By Hugo Steinhaus
Problems-with instructive solutions-on numbers, equations, polygons, polyhedra, and plenty of different themes. Very demanding. extra thirteen difficulties with out strategies.
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Additional resources for One Hundred Problems in Elementary Mathematics
318]. G/ on the set Œ1; : : : ; m. In particular, the Galois group of an irreducible factor of Rf;T can be determined by a purely group-theoretic computation. Our linear resolvent is constructed as follows. x i D1 j D2 ri rj /; A Linear Resolvent for Degree 14 Polynomials 49 where ri are the roots of the degree 14 polynomial f . However, since T is linear, it can also be computed as a resultant (as in ). x/ D g. x/. The list of the irreducible factors of F91 is enough to distinguish 4 of the 9 remaining Galois groups (14T4, 14T7, 14T12, and 14T23), as seen in Table 1.
Strosnider 3 Stem Field Invariants As before, let f be a degree 14 polynomial defined over Q7 , and let G be its Galois group. Our aim in this section is to introduce three field-theoretic invariants, related to the stem field of f , that will aid in our computation of G. First, we consider the stem field of f and its corresponding subgroup H (under the Galois correspondence). Thus H is isomorphic to G \ S13 , the point stabilizer of 1 in G. H / represents the normalizer of H in G), which is in turn isomorphic to the centralizer of G in S14 .
Theorem 1. Rf;T /. G/, where is the natural group homomorphism from S5 to Sm given by the natural right action of S5 on S5 =H . Note that we can always ensure Rf;T is squarefree by taking a suitable Tschirnhaus transformation of f [9, p. 324]. G/ on the set Œ1; : : : ; m. In particular, Rf;T has a root in F if and only if G is conjugate under S5 to a subgroup of H . 3 Example: Discriminant Perhaps the most well-known example of a resolvent polynomial is the discriminant. ˛i ˛j /2 ; 1Äi