I do not know how to correctly interpret Hilbert's Irreducibility theorem with Galois group as my aim.

Here $K$ is a number field (or simply $\mathbf{Q}$).

Scenario 1: Take a field $L$ that is a finite Galois extension of $K(t)$ ($t$ an indeterminate) with Galois group $G$. Writing $L=K(t)[X]/(f(t,X))$ for an irreducible polynomial $f(t,X)\in K(t)[X]$, and taking a specialization $t=a\in K$ guaranteed by Hilbert we can see the Galois group descends and we get a $G$-Galois extension over the number field $K$ as $K[X]/(f(a,X))$.

I understand this situation well.

Scenario 2: Instead of a $G$-Galois extension we are merely provided with an irreducible polynomial whose SPLITTING FIELD has $G$ as Galois group, so the {\it degree of the polynomial can be less than the order of $G$.}

I took the following example from Malle and Matzat's book on Inverse Galois Theory. (Page 88, attributed to Beckman). (Instead of a general degree $n$ I take $n=3)$.

He claims $f(t,X) = X^3-3tX +2t \in \mathbf{Q}(t)[X]$ is irreducible with $S_3$ as Galois group. (of its splitting field).

For the special value $t=4$ we get the irreducible polynomial $X^3-12X+8$, but the discriminant is a square (of 72) and we get a cubic number field as splitting field and not the expected $S_3$ extension of $\mathbf Q$.

What mistake am I making in this scenario?

Instead of giving a degree $n$-polynomial in $K(t)[X]$ with $S_n$ as Galois group I would be more comfortable with an irreducible polynomial of degree $n!$ with $S_n$ as Galois group so that I can specialize that polynomial. Perhaps it is expecting too much.

  • 2
    $\begingroup$ I don't think you are making any mistake. Hilbert irreducibility implies that, for your cubic $f(t,X)$, $f(a,X)$ is irreducible for most values of $a$ and that the Galois group of the splitting field of $f(a,X)$ is $S_3$ for most values of $a$, but the set of $a$ in the first statement is not the same (as you have discovered with $a=4$) as in the second statement. $\endgroup$ – Felipe Voloch Jan 2 '13 at 2:41
  • $\begingroup$ 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. $\endgroup$ – P Vanchinathan Jan 2 '13 at 4:46

In ``Scenario 2'', you have to take a minimal polynomial $g(t,X)$ of a primitive element of the splitting field of $f(t,X)$ over $K(t)$. Then $g(a,X)$ is irreducible for infinitely many $a\in K$ by Hilbert's irreducibility theorem, and $g(a,X)$ and $g(t,X)$ have the same Galois group over $K$ and $K(t)$, respectively.

So preservation of Galois groups follows indeed from Hilbert's irreducibility theorem, but one has to apply it to the correct polynomial.

As to the last paragraph of the question: I doubt that there is a simply shaped polynomial of degree $n!$ with Galois group $S_n$ over the rationals. But maybe this one is good enough for the OP's purpose: $X^n-X-1$ has Galois group $S_n$ over $\mathbb Q$, see page 42 in Serre's Topics in Galois Theory. The proof isn't easy, even seeing that this polynomial is irreducible (as proved by Selmer in the 50s) requires a trick.


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