User peter mueller - MathOverflow

Answer by Peter Mueller for A generalisation of the theorem of Maschke

2013-05-14T13:24:01Z

Your expected generalization is true. Without loss of generality, you may assume that $G$ is a subgroup of $GL(V)$. A linear group whose elements have finite orders is locally finite, that's an old Theorem by Schur, and contained for instance in Wehrfritz's book Infinite Linear Groups. Now take a maximal linearly independent subset $S$ of $G$, so $\langle S\rangle$ and $G$ have the same invariant subspaces on $V$. By Schur, $\langle S\rangle$ is finite, so your generalization holds by Maschke.

Answer by Peter Mueller for A question from complex analysis

2013-04-19T16:21:06Z

The answer is no. Take the polynomials \begin{align} f(x) &= x^6 + x^5 + x^4 + x^3 + x^2 + x\\ g(x) &= x^8 + x^6 + x^5 + x^4 + x^3 + x. \end{align} From \begin{align} f(x) &= x(x+1)(x^2+x+1)(x^2-x+1)\\ g(x) &= x(x+1)(x^2+x+1)(x^2-x+1)^2 \end{align} wee see that $f(x)$ and $g(x)$ have the same complex roots. Upon setting $x=s^{1/10}$ we see that \begin{align} \alpha_6,\dots,\alpha_1 &= 1/10,\; 2/10,\; 3/10,\; 4/10,\; 5/10,\; 6/10\\ \beta_6,\dots,\beta_1 &= 1/10,\; 3/10,\; 4/10,\; 5/10,\; 6/10,\; 8/10 \end{align} is a counterexample for $n=6$.

Added: More examples, with $n=4$, can be constructed using \begin{align} f(x) &= x(1+x+x^v+x^{v+1})\\ g(x) &= x(1+x^u+x^v+x^{u+v}) \end{align} if $v$ is odd and $1\lt u\lt v$ is a divisor of $u$. In this case, $f(x)$ and $g(x)$ again have the same complex roots.

Answer by Peter Mueller for Is there a nice criterion for when the splitting fields of two irreducible polynomials are equal?

2013-04-03T19:25:40Z

Without loss of generality we may assume that $f_1$ and $f_2$ are monic with integral coefficients.

Let $P_i$ be the set of primes $p$ such that the image of $f_i$ in $\mathbb F_p[x]$ factors into linear factors.

Then, as a consequence of Chebotarev's density theorem (actually the weaker Frobenius density theorem is good enough here) the splitting fields of $f_1$ and $f_2$ are the same if and only if the sets $P_1$ and $P_2$ differ only by finitely many elements.

Using effective versions of Chebotarev, one can make a finite (yet impracticable) criterion from that.

Answer by Peter Mueller for Simple field extension and rational points

2013-03-31T19:28:35Z

I believe the following is a negative example: Let $s,t,u,v$ be variables over $\mathbb F_p$. Set $F=\mathbb F_p(s,t,u,v)$ and $F'=F(\sigma,\tau)$ with $\sigma^p=s$, $\tau^p=t$.

Set $$f(X,Y,Z)=(X^p-sZ^p)u+(Y^p-tZ^p)v.$$

Then $f(\sigma,\tau,1)=0$.

We show that any solution of $f=0$ over $F'$ has this form up to a scalar factor: Let $x,y,z\in F'$ with $f(x,y,z)=0$. As $F'=\mathbb F_p(u,v,\sigma,\tau)$, we get $$x^p-sz^p, y^p-tz^p\in\mathbb F_p(u^p,v^p,\sigma^p,\tau^p)=\mathbb F_p(u^p,v^p,s,t),$$ hence $$A(u^p,v^p,s,t)u+B(u^p,v^p,s,t)v=0$$ for rational functions $A,B$ over $\mathbb F_p$ with $x^p-sz^p=A(u^p,v^p,s,t)$ and $y^p-tz^p=B(u^p,v^p,s,t)$.

This implies $A(u^p,v^p,s,t)=0$, for otherwise $u$ were a rational function in $u^p$. For the same reason $B(u^p,v^p,s,t)=0$. We get $x^p-tz^p=0=y^p-tz^p$, and the claim follows.

Answer by Peter Mueller for Fixed points of group action

2013-03-26T20:55:28Z

The answer is no for every prime $p$. Set $\alpha=\pi/p$. Then the image of $\begin{pmatrix}\cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha\end{pmatrix}$ in $PGL(2,\mathbb R)$ has order $p$, but no fixed points.

