Galois Groups of a family of polynomials - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T15:12:52Z http://mathoverflow.net/feeds/question/44844 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44844/galois-groups-of-a-family-of-polynomials Galois Groups of a family of polynomials Victor Miller 2010-11-04T16:48:49Z 2010-11-09T06:36:14Z <p>I've stumbled across the family of polynomials $f_p(x) = x^{p-1} + 2 x^{p-2} + \cdots + (p-1) x + p$, where $p$ is an odd prime. It's not too hard to show that $f_p(x)$ is irreducible over $\mathbb{Q}$ -- look at the Newton polygon of $f_p(x+1)$ over $\mathbb{Q}_p$ and you see that it factors as the product of an irreducible polynomial of degree $p-2$ and a linear. Since $f_p(x)$ has no real roots (look at the derivative of $f_p(x) (x-1)^2$) it must be irreducible over $\mathbb{Q}$. It's also not hard to see that the only primes dividing the discriminant are $2, p$ and primes dividing $p+1$. I would expect that the Galois group of a randomish polynomial would be the full symmetric group. Indeed, according to Magma this is true for $f_p(x)$ for $p=3,5, \dots, 61$ with the exception of $p=7,17$. So my question is -- are these the only exceptions?</p> <p>Added later: I've had Magma find the Galois group for primes through 101 and it found another exception: $p=97$. So my initial guess was wrong.</p> <p>Another addition: If one looks at odd $p$ (not just prime) for $p &lt; 100$ there is another exception, 49. Also 241 is <strike> not </strike> an exception (misread magma's output).</p> <p>The ideas in the following two papers may be of help:</p> <p>"On the Galois Groups of the exponential Taylor polynomials" by Robert Coleman, in L'Enseignement Mathematique, v 33 (1987) pp 183-189</p> <p>and</p> <p>"On the Galois Group of generalized Laguerre polynomials" by Farshid Hajir, J. Th´eor. Nombres Bordeaux 17 (2005), no. 2, 517–525 (also available on the author's web page).</p> http://mathoverflow.net/questions/44844/galois-groups-of-a-family-of-polynomials/44874#44874 Answer by David Speyer for Galois Groups of a family of polynomials David Speyer 2010-11-04T20:36:41Z 2010-11-05T00:59:38Z <p>The discriminant of this polynomial is $(-1)^{\binom{p}{2}} 2 p^{p-3} (p+1)^{p-2}$. Thus, whenever $p$ is of the form $2k^2-1$ with $k$ odd, the discriminant will be a perfect square and the Galois group will be a subgroup of the alternating group. That explains all the counterexamples we have found. We now verify that this is the discriminant. </p> <hr> <p>I'm finding signs difficult tonight, so I'll only get my formulas right up to a sign.</p> <p>We start with the following observation. Let $f$ be a polynomial of degree $n$. Then <code>$$\mathrm{Disc} ( f(x) (x-1)) = f(1)^2 \mathrm{Disc} (f(x)) \quad (*)$$</code> To see this, let $r_1$, $r_2$, ..., $r_n$ be the roots of the polynomial $f$ and remember that $\mathrm{Disc}(f) = \prod_{i &lt; j} (r_i-r_j)^2.$</p> <p>Now, recall the identity <code>$$\mathrm{Disc}(x^{n+1}-ax+b) = \left( (n+1)^{n+1} b^n - n^n a^{n+1} \right).$$</code> It is easy to check that the right hand side vanishes if and only if $x^{n+1}-ax+b$ has a root in common with its derivative $(n+1)x^n-a$, so the two sides agree up to a constant. Checking the constant is left as an exercise.</p> <p>Plugging in $a=b+1$, we get <code>$$\mathrm{Disc}(x^{n+1}-(b+1)x+b) = \left( (n+1)^{n+1} b^n - n^n (b+1)^{n+1} \right).$$</code> Applying $(*)$ to the identity $x^{n+1}-(b+1)x+b = (x-1)(x^n+x^{n-1}+\cdots+x^2+x-b)$, we get <code>$$\mathrm{Disc}(x^n+x^{n-1}+\cdots+x^2+x-b) = \frac{ (n+1)^{n+1} b^n - n^n (b+1)^{n+1} }{(n-b)^2}.$$</code></p> <p>Taking the limit as $b \to n$, and using l'Hospital twice, we get <code>$$\mathrm{Disc}(x^n+x^{n-1}+\cdots+x^2+x-n) =$$</code> <code>$$(-1)^{\binom{n+1}{2}} \frac{ (n+1)^{n+1} n (n-1) n^{n-2} - n^n (n+1) n (n+1)^{n-1} }{2}$$</code> <code>$$= \frac{ (n+1)^n n^{n-1} \left( (n+1)(n-1) - n^2 \right) }{2} = \frac{ (n+1)^n n^{n-1} }{2}$$</code></p> <p>Now, applying $(*)$ once more, we get <code>$$\mathrm{Disc}(x^{n-1} + 2 x^{n-2} + \cdots + (n-1) x + n )$$</code> <code>$$\frac{ (n+1)^n n^{n-1} }{2} \ \left( \frac{n(n+1)}{2} \right)^{-2} = 2 (n+1)^{n-2} n^{n-3}.