bio  website  mathematik.uniwuerzburg.de/… 

location  Würzburg  
age  48  
visits  member for  3 years, 1 month 
seen  14 mins ago  
stats  profile views  2,097 
I'm a professor of mathematics at the University of Würzburg (Germany). For further infos see my web page.
2d

awarded  Enlightened 
2d

awarded  Nice Answer 
Dec 4 
comment 
A generalized meanvalue theorem
By stated result I meant the OP's $3$point version, not the one from your answer which is a different formulation of the Schwarz version. 
Dec 3 
answered  A generalized meanvalue theorem 
Dec 1 
comment 
Algebraic dependency over $\mathbb{F}_{2}$
@David: If $h(g(x))$ is a polynomial, where $g$ is a polynomial and $h$ is a rational function, then $h$ is actually a polynomial. (Write $h(z)=p(z)/q(z)$, then $1=u(z)p(z)+v(z)q(z)$ by Bezout, so $1=u(g(x))p(g(x))+v(g(x))q(g(x))$, thus the numerator and denominator of $p(g(x))/q(g(x))$ are relatively prime, so $q$ is a constant.) 
Dec 1 
comment 
Algebraic dependency over $\mathbb{F}_{2}$
... continued: The proof of the background result in Schinzel is quite elementary, like the direct field theoretic proofs of Lüroth's Theorem. 
Dec 1 
comment 
Algebraic dependency over $\mathbb{F}_{2}$
This proof can be elementarized quite a bit: If $f_1$ and $f_2$ are algebraically dependent, then $\mathbb F_2(f_1,f_2)\subseteq\mathbb F_2(x_1,x_2)$ has transcendence degree $1$ over $\mathbb F_2$. By Theorem 4, Chapter 1 in Schinzel's book `Polynomials with special regard to reducibility', there is a polynomial $g\in\mathbb F_2[x_1,x_2]$ with $\mathbb F_2(f_1,f_2)=\mathbb F_2(g)$. So $f_i=h_i(g(x_1,x_2))$ for univariate polynomials $h_i$. In particular, the pairs $(f_1(a),f_2(a))$ assume at most $2$ different values, while the assumption requires $4$ values. 
Nov 28 
comment 
Generalization of a property of $A_n; n\geq 5$
I've enhanced the answer. 
Nov 28 
revised 
Generalization of a property of $A_n; n\geq 5$
added 839 characters in body 
Nov 27 
comment 
Generalization of a property of $A_n; n\geq 5$
@Derek Holt: Why does it hold for $2$transitive actions? A $2$set stabilizer in a $2$transitive simple group need not be a maximal subgroup. 
Nov 27 
answered  Generalization of a property of $A_n; n\geq 5$ 
Nov 19 
answered  not Gauss sum with the same magnitude 
Nov 8 
comment 
Restriction of scalars for the adjoint representation of $SL_2(\mathbb F_q)$
@Venkataramana: So do I. The matrices $A$ are clearly elements from $V$, and as I said in the proof, an orbit of $G$ through $A$ is the conjugacy class $A^G$ ($=\{g^{1}Ag\;\;g\in G\}$). If $A$ lies in a submodule $W$, then $A^G\subseteq W$, and then also the $\mathbb F_p$ span of $A^G$ is contained in $W$. But the latter is $V$ for the given matrices $A$. 
Nov 7 
revised 
Restriction of scalars for the adjoint representation of $SL_2(\mathbb F_q)$
added 104 characters in body 
Nov 7 
answered  Restriction of scalars for the adjoint representation of $SL_2(\mathbb F_q)$ 
Oct 31 
awarded  Good Answer 
Oct 31 
answered  Sylowsubgroups of the group of units of a finite field 
Oct 28 
comment 
Is the set of certain polynomials finite or infinite?
I'm not sure which problem @user64494 has with the solution, 1. and 2. clearly yield his claim. MO is a site for math on research level, so the answers should be like that too. The only thing one might add to the proof is why $p_n$ has integral coefficients. That follows from the theorem about symmetric polynomials, together with the fact that $2\cos(n x)$ is an integral polynomial in $2\cos(x)$. 
Oct 26 
reviewed  Approve Composite families of formal power series over $\mathbb C$ as algebraic variety 
Oct 24 
awarded  Nice Answer 