Answer by Peter Mueller for Hurwitz's construction of simple covers

2013-03-18T09:35:56Z

I think most people understand under this term the following: It is a branched cover $X\to\mathbb P^1(\mathbb C)$ of a compact connected Riemann surface $X$ to the Riemann sphere $\mathbb P^1(\mathbb C)$ such that the monodromy generators belonging to the branch points are transpositions. Or equivalently: If $n$ is the degree of the cover $f$, then each fiber $f^{-1}(x)$ for $x\in \mathbb P^1(\mathbb C)$ has at least $n-1$ distinct points.

Answer by Peter Mueller for All and the only algebraically closed fields s.t. any regular n-by-n matrix has a k-th root for every k

2013-03-17T19:45:29Z

wccanard's comment gives one direction: If the field has characteristic $p$, then there is no matrix $A$ with $A^p=\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}$ .

For the other direction, let the field have characteristic $0$. We want to show that $A^k=B$ has a solution $A$ for any $B$. Without loss of generality, $B$ is a Jordan block with eigenvalue $1$, so $B-1$ is nilpotent. Then $A=\sum_{i\ge0}\binom{1/k}{i}(B-1)^i$ is a finite sum with $A^k=1+(B-1)=B$.

Answer by Peter Mueller for Subgroups of the general linear groups

2013-03-16T18:26:11Z

No, not in general. For instance $GL_4(2)$ is isomorphic to the alternating group $A_8$, and the direct product of the two Klein four groups acting regularly on $1,2,3,4$ and $5,6,7,8$, respectively, is abelian of order $16\gt 2^4-1$.

Actually, for any prime $p$, there is an abelian subgroup of order $p^4$ in $GL_4(p)$, just take the matrices of the form $\begin{pmatrix} E & A\\ 0 & E\end{pmatrix}$ , where $E$ is the $2\times 2$ identity matrix, and $A$ an arbitrary $2\times 2$ matrix.

Generalizing to bigger $n$, there are more drastic counterexamples.

I would expect your result to be true if $M$ is a $p'$-group. At ant rate, Maschke + Schur easily show that $\lvert M\rvert\le p^n-1$ when $M$ is an abelian $p'$-group.

Answer by Peter Mueller for Bolza curve admits no anticonformal fixedpointfree involution

2013-03-07T22:56:54Z

Let $\mathbb C(x,y)$ be the function field of the curve $B$. From $y^2=x^5-x$ we obtain that each element in $\mathbb C(x,y)$ has the form $A(x)+yB(x)$ with rational functions $A$ and $B$. Identify the conformal automorphisms of $B$ with the $\mathbb C$-automorphisms of $\mathbb C(x,y)$.

Thus any conformal automorphism of $B$ has the form $(x,y)\mapsto (R(x)+yU(x),V(x)+yS(x))$ with rational functions $R,S,U,V$. Now it is well known that the hyperelliptic involution $(x,y)\mapsto (x,-y)$ of a hyperelliptic curve commutes with all conformal automorphisms. (The reason for that is that $\mathbb C(x)$ is the unique degree $2$ rational subfield of $\mathbb C(x,y)$.) Writing out what it means that the hyperelliptic involution commutes with conformal automorphisms yields $U=V=0$.

An anti-conformal automorphism is the composition of a map as above and complex conjugation, so it has the form $$\sigma: (x,y)\mapsto (R(\bar x),\bar yS(\bar x))$$ Assuming that $\sigma$ is an involution yields the necessary condition $$R(\bar R(x))=x,$$ where $\bar R$ arises from $R$ by complex conjugation of the coefficients.

In particular, $R$ has degree $1$, so it is a linear fractional map.

As $(R(x),yS(x))$ is on the curve when $(x,y)$ is on the curve, we get $$(x^5-x)S(x)^2=R(x)^5-R(x).$$

Looking at degrees and poles shows that either $R(x)=\gamma(x-\alpha)$ or $R(x)=\gamma/(x-\alpha)$ for some $\alpha$ with $\alpha^5=\alpha$ and $\gamma\in\mathbb C$.

In the first case, degrees tell us that $S(x)$ is a constant $\delta$. Upon replacing $x$ by $x+\alpha$ we get $$((x+\alpha)^5-(x+\alpha))\delta^2=\gamma^5 x^5-\gamma x.$$ Comparing at $x^4$ yields $\alpha=0$. So $R(x)=\gamma x$. But then $R(0)=S(0)=0$, and $(0,0)$ is a fixed point of $\sigma$.

