Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
2  
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. –  Felipe Voloch Jan 2 '13 at 2:41
    
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. –  P Vanchinathan Jan 2 '13 at 4:46
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.