$$</code></p> http://mathoverflow.net/questions/44844/galois-groups-of-a-family-of-polynomials/44887#44887 Answer by Dror Speiser for Galois Groups of a family of polynomials Dror Speiser 2010-11-04T22:42:44Z 2010-11-04T22:42:44Z <p>The following proves David's statement on the discriminant.</p> <p>$f_n(x) := x^{n-1}\ +\ 2x^{n-2}\ +\ ...\ +\ n$</p> <p>$f_n(x) = x\frac{x^n-1}{(x-1)^2}-\frac{n}{x-1}$</p> <p>Say $f_n(\alpha)=0$,</p> <p>$f'_n(x) = \frac{x^n-1}{(x-1)^2} + x\frac{nx^{n-1}(x-1)^2-2(x-1)(x^n-1)}{(x-1)^4}+\frac{n}{(x-1)^2}$</p> <p>$f'_n(\alpha) = \frac{n}{\alpha(\alpha-1)}+(n(\frac{n}{\alpha(\alpha-1)}+\frac{1}{(\alpha-1)^2})-\frac{2\alpha}{\alpha-1}\frac{n}{\alpha(\alpha-1)})+\frac{n}{(\alpha-1)^2}$</p> <p>$=\frac{n}{\alpha(\alpha-1)^2}((\alpha-1)+(n(\alpha-1)+\alpha)-2\alpha+\alpha)$</p> <p>$=\frac{n(n+1)(\alpha-1)}{\alpha(\alpha-1)^2}=\frac{n(n+1)}{\alpha(\alpha-1)}$</p> <p>$\Delta_{f_n} = (-1)^{\frac{(n-1)(n-2)}{2}} \text{Nm}(f'_n(\alpha)) = (-1)^{\frac{(n-1)(n-2)}{2}}\frac{(n(n+1))^{n-1}}{nf(1)} = (-1)^{\frac{(n-1)(n-2)}{2}}2n^{n-3}(n+1)^{n-2}$</p> <p>This shows that whenever $n+1$ is twice an odd square the discriminant is a square.</p> http://mathoverflow.net/questions/44844/galois-groups-of-a-family-of-polynomials/45412#45412 Answer by Lavender Honey for Galois Groups of a family of polynomials Lavender Honey 2010-11-09T06:36:14Z 2010-11-09T06:36:14Z <p>Let $\alpha$ be a root of a polynomial $f(x) \in \mathbf{Q}[x]$ of degree $n$, let $K = \mathbf{Q}(\alpha)$, $L$ be the Galois closure of $K$, and $G = \mathrm{Gal}(L/\mathbf{Q}) \subset S_n$. How does one prove that a permutation group contains $A_n$? Following Jordan, the usual method is to show that it is sufficiently highly transitive. Also following Jordan, to do this it suffices to construct subgroups of $G$ which act faithfully and transitively on $n-k$ points and trivially on the other $k$ points (for $k$ large, $\ge 6$ using CFSG), and to show that $G$ is primitive. (The standard method for doing this is to find $l$-cycles for a prime $l$.) In the context of a Galois group, the most obvious place to look for "elements" is to consider the decomposition groups $D$ at places of $\mathbf{Q}$. If $l$ is unramified in $K$, this corresponds to looking at a Frobenius element (conjugacy class). In practice (as far as a computation goes) this is quite useful, but theoretically it is not so great unless there is a prime $l$ for which the factorization is particularly clean. This leaves the places which ramify in $L$. For example, if $v = \infty$, one is considering the action of complex conjugation; if there are exactly two complex roots then $c$ is a $2$-cycle, and from Jordan's theorem (easy in this case) we see that if $G$ is primitive then $G$ is $S_n$. </p> <p>The proposed method (following Coleman et. al.) for proving that $G$ contains $A_n$ is somewhat misguided, I think. The key point about the polynomial $\sum_{k=0}^{n} x^k/k!$ is that the corresponding field is ramified at many primes, and the decomposition groups at these primes give the requisite elements. Conversely, the polynomial considered in this problem corresponds to a field with somewhat limited ramification - as has been noted, the only primes which ramify divide $p(p+1)$.</p> <p>It can be hard to compute Galois groups of random families of polynomials in general. I do not know if this is true in the present case, but given the lack of motivation I won't spend any more time thinking about it than the last hour or two, and instead give some partial results. However, the methods given here may well apply more generally. Let $n = p - 1$.</p> <p>CLAIM: Suppose that $p+1$ is exactly divisible by a prime $l > 3$. Then $G$ contains $A_{n}$. (This applies to a set $p$ of relative density one inside the primes.)