In the second case we have $R(x)=\gamma/(x-\alpha)$. This forces $S(x)=\delta/(x-\alpha)^3$ for some $\delta$. Again, replacing $x$ with $x+\alpha$ and clearing denominators yields $\alpha=0$. We obtain $R(x)=\gamma/x$. From $R(\bar R(x))=x$ we get $\gamma/\bar\gamma=1$, so $\gamma$ is real.

Easy calculations yield $\gamma^4=1$, so $\gamma=\pm1$, and $S(x)=\delta/x^3$ with $\delta^2=-\gamma$. So $$ \sigma: (x,y)\mapsto (-\delta^2/\bar x,\delta\bar y/\bar x^3)$$ with $\delta^2=\pm1$.

We compute $$\sigma^2:(x,y)\mapsto (x,-\frac{\delta}{\bar\delta}y).$$

As $\sigma^2=1$, we get $\delta^2=-1$. But then $(1,0)$ is a fixed point of $\sigma$.

Answer by Peter Mueller for From reducible polynomial to an irreducible one

2013-03-07T17:24:01Z

As a more elementary variant of Joel's answer, you may also consider something like this: Let $f(X)\in\mathbb Z[X]$, and let $a_n$ be the leading coefficient. Then $f_p(X)=f(X)+\frac{1}{p}$ is irreducible for every prime $p$ not dividing $a_n$, because $pf_p(X)$ is the reciprocal of an Eisenstein polynomial.

So irreducible polynomials become arbitrarily close to the polynomial you start with.

Answer by Peter Mueller for Branched Regular Cover over 4-times punctured sphere

2013-03-05T14:44:31Z

The answer is no:

The monodromy group of the cover, as a permutation group on the fiber $f^{-1}(b)$ of a non-branch point $b$, is generated by four elements $s_1,s_2,s_3,s_4$ with $s_1s_2s_3s_4=1$. By Riemann's existence theorem, the question is equivalent to the following, with $n=g+1$:

Let $s_1,s_2,s_3,s_4$ be $n$-cycles in the symmetric group on $n$ letters with $s_1s_2s_3s_4=1$. Does this imply that the group generated by $s_1, s_2,s_3, s_4$ has order $n$?

This does not hold for $n=4$: Take for example $s_1=(1, 2, 3, 4)$, $s_2=(1, 3, 4, 2)$, $s_3=s_2^{-1}$, $s_4=s_1^{-1}$. Then $s_1s_2s_3s_4=1$, but $s_1s_2=(2, 4, 3)$.

This works for any $n\ge4$: Simply take $s_1$ an $n$-cycle and $s_2$ another $n$-cycle which is not a power of $s_1$, and $s_3$, $s_4$ the inverses of $s_2$ and $s_1$ as above.

(Note: I use the right action of permutations, so in $s_1s_2$, one first applies $s_1$, and then $s_2$ afterwards.)

Answer by Peter Mueller for Reducing system of polynomials with symbolic factors

2013-02-22T12:41:07Z

Added more details: From specializing $a,b,c,d$ to random integers it seems that your system is $0$-dimensional, and $z$ is the root of a degree $16$ polynomial $f_{a,b,c,d}(z)$. You might compute these degree $16$ polynomials for many integer tuples $a,b,c,d$, and try to guess/interpolate the dependencies of the coefficients of $f_{a,b,c,d}(z)$ on $a,b,c,d$. Once you know $f_{a,b,c,d}(z)$, the rest of the computation should be cheap.

If one fixes all but one of the parameters $a,b,c,d$, and lets the remaining one run through some possibilities, one can guess the degrees of the dependencies of the coefficients of $f_{a,b,c,d}$ in $a,b,c,d$, and then solve the corresponding interpolation problem. For instance, the coefficient of $z^{14}$ seems to be $-3a/4+d-13/4$. The highest degree dependency seems to occur for the coefficient of $z^4$, of degrees $5,3,3,4$ in $a,b,c,d$, respectively. So in order to compute this coefficient, one has to compute $6\cdot 4\cdot 4\cdot 5=480$ examples, and solve the corresponding linear system of equations in $480$ unknowns. Don't know if that is possible, but I would expect it can be done.

