bio | website | mathematik.uni-wuerzburg.de/… |
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location | Würzburg | |
age | 49 | |
visits | member for | 3 years, 5 months |
seen | 6 hours ago | |
stats | profile views | 2,170 |
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 $A$. |