P Vanchinathan
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Registered User
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Jun 3 |
answered | General bound for the number of subgroups of a finite group |
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May 20 |
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Permutations of $(Z/pZ)^*$ What is the meaning of $a(i)-a(j)$ in the second paragraph of your question? Does it mean the function whose value at $k$ is $a(i)(k)-a(j)(k)$, using the subtraction in the ring $\mathbf{Z}/p\mathbf{Z}$. In that case,your condition (A) forces this to be a permutation. Then $a(i)$ and $a(j)$ are permutations agreeing nowhere. Is the map $a$ a group homomorphism? Then $a$ specifies a free action of $\mathbf{Z}/p\mathbf{Z}^*$ on itself. |
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May 19 |
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Existence of nontrivial roots of a homogeneous polynomial over a finite field in extension fields My answer given yesteray is wrong. One can choose $p$ variables to have non-zero value, and set all others to be zero to get a non-trivial solution. Should I delete it? |
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May 18 |
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Existence of nontrivial roots of a homogeneous polynomial over a finite field in extension fields The above is wrong. One can choose $p$ variables to have non-zero value, and set all others to be zero to get a non-trivial solution. I am sorry to have rushed in without fully verifying. |
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May 17 |
answered | Existence of nontrivial roots of a homogeneous polynomial over a finite field in extension fields |
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May 15 |
awarded | ● Necromancer |
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May 15 |
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Possible ratios of Pythagorean fractions Am I misreading the question? Do you want to know if $4/9$ is a RATIO of two Pythogorean fractions, as opposed to being a Pythagorean fraction by itself? The wording of your question means an expression for $4/9$ as $\frac{p_1q_1/(p_1^2-q_1^2)}{p_2q_2/(p_2^2-q_2^2)}$ is what is desired (using Barry Cipra's suggestion). |
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May 15 |
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polynomial zero within a square I see the snag. Can't we, with hindsight, specify values to be even higher at 1, $1+i$ so that scaling constant can't bring down. Of course this depends on the least absolute value of the zero. I'll try to rework again. Thanks, Johan! |
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May 14 |
answered | polynomial zero within a square |
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May 14 |
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Simplifying an algebraic integer expression Perhaps you have more info that could reduce the degree. For example, the fields generated individually by $p_{41}$ and $p_{21}$ could intersect in a field of degree 16. and similar things.Also possibly a a convenient automorphism (a Galois group element) might send your number $p_4$ to a number more amenable for computations. |
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May 3 |
answered | Does there exist a polar decomposition of matrices over finite fields? |
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May 1 |
answered | Generate a higher degree symmetric polynomial from an existing one |
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Apr 29 |
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The largest number of irreducible characters of the same degree in a finite group Have no idea how to solve this problem. But do remember that M. Isaacs (author of a book on Character Theory) has written papers that have a lot to do with the set of character degrees. |
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Apr 14 |
awarded | ● Yearling |
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Apr 11 |
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Irreducible Degrees and the Order of a Finite Group Thanks, John for a clear writing. |
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Apr 11 |
answered | Combined standard deviation |
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Apr 11 |
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Is the Segre embedding projectively normal? More along this line: The cone over the Grassmannian for the Pl\"ucker embedding is even a factorial variety. (And quadratic relations among Pl\'ucker co-ordinates holds.) Book by Lakshmibai on 'Standard Monomial Theory' has more generalisations. Popov's generalisation: same is true for orbits of highest weight vectors in an irreducible representation of simply-connected semisimple algebraic group $G$ (over C?). |
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Apr 9 |
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Irreducible Degrees and the Order of a Finite Group Correction to my earlier comment: I meant real two-dimensional representation for odd cyclic groups. |
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Apr 9 |
awarded | ● Nice Question |
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Apr 8 |
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Irreducible Degrees and the Order of a Finite Group @David Speyer: Thanks for the detailed answer here and at SE. I am teaching rep theory without using group algebras and fumbling. In Lagrange's theorem we see cosets are of same cardinality and provide a partition of $G$. About degree of intermediate fields in finite extensions also we have a transparent proof. I am looking for such a simple underlying idea. As odd ordered cyclic groups have 2-dimensional irreps as symmetries of regular polygons we need to bring the dependence on complex numbers (via Schur's lemma). Not that easy perhaps. :-( |
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Apr 8 |
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A question from Otto Forster’s book on Riemann surfaces (TeX comment) Insert a missing dollar sign to make the question readable. |
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Apr 8 |
asked | Irreducible Degrees and the Order of a Finite Group |
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Apr 1 |
answered | Parabolic-type subgroups of GL(V) |
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Mar 26 |
answered | Decomposition of solvable Lie group |
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Mar 21 |
answered | 3D Rotation Representation for Multiple Turns |
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Mar 15 |
answered | Subfield of rational function field and which is not a rational function field |
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Mar 9 |
awarded | ● Critic |
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Jan 20 |
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Minimum sum among fixed length factors of a number The equality connecting arithmetic mean and geometric mean should be relevant here. |
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Jan 19 |
awarded | ● Enthusiast |
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Jan 12 |
answered | The inverse Galois problem, what is it good for? |
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Jan 10 |
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Quotients of rational surfaces For $k=\mathbf{Q}$ work of Saltman,(retract rational) on Noether's problem could be relevant, and Swan, Lenstra for abelian groups. But all of them talk of higher dimensional varieties. Specifically cyclically permuting $n$ variables over $\mathbf{Q}$ does not give a fixed field that is purely transcendental for $n=47$ (Swan) and $n=8$ (Lenstra). |
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Jan 6 |
answered | Does every polynomial diophantine equation have solutions modulo p? |
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Jan 3 |
awarded | ● Commentator |
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Jan 3 |
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Primitive Elements for $S_n$ Galois Extensions? Thanks to Elkies, and Swain. Your answers are clear and explicit and readily usable to construct. I have now my work cut out getting that polynomial of degree 24 giving $S_4$ as Galois extension. |
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Jan 3 |
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Primitive Elements for $S_n$ Galois Extensions? Thanks Qiaochu: the opening paragraph was missed out during editing and I have restored it now. |
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Jan 3 |
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Primitive Elements for $S_n$ Galois Extensions? Inserted part missed out in cut and paste |
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Jan 3 |
asked | Primitive Elements for $S_n$ Galois Extensions? |
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Jan 3 |
awarded | ● Scholar |
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Jan 2 |
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How to apply Hilbert’s Irreducibilty theorem? Successful specialisation is ok. When a specialisation fails what exactly is the meaning? The polynomial fails to be irreducible? Or it remains irreducible but with different Galois group for the splitting field? I was under the impression that if a specilisation is irreducible then Galois group is the same. |
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Jan 2 |
asked | How to apply Hilbert’s Irreducibilty theorem? |
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Dec 28 |
answered | Is there a “geometric” intuition underlying the notion of normal varieties? |
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Dec 27 |
answered | parabolic subgroup |