The Sage code for the computation of $f_{a,b,c,d}$ is:

while True:
    a,b,c,d=(floor(20*random()-10) for i in range(4))
    R.<u,v,w,x,y,z>=PolynomialRing(QQ,6,order='lex')
    l=[6-c*v*u+ ... SNIP!!! ... +4*y*z*u*v-b]
    i=ideal(l)
    g=i.groebner_basis()
    zz=g[5]
    print (a,b,c,d),zz/zz.coefficient({z:16})

produces

(9, 4, 9, 5) z^16 - 5*z^14 + 47/4*z^12 - 833/64*z^10 + 2263/256*z^8 + 3973/1024*z^6 - 27697/4096*z^4 + 5775/4096*z^2 + 625/4096
(-7, 7, 5, -1) z^16 + z^14 + 13/8*z^12 - 23/8*z^10 + 3011/256*z^8 - 1193/128*z^6 + 16581/2048*z^4 + 2303/4096*z^2 + 2401/65536
(2, -7, 3, 1) z^16 - 15/4*z^14 + 29/4*z^12 - 305/32*z^10 + 75/8*z^8 - 1703/256*z^6 + 453/128*z^4 - 1233/1024*z^2 + 81/256
(-2, -5, -10, 2) z^16 + 1/4*z^14 - 17/8*z^12 - 41/32*z^10 + 457/256*z^8 + 1133/1024*z^6 - 605/1024*z^4 - 21/64*z^2 + 1/16
(-10, -3, -9, 4) z^16 + 33/4*z^14 + 219/8*z^12 + 1023/64*z^10 + 4017/256*z^8 - 6919/1024*z^6 - 1301/1024*z^4 - 138985/16384*z^2 + 130321/65536
(-4, 5, -4, 7) z^16 + 27/4*z^14 + 61/4*z^12 + 79/64*z^10 - 1293/64*z^8 + 4333/1024*z^6 + 24929/2048*z^4 - 40071/4096*z^2 + 130321/65536
(-7, -6, -9, 0) z^16 + 2*z^14 + z^12 - 23/4*z^10 + 643/128*z^8 - 331/128*z^6 + 171/256*z^4 - 243/256*z^2 + 6561/65536
... snip ...

Answer by Peter Mueller for Expression for the sum of square roots of zeros of a polynomial

2013-02-27T10:46:30Z

I don't think that anything better than what you figured out already can be done. From the algebraic point of view, you cannot distinguish the $2^n$ siblings of $r$, corresponding to the different choices of the signs of the square roots. So you get a degree $2^n$ polynomial where $r$ is one of the roots. I believe it is not possible to take positivity into account.

Answer by Peter Mueller for Transformation of a bivariate polynomial into a homogeneous one

2013-02-27T10:33:09Z

Set $u=S(x,y)$, $v=T(x,y)$, and let $x=K(u,v)$, $y=L(u,v)$ be the inverse map with $H(u,v)=P(x,y)$ homogeneous in $u,v$. Over an algebraic closure of the (unspecified) base field, $H(u,v)$ splits into linear factors. Suppose that $v-\alpha u$ is such a factor. Then $0=P(K(u,\alpha u),L(u,\alpha u))$, so the algebraic curve given by $P(X,Y)=0$ has a component with a polynomial parameterization.

If the substitutions are allowed to be rational maps, then we analogously get that the curve has a component which admits a rational parameterization. More explicitly, let $Q(X,Y)$ be an irreducible divisor of $P(X,Y)$ such that $Q(K(u,\alpha u),L(u,\alpha u))=0$. This is equivalent to the following: The genus of the function field of the curve $Q(X,Y)=0$ is $0$. This can be checked by hand in small cases, or some computer algebra system.

However, even if the genus is $0$, it might still be difficult to find the parametrization.

Answer by Peter Mueller for On the group actions on Hurwitz surfaces

2013-02-17T10:15:05Z

I think that Klim wants to talk about proper normal subgroups $N$ of $G$. In that case, $N$ cannot contain an inertia generator: $G$ is generated by $a,b,c$ with relatively prime orders $2,3,7$ and $abc=1$. So if for instance $a\in N$, then modulo $N$ we have $bc=1$, and the order of $b$ divides $3$ and $7$. So $a,b,c\in N$, hence $G=N$.

Answer by Peter Mueller for Polynomials over $\mathbb F_2$ without zeros in $\mathbb F_2$ having an inverse series with support of large density.

