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The following Cubics have 3 real roots but the first has Galois group $C_3$ and the second $S_3$

  • $x^3 - 3x + 1$ (red)
  • $x^3 - 4x + 2$ (green)

Is there any geometric way to distinguish between the two cases? Obviously graphing this onto the real line does not help.

enter image description here

It is not clear to me why you cannot transpose the red dots but you can transpose the green ones.

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The discriminant tells the story; if the discriminant is a square, then the Galois group is $C_3$; if the discriminant is not a square, then the Galois group is $S_3$. –  Arturo Magidin May 11 '11 at 21:03
    
I think it largely comes down to what counts as "geometric". I could very well imagine there being some sort of high-powered arithmetic geometry explanation for the difference, but I don't think such an explanation (if there is one) would give any sort of visualizable clues as to whether a given cubic has $C_3$ or $S_3$ –  Zev Chonoles May 11 '11 at 21:05
    
If the discriminant is square does that correspond to some lattice being square? I don't think so. I want to construct a geometric object from an arbitrary cubic (maybe some restrictions, but certainly not knowing their Galois group beforehand) that will show what is happening. –  quanta May 11 '11 at 21:07

2 Answers 2

There's no reason to expect that the set of real points tells you the full story in an arithmetic situation. For example, can you tell that $\pi$ is transcendental but that $\sqrt{10}$ isn't from looking at their relative positions on the number line?

One thing you can do which (depending on your tastes) could count as geometric is looking at Frobenius elements.

Proposition: Let $f(x) \in \mathbb{Z}[x]$ be a monic irreducible polynomial. Suppose that for some prime $p$ not dividing the discriminant, $f$ splits into irreducible factors of degrees $d_1, d_2, ... d_k$. Then there is an element of cycle type $(d_1, d_2, ... d_k)$ in the Galois group of $f$.

Hence you can find a transposition in the Galois group (and show that the Galois group is $S_3$) by finding a prime $p$ relative to which $f$ splits as the product of a linear and an irreducible quadratic factor. (This is completely analogous to the situation over $\mathbb{R}$: one might say that complex conjugation is the "Frobenius element at infinity.")

Geometrically, instead of looking at the real points of the scheme $\text{Spec } \mathbb{Z}[x]/f(x)$, we look at points over finite fields. But from a scheme-theoretic point of view all of these points are included in the geometry of the "arithmetic curve" $\text{Spec } \mathbb{Z}[x]/f(x)$.

The Frobenius density theorem even guarantees that the converse holds: for every cycle type in the Galois group there is a prime $p$ realizing that cycle type.

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so because $x^3 - 4x + 2 \equiv (x + 1)(x^2 + 4x + 2) \pmod 5$ this polynomial has a transpose (and thus Galois group $S3$) and presumably $x^3 - 3x + 1$ modulo every prime always factors completely or doesn't factor at all (I guess it would never factor until I found) $x^3 - 3x + 1 \equiv (x + 3)(x + 4)(x + 10) \pmod {17}$. On the other hand $x^3 - 4x + 2 \equiv (x+39)(x+46)(x+49) \pmod {67}$. So it seems like you have to actually check every prime until you find one.. and that will never end if your polynomial is cyclic. –  quanta May 12 '11 at 0:36
    
@quanta: yes, if you can't find a transposition then that doesn't give a proof that the Galois group is $C_3$, just evidence. But as the comments say, that's what the discriminant is for. –  Qiaochu Yuan May 12 '11 at 0:38
    
Although I don't know what factoring over finite fields has to do with geometry, I found the term "Galois geometry" which I will try to read about hopefully to understand this. –  quanta May 12 '11 at 0:39
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@quanta: it's arithmetic geometry. The idea is that we have a natural inclusion $\mathbb{Z} \to \mathbb{Z}[x]/f(x)$ which scheme-theoretically gives a branched cover $\text{Spec } \mathbb{Z}[x]f(x) \to \text{Spec } \mathbb{Z}$. The fiber of this map over $p$ is $\text{Spec } \mathbb{F}_p[x]/f(x)$ which describes the factorization of $f \bmod p$ and one is studying the "monodromy action" (en.wikipedia.org/wiki/Monodromy) on this fiber when one studies Frobenius elements. In other words, it's an arithmetic analogue of a geometric (and topological) situation. Again, I think it is too... –  Qiaochu Yuan May 12 '11 at 0:47
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...much to expect a classically geometric answer to an arithmetic question. You have to broaden your definition of geometry to see the geometry in this situation. –  Qiaochu Yuan May 12 '11 at 0:49

Almost all cubics (with integer coefficients and three real roots) have Galois group $S_3$. What exactly is meant by "almost all" is a little technical, but the phrase can be made precise, and the result rigorously proved. One consequence is that if you start with a $C_3$ cubic and perturb the roots the tiniest little bit then with probability $1$ you now have an $S_3$ cubic. So just looking at the red dots can't help you: it's guaranteed that there is a set of green dots so close by that you wouldn't be able to distinguish them with an electron microscope.

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