7,737 reputation
12639
bio website mathematik.uni-wuerzburg.de/…
location Würzburg
age 49
visits member for 3 years, 5 months
seen 6 hours ago
I'm a professor of mathematics at the University of Würzburg (Germany). For further infos see my web page.

Mar
16
answered Invariant subspaces of an $F_2$-representation of the affine linear group of dimension 1
Mar
10
answered Permutation polynomials mod $p$ of the form $(x+1)^n-x^n$
Feb
28
comment If d(“G/H”) < d(G) = 2, must H contain a primitive element?
@Pablo: Indeed, my comment wasn't correct. Now that Derek Holt provided the details, there is no need to fix it.
Feb
26
awarded  Good Answer
Feb
26
revised Circles and rational functions
added 248 characters in body
Feb
26
revised Circles and rational functions
improved the answer, provided a link to a preprint with more details
Dec
27
answered Measuring how closely a missing projective plane can be approached by an equivalent structure
Dec
17
awarded  Enlightened
Dec
17
awarded  Nice Answer
Dec
4
comment A generalized mean-value 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 mean-value 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 $A$.