2013-02-14T18:01:03Z

This is not an answer, rather a possible suggestion on how to deal with irreducible polynomials $A(x)$: Let $A(x)\in\mathbb F_2[x]$ be irreducible of degree $n$. Then \begin{equation} A(x)=\prod_{i=0}^{n-1}(1+\lambda^{2^i}x) \end{equation} for some $\lambda\in\mathbb F_{2^n}$. The partial fraction decomposition and geometric series yield \begin{equation} \frac{1}{A(x)}=\sum_{i=0}^{n-1}\frac{\alpha^{2^i}}{1+\lambda^{2^i}x} = \sum_{m=0}^\infty\sum_{i=0}^{n-1}\alpha^{2^i}(\lambda^{2^i}x)^m = \sum_{m=0}^\infty T(\alpha\lambda^m)x^m, \end{equation} where $\alpha=\lambda/f'(1/\lambda)$ and $T$ is the trace map from $\mathbb F_{2^n}$ to $\mathbb F_2$.

Note that the power series is periodic with period $e$, where $e$ is the multiplicative order of $\lambda$. Thus if $U$ is the subgroup of order $e$ of $\mathbb F_{2^n}^\star$, then the density of $1$'s is the number of $u\in U$ with $T(\alpha u)=1$ divided by $\lvert U\rvert$.

An easy case is when $e=2^n-1$, so $U=\mathbb F_{2^n}^\star$. Half of the elements of $\mathbb F_{2^n}$ have trace $0$, so the density of $1$'s is $2^{n-1}/(2^n-1)$.

So when not only $n$ is prime, but even $2^n-1$ is prime, then we have this case and the density is only slightly bigger than $1/2$.

The general case seems to be more challenging. It is always difficult to relate an additive function like the trace map with subgroups of the multiplicative group of fields.

Answer by Peter Mueller for Order of an element in a finite field

2013-02-13T17:04:00Z

Let $n$ be the degree of $f$, so the order is a divisor of $p^n-1$. As the computation of powers in $E$ is a cheap operation, I believe the following could be suitable: If $\alpha^m=1$, then check if there is a prime divisor $r$ of $m$ with $\alpha^{m/r}=1$. If not, then $m$ is the order, and if yes, then set $m=m/r$ and repeat the process.

Start the whole thing with $m=p^n-1$.

Probably there are better ideas, but I cannot think of one.

Answer by Peter Mueller for Decomposition of primes in Galois closures of number fields

2013-02-13T12:40:49Z

Changed answer: The answer is no: Let $L$ be a root field of $X^3-2$ over $\mathbb Q$, and $M$ the Galois closure of $L/\mathbb Q$. Then the primes $2$ and $3$ are both totally ramified in $L$, yet in both cases there is only one prime $P$ above $p=2$ or $p=3$, with $e(P)=3$ if $p=2$, and $e(P)=6$ if $p=3$.

Answer by Peter Mueller for Specialization of curves defined over function field

2013-02-13T12:46:20Z

Writing out the isomorphism and its inverse map, all one has to do is to care that the denominators of the finitely many coefficients from $K(t)$ which appear in these two maps don't vanish under the specialization.

Answer by Peter Mueller for Inequality with Euler's totient function

2013-02-07T17:46:26Z

Take $n=382315009082231724951830011$. Then $3^n-2$ is divisible by the primes $5, 19, 23, 47, 71, 97, 149, 167, 173, 263, 359, 383, 389, 461, 479, 503, 557$. Furthermore, $\varphi(3^n-2)<2/3\cdot(3^n-2)<2\cdot 3^{n-1}$.

The following Sage code verifies this examples. I believe that this is close to a minimal counterexample.

sage: n=382315009082231724951830011
sage: l=[5, 19, 23, 47, 71, 97, 149, 167, 173, 263, 359, 383, 389, 461, 479, 503, 557]
sage: set(3.powermod(n,p) for p in l)
set([2])
sage: prod(1-1/p for p in l).n()
0.666250824539016

As there are some speculations about how to find such an example, here is the (not really clever) Sage code which greedily collects the congruences for $n$ which do not contradict each other:

p,n,Q = 5,3,4
lp = [p]
s = 1-1/p
while True:
    p=p.next_prime()
    e=IntegerModRing(p)(3).multiplicative_order()
    l=[z for z in [1..e-1] if 3.powermod(z,p) == 2]
    if len(l) == 0:
        continue
    b=l[0]
    if (b-n) % e.gcd(Q) != 0:
        continue
    QQ=Q.lcm(e)
    n=CRT_list([n,b],[Q,e]) % QQ
    Q=QQ
    s*=(1-1/p)
    lp.append(p)
    print p,s.n()
    if 2/3>s:
        print n,lp
        break

