly, if $f \pmod{\mathfrak p} = f_1 \cdots f_r$ for (by hypothesis) distinct monic $f_i$ over $\mathcal{O}_K/\mathfrak{p}$, then $\mathcal{O}_L/\mathfrak{P}$ is the splitting field of $f \pmod{\mathfrak{p}}$ over $\mathcal{O}_K/\mathfrak{p}$, it follows from generalities about Galois extensions that (under appropriate/obvious assumptions regarding the map into $S_n$) $\sigma$ corresponds to the permutation $(1,\ldots,d_1)(d_1+1,\ldots,d_2)\cdots(d_{r-1}+1,\ldots,d_r)$ where $d_i = \deg(f_i)$ for $1 \leq i \leq r$. Pullback $\sigma$ to an element $\text{Gal}(L/K)$ along the surjective homomorphisms $\text{Gal}(L/K) \to D_{\mathfrak{P}\mid\mathfrak{p}} \to H$ to reach the desired conclusion.

$\endgroup$
  • $\begingroup$ What does it mean to say that a prime is "above" another, and what does it translate to in terms of $\mathbb Z[x]$? $\endgroup$ – Nishant Jun 17 '14 at 5:18
  • $\begingroup$ @Nishant: Setting $K = \mathbb{Q}$ and (hence) $\mathcal{O}_K = \mathbb{Z}$ above, a prime $\mathfrak{p}$ of $K$ is a prime ideal in $\mathbb{Z}$, which in this case simply means that $\mathfrak{p} = (p) = \{rp : r \in \mathbb{Z}\}$ for some prime number $p$. (Since Dedekind's theorem is usually studied in courses on algebraic number theory, I took for granted that you are familiar with concepts like rings and fields in my answer. Hopefully that's alright with you.) Similarly, a prime $\mathfrak{P}$ in the splitting field $L$ of $f(X)$ over $\mathbb{Q}$ is any prime ideal of (ct'd ...) $\endgroup$ – Dan Jun 17 '14 at 22:33
  • $\begingroup$ (... ct'd) the ring of algebraic integers $\mathcal{O}_L$ in $L$ (a ring in $L$ containing $\mathbb{Z}$ as a subring). Multiplying ideals of $\mathbb{Z}$ by $\mathcal{O}_L$ give us ideals in $\mathcal{O}_L$, and in particular $(p)\cdot \mathcal{O}_L = \{rp \cdot \alpha : r \in \mathbb{Z}, \alpha \in \mathcal{O}_L\} = \{p\alpha : \alpha \in \mathcal{O}_L\} = p\mathcal{O}_L$ is an ideal in $\mathcal{O}_L$ which may no longer be prime! (This study of how $(p)$ decomposes when "lifted" like this to a prime in $\mathcal{O}_L$ is a central topic in algebraic number theory.) (ct'd ...) $\endgroup$ – Dan Jun 17 '14 at 22:38
  • $\begingroup$ (... ct'd) That said, as an ideal in the ring of integers $\mathcal{O}_L$ of a number field $L$, we know that $p\mathcal{O}_L$ has a unique (up to reordering) factorization into a product $p\mathcal{O}_L = \mathfrak{P_1}^{e_1} \cdots \mathfrak{P_g}^{e_g}$ of power of distinct prime ideals $\mathfrak{P_i}$ of $\mathcal{O}_L$ (where $e_i > 0$ for each $i$). A prime ideal $\mathfrak{P}$ in $\mathcal{O}_L$ is said to lie over or be above the ideal $(p)$ in $\mathbb{Z}$ if it occurs in this factorization, i.e., if $\mathfrak{P} = \mathfrak{P_i}$ for some $i$. $\endgroup$ – Dan Jun 17 '14 at 22:43
  • $\begingroup$ Oh, okay, thanks! Maybe I'll understand this a bit more after I start studying algebraic geometry... $\endgroup$ – Nishant Jun 18 '14 at 4:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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