Gen Poly has Sn group Hadlock

[Ray]

Can anyone explain the idea behind Hadlock's proof that there is an Sn for every poly of degree n? Theorem 37 page 217
I can follow how to build up G from F using symmetric functions and the primitive element theorem. A lso I get the idea of constructing a poly of deg n! from one of deg n. But he starts with rationals beta1 etc to make G irreducible and I don't see the connection back down to F.

 

[Greg Bernhardt]

I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

 

[mathwonk]

Your question is a little vague to me, especially since I do not have the book, and Amazon does not allow searching Theorem 37 on page 217. They do show p. 216, lemma 37f, where Hadlock proves irreducibility of F however. What is it about F that you want to know?

Your question is also a bit muddled. Hadlock is apparently proving that for every positive integer n, there is a polynomial of degree n, which is irreducible over Q and has Galois group S(n).

The idea is apparently to find such a polynomial with variable coefficients and then specialize the coefficients carefully so the result remains irreducible.

 

[Ray]

Hope I can make my question clearer: Why does the first line of Hadlock's proof begin "By Hilbert's irreducibility theorem..."?

ie Hadlock starts with rationals beta1 etc to make G irreducible and I don't see the connection back down to F.

 

[blue_raver22]

Hadlock's proof is based on the concept of Galois groups, which are groups of automorphisms that preserve the field structure of a given field extension. In this case, the field extension is from the field of rational numbers (F) to the field of symmetric functions (G).

The idea behind Hadlock's proof is to show that for any given polynomial of degree n over F, there exists a corresponding element in G that has a Galois group isomorphic to the symmetric group Sn. This means that the Galois group of G over F is Sn, and thus there is an Sn for every poly of degree n.

To prove this, Hadlock first constructs G from F using symmetric functions and the primitive element theorem. This means that every element in G can be expressed as a polynomial of degree n! over F.

Next, Hadlock introduces the concept of irreducible elements in G. An element in G is irreducible if it cannot be expressed as a product of two non-constant elements in G. Hadlock then shows that by choosing certain rational numbers beta1, beta2, ..., beta(n-1), we can construct an irreducible element in G. This is important because it allows us to create a Galois extension from F to G, with G being an irreducible polynomial over F.

Finally, Hadlock shows that this Galois extension has a Galois group isomorphic to Sn. This is done by showing that the Galois group of G over F is generated by the automorphisms that permute the roots of the irreducible polynomial. Since this polynomial has n! roots, the Galois group must be isomorphic to Sn.

In summary, Hadlock's proof uses the concept of Galois groups to show that for any polynomial of degree n over F, there exists a corresponding element in G with a Galois group isomorphic to Sn. This proves that there is an Sn for every poly of degree n, as stated in Theorem 37 on page 217.