Answer by Peter Mueller for Maximal order of a metacyclic transitive permutation group of degree $n$

2013-02-06T16:34:37Z

The following complements Derek's proof by showing the upper bound for element orders in the holomorph of $C_n$: Set $R=\mathbb Z/n\mathbb Z$. The holomorph of $C_n$ is $R\rtimes R^\times$. Pick $x=(a,b)\in R\rtimes R^\times$. Then $x^k=(a(1+b+\dots+b^{k-1}),b^k)$. Write $s_k=1+b+\dots+b^{k-1}$. Look at the $n$ terms $s_k$ for $k=1,\dots,n$. If it happens that two of them are equal, say $s_u=s_v$ for $1\le u\lt v\le n$, then $0=s_v-s_u=b^us_{v-u}$, and therefore $s_{v-u}=0$. If however these terms are pairwise distinct, then one of them must be $0$. So at any rate, there is $1\le k\le n$ with $s_k=0$. Furthermore, $0=(b-1)s_k=b^k-1$, so $x^k=1$.

Answer by Peter Mueller for Lower bound on the dimension of a subspace of $\mathbb Z_2^r$?

2013-02-05T12:34:30Z

Let $\pi_i:V\to\mathbb Z_2$ be the projection to the $i$-th coordinate. The $\pi_i$ are elements of the dual space $V^\star$, and by assumption, any $k$ of these $\pi_i$ are linearly independent. Let $C$ be subspace of $\mathbb Z_2^r$ consisting of those tuples $(a_1,\dots,a_r)$ such that $\sum a_i\pi_i=0$. So $C$ is a code of minimal Hamming distance $d\ge k$. On the other hand, $\dim C=r-\dim\langle\pi_i|i=1,2,\dots,r\rangle\ge r-\dim V^\star=r-\dim V$.

Thus $\dim V\ge r-\dim C$. Now use some good bounds from error correcting codes. If $k$ is bigger than $r/2$, then the Plotkin bound gives good results, here $2^{\dim C}\le k/(k-r/2)$, which is much better than $\dim V\ge k$.

If $k=r/2$, or $k\lt r/2$ then one can use the Griesmer bound, or other bounds if $k$ is small compared to $r$.

Answer by Peter Mueller for why are subextensions of Galois extensions also Galois?

2013-01-31T18:36:06Z

Edited (in view of new comments to this answer and the original question): I believe that Delgado missed the point that $M=Fix(Aut_M(L))$ isn't a formal consequence of $K=Fix(Aut_K(L))$ for algebraic extensions $K\subseteq M\subseteq L$. I took a closer look into the (master?) thesis. It doesn't claim to contain anything new, This work is a journey through the main ideas and sucessive [sic] generalizations of Galois Theory, towards the origins of Grothendieck’s theory of Dessins d’Enfants ... as the author puts it in his abstract.

The chapter on Galois theory just repeats well-known text book material, mostly without proofs. Considering the verbose character of this chapter, I'm sure the author would have said more than We immediately conclude that ... if there had been a novel aspect. To me it appears that he simply missed an essential aspect of Galois theory.

At any rate, from $K=Fix(Aut_K(L))$ alone we cannot conclude much, one somehow has to use the fact that $L/K$ is algebraic too as the following example shows: If $L=K(x)$ for a transcendental $x$, and if $K$ is infinite, then $K$ is the fixed field of $Aut_K(L)$, but for most rational functions $r(x)$ the extension $L/K(r(x))$ isn't Galois in either sense.

So if we want to show that $M=Fix(Aut_M(L))$ for an algebraic extension $L/K$ with $K=Fix(Aut_K(L))$, then I believe that one is automatically lead to the usual kind of arguments, which by the are also listed in this thesis.

Answer by Peter Mueller for When is PSU(2,q^2) = PSL(2,q) ?

2013-01-29T22:39:46Z

It is a fairly standard result that $SU(2,q^2)$ and $SL(2,q)$ are isomorphic, see e.g. II.8.8. in Huppert's Endliche Gruppen. I would expect that it is also in the third volume of The Classification of the Finite Simple Groups by Gorenstein-Lyons-Solomon, but I don't have the volume at hand right now.