</p> <p>STEP I: Factorization of $p$; $G$ is primitive. Let $f(x) = x^{p-1} + 2 x^{p-2} + \ldots + p$. Note that $$(x-1)^2 f(x) = x(x^{p} - 1) - p(x-1) = x^{p+1} - 1 - (p+1)(x-1).$$ We deduce that $f(x) \equiv x(x-1)^{p-2} \mod p$, and that $$p = \mathfrak{p} \mathfrak{q}^{p-2}$$ for primes $\mathfrak{p}$ and $\mathfrak{q}$ in the ring of integers $O_K$ of $K$ both of norm $p$. (To show this one needs to check that $[O_K:\mathbf{Z}[\alpha]]$ is co-prime to $p$ - one can do this by considering the Newton Polygon of $f(x+1)$.) Let $D \subset G$ be a decomposition group at $p$. This corresponds to choosing a simultaneous embedding of the roots of $f(x)$ into an algebraic closure of the $p$-adic numbers. We see that we may write $f(x) = a(x) b(x)$ as polynomials over the $p$-adic numbers (which I can't latex at this point for some reason), where $a(x) \equiv x \mod p$ has degree one and $b(x) \equiv (x-1)^{p-2}$ is irreducible of degree $p-2$ and corresponds to a totally ramified extension. Clearly $D$ acts transitively on the $p-2 = n-1$ roots of $b(x)$ and fixes the roots of $a(x)$. Since $D \subset G \cap S_{n-1}$, we see that $G \cap S_{n-1}$ is transitive in $S_{n-1}$ and so $G$ is $2$-transitive (and hence primitive).</p> <p>Step II: Factorization of $l$: Let $l$ be a prime dividing $p+1$. We assume that $l \ge 5$ and $l$ exactly divides $p+1$. We see that $$f(x) \equiv (x-1)^{l-2} \prod_{i=1}^{k-1} (x-\zeta^i)^{l}$$ where $\zeta$ is a $k$th root of unity and $kl=p+1$. This suggests that: $$l = \mathfrak{p}^{l-2} \prod_{i=1}^{k} \mathfrak{q}^l.$$ This also follows from a Newton polygon argument applied to $f(x - \zeta^i)$. (Warning, this uses that $l$ exactly divides $p+1$.)</p> <p>Step III: Some basic facts about local extensions:</p> <p>Lemma 1. Suppose the ramification degree of $E/\mathbb{Q}_l$ is $l^m$. Then the ramification degree of the Galois closure of $E$ is only divisible by primes dividing $l(l^m-1)$. Proof. Kummer Theory.</p> <p>Lemma 2. Suppose that $h(x) \in \mathbf{Q}_l[x]$ is an irreducible polynomial of degree $k$ with $(k,l) = 1$, such that the corresponding field $E/\mathbf{Q}_l$ is totally ramified. If $F$ is the splitting field of $h(x)$, then $\mathrm{Gal}(F/\mathbf{Q}_l) \subset S_k$ contains a $k$-cycle. Proof: From a classification of tamely ramified extensions, there exists an unramified extension $A$ such that $[EA:A] = [E:\mathbf{Q}_l]$ and $EA/A$ is cyclic and Galois. It follows that $\mathrm{Gal}(EA/A)$ acts transitively and faithfully on the roots of $h(x)$, and is thus generated by a $k$-cycle.</p> <p>Step IV: $G$ contains an $l-2$-cycle. Consider the decomposition group $D$ at $l$. The orbits of $D$ correspond to the factorization of $l$ in $O_K$. On the factors corresponding to primes of the form $\mathfrak{q}^p_i$, the image of $D$ factors through a group whose inertia has degree divisible only by primes dividing $l(l-1)$, by Lemma 1. On the other hand, on the factor corresponding to $\mathfrak{p}^{l-2}$, the image of inertia contains an $l-2$ cycle, by Lemma 2. Since $(l(l-1),l-2) = 1$, we see that $D \subset G$ contains an $l-2$ cycle.</p> <p>Step V: Jordan's Theorem. Since $G$ is primitive, and $G$ contains a subgroup that acts transitively and faithfully on $l-2$ points (and trivially on all other points), we deduce (from the standard proof of Jordan's theorem) that $G$ is $n-(l-2)+1 = n+3-l$ transitive. This is at least $6$ (since $n+2$ is at least $2l$) and so $G$ contains $A_n$ (by CFSG).</p> <p>STEP VI: (for you, dear reader) Find the analogous argument when $p+1$ is exactly divisible by $l^k$ for some $k \ge 2$ --- try to construct a cycle of degree $l^k - 2$, although be careful as it will no longer be the case (as it was above) that $[O_K:\mathbf{Z}[\alpha]]$ was co-prime to $l$. This still leaves $p-1$ either a power of $2$ or a power of $2$ times $3$, which might be annoying --- one would have to think hard about the structure of the decomposition group at $2$ in those cases.</p>