Answer by Peter Mueller for Selecting $k$ integers from an interval $[0, N]$ to maximize the minimum difference between pairwise sums

2013-01-29T14:35:32Z

I think it could be interesting to look at the continuous analogue of the question, where you look for the reals in the interval $[0,1]$ instead of the integers in $[0,N]$, because multiplying by $N$ and rounding gives solutions for you original problem, provided that $N$ is big enough.

I made a Monte-Carlo test for small values of $k$. It seems to be the case that the optimal solutions arise from $k$-element Sidon-subsets of ${0,1,\dots,n}$ divided by $n$, with minimal $n$. I'm wondering if that always holds.

Answer by Peter Mueller for Metacyclic groups in $AGL(4,3)$

2013-01-27T12:31:59Z

There is no such group. Let $G$ be a transitive metacyclic subgroup of $\text{AGL}(4,3)$. Let $C$ be a cyclic normal subgroup of $G$ with $G/C$ cyclic.

As $G$ permutes transitively the orbits of $C$, the kernels of the action of $C$ on its orbits all have the same size, thus they are equal because $C$ is cyclic. But $C$ acts faithfully on the union of the orbits. We infer that $C$ acts regularly on each orbit. In particular, $\lvert C\rvert$ divides $3^4$.

Note that the Sylow $3$-subgroups of $G$ are transitive too, and subgroups of metacyclic groups are metacyclic too. So we may assume that $G$ is a $3$-group.

On the other hand, $9$ is the maximal order of a $3$-element in $\text{AGL}(4,3)$. (One can see that most easily from the embedding $\text{AGL}(4,3)\le\text{GL}(5,3)$.)

From that we see that $C$ has order $9$, and $G$ is a semidirect product of $C$ with another cyclic group $D$ of order $9$.

View $G=C\rtimes D$ as a subgroup of $\text{GL}(5,3)$. From Jordan's normal form theorem, we see that $\text{GL}(5,3)$ contains two conjugacy classes of elements of order $9$. Let $U$ be the group of upper triangular matrices of $\text{GL}(5,3)$. Consider the two cases of $C$ (corresponding to the Jordan block sizes $4+1$ and $5$, respectively). In both cases, one computes that the exponent of $N_U(C)/C$ is $3$, so there is no room for the cyclic group $D$ of order $9$.

Answer by Peter Mueller for Algebraic characterization of points constructible by compass and straightedge

2013-01-20T16:30:14Z

Your question is about showing that the normal hull of $K_n$ over $K_0$ has $2$-power degree, if $[K_i:K_{i-1}]=2$ for all $i$. But that follows be induction: Let $L$ be the normal hull of $K_{n-1}$ over $K_0$, so $[L:K_0]$ is a $2$-power.

The normal hull $N$ of $K_n$ over $K_0$ is the composite of the conjugates of $K_n$ over $K_0$. But all these conjugates are extensions of $L$ of degree $2$ (or $1$, if $K_n\subseteq L$), and from that the claim follows.

Answer by Peter Mueller for Automorphism of finite groups and Hurwitz spaces

2013-01-20T11:20:41Z

While the answers by Eric and ARupinski give negative examples for your question, here is the precise characterization for when the answer is yes: Let $\alpha$ be an automorphism of the transitive subgroup $G\le S_n$, and $G_1$ be the stabilizer of $1$ in $G$. Then $\alpha$ extends to an automorphism of $S_n$ if and only if $G_1$ and $\alpha(G_1)$ are conjugate in $S_n$.

The necessity of the condition is clear, and the sufficiency is a nice exercise. I believe the result is also in the permutation groups book by Dixon-Mortimer.

Answer by Peter Mueller for finite abelian p-groups with solvable automorphism group

2013-01-13T17:13:14Z

I believe that for $p\ge5$, these are the direct products of cyclic groups of pairwise distinct orders, for if $C_{p^e}\times C_{p^e}$ is a direct factor of $G$, then $GL(2,p)$ is a homomorphic image of a subgroup of $A=\text{Aut}(G)$.

Similarly for $p=2$ or $3$, I expect the groups $G$ you are looking for are those where no order in the direct product appears more than $2$ times.

In order to show that these groups indeed have a solvable automorphism group, you probably may apply induction: Let $N$ be the subgroup of $G$ generated by the $p$-th powers in $G$. Then $N$ has the same shape as above (with the order of each direct factor divided by $p$). So $\text{Aut}(N)$ is solvable by induction. As $C_A(N)$ is the kernel of the restriction of $A$ to $N$, all what remain to show is that $C_A(N)$ is solvable too.

Alternatively, one could try to use the induction hypothesis for $G/K$, where $K$ is generated by the elements of order $p$. Note that $N$ is (non-canonically) isomorphic to $G/K$.

Answer by Peter Mueller for Integral solutions to $a_1 \times a_2 \times ... \times a_k = N$

2013-01-10T18:50:29Z

Write $N=2^a3^b5^c7^d$. (If $N$ has not this shape, there are no solutions.) In a solution, let $m_i$ be the number of occurrences of the factor $i$. So $m_1+m_2+\dots+m_9=k$, $m_2+2m_4+m_6+3m_8=a$, $m_3+m_6+2m_9=b$, $m_5=c$, and $m_7=d$. The tuples $m_i$ can be computed in terms of $a$, $b$, $c$ and $d$. For each such tuple, the number of solutions equals the multinomial coefficient $\binom{k}{m_1,m_2,\dots,m_9}$ (as the $a_i$'s are not ordered). I doubt that a more precise answer or a closed formula can be obtained.

Comment by Peter Mueller

2013-05-17T14:47:46Z

If nobody here has an idea or reference, you might ask Robert Boltje (boltje.math.ucsc.edu), he did some work on various aspects of Brauer's induction theorem.

Comment by Peter Mueller

2013-05-14T16:05:12Z

@Randall: Isn't your question essentially the same one as your question mathoverflow.net/questions/24513/… from 3 years ago? How are your numbers $pxy$ given, by minimal polynomials or complex floats up to some precision?

Comment by Peter Mueller

2013-04-26T14:11:20Z

@Anixx: Maybe because it is not clear at all what you are suggesting. (I didn't downvote though.) It seems to me that your approach more or less only works if you can compute a closed expression for the higher iterates - which is trivial in your examples - but impossible in general.

Comment by Peter Mueller

2013-04-22T13:41:37Z

@bo.gu: I do not understand your question. Are you asking how I did find the counter-examples, or do you want to know why these polynomials yield counter-examples?

Comment by Peter Mueller

2013-04-19T17:14:27Z

Sorry, I fixed the typo.

Comment by Peter Mueller

2013-04-16T09:29:45Z

@Will: $S^4$ is meant to be the symmetric group on $4$ letters? Anyway, in this characteristic $2$ context without Riemann's existence theorem, it isn't clear to me that this cover exists. I would be more convinced if one would write it down explicitly for an elliptic curve in Weierstrass form.

Comment by Peter Mueller

2013-04-12T21:12:58Z

@Will Sawin: See en.wikipedia.org/wiki/…

Comment by Peter Mueller

2013-04-04T14:52:11Z

This site is not for homework help, see mathoverflow.net/faq

Comment by Peter Mueller

2013-04-04T12:52:03Z

@James D. Taylor: I changed my answer, so your comment probably does not apply anymore.

Comment by Peter Mueller

2013-04-03T22:03:18Z

How could it help here? The base fields are ^number fields^. Even over $p$-adic fields, the Krasner lemma only gives a sufficient, but not a necessary criterion for equality of fields.

Comment by Peter Mueller

2013-04-03T19:11:06Z

I think it is not quite that simple. The factors $i/N$ on the right hand side of the recursion shows that the identity for the generating polynomial $F$ involves the partial derivative with respect to $x$. Still, this generating function approach seems to be natural thing to try.

Comment by Peter Mueller

2013-04-03T13:24:55Z

Look at the list of books given in en.wikipedia.org/wiki/Inverse_Galois_theory In addition to that, also the book Field Arithmetic by Fried and Jarden contains quite a bit of (infinite) Galois theory. Furthermore, Algebraic Patching by Jarden is mostly about a relatively recent technique in Galois theory.

Comment by Peter Mueller

2013-04-01T08:12:30Z

I extended the answer, hope it is clearer now.

Comment by Peter Mueller

2013-03-30T22:47:07Z

As stated the claim is wrong, and certainly not appropriate for this site. See the mathoverflow.net/faq.

Comment by Peter Mueller

2013-03-29T20:19:55Z

@Aaron: Of course, there are many obvious examples like this: If $x-x_0$ is to be a factor of $f_{a,b}(x)=\sum_{k=0}^n(a+bk)x^k$, then there is a linear equation for $a,b$ from $f_{a,b}(x_0)=0$.