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I have been using the following result:

Given a polynomial $f(x,t)$ of degree $n$ in $\mathbb{Q}[x,t]$, if a rational specialization of $t$ results in a separable polynomial $g(x)$ of the same degree, then the Galois group of $g$ over $\mathbb{Q}$ is a subgroup of that of $f$ over $\mathbb{Q}(t)$.

However, I have been unable to prove this for myself, and cannot seem to find a proof of it anywhere. Is there an elementary proof? And if not, can anyone direct me to a source containing one, or at least explain the general principle?

My need to understand the result arose from considering the following:

If I specialize $t$ such that $g$ factorizes as $x^k.h(x)$, where $h$ is an irreducible polynomial of degree $n-k$, is it legitimate to surmise that the Galois group of $h$ over $\mathbb{Q}$ is a subgroup of the original?

If nothing else, I'd be very grateful for an answer to this!

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2 Answers 2

up vote 16 down vote accepted

Here is a broader setup for your question. Let $A$ be a Dedekind domain with fraction field $F$, $E/F$ be a finite Galois extension, and $B$ be the integral closure of $A$ in $E$. Pick a prime $\mathfrak p$ in $A$ and a prime $\mathfrak P$ in $B$ lying over $\mathfrak p$. The decomposition group $D(\mathfrak P|\mathfrak p)$ naturally maps by reduction mod $\mathfrak P$ to the automorphism group $\text{Aut}((B/\mathfrak P)/(A/\mathfrak p))$ and Frobenius showed this is surjective. The kernel is the inertia group, so if $\mathfrak p$ is unramified in $B$ then we get an isomorphism from $D(\mathfrak P|\mathfrak p)$ to $\text{Aut}((B/\mathfrak P)/(A/\mathfrak p))$, whose inverse is an embedding of the automorphism group of the residue field extension into $\text{Gal}(E/F)$.

If we take $A = {\mathbf Z}$ then we're in the number field situation and this is where Frobenius elements in Galois groups come from.

In your case you want to take $A = {\mathbf Q}[t]$, so $F = {\mathbf Q}(t)$. You did not give any assumptions about $f(x,t)$ as a polynomial in ${\mathbf Q}[x,t]$. (Stylistic quibble: I think it is better to write the polynomial as $f(t,x)$, specializing the first variable, but I'll use your notation.) Let's assume $f(x,t)$ is absolutely irreducible, so the ring $A' = {\mathbf Q}[x,t]/(f)$ is integrally closed. [EDIT: I should have included the assumption that $f$ is smooth, as otherwise $A'$ will not be integrally closed, but this "global" int. closed business is actually not so important. See comments below.] Write $F'$ for the fraction field of $A'$. After a linear change of variables we can assume $f(x,t)$ has a constant coefficient for the highest power of $x$, so $A'$ is the integral closure of $A$ in $F'$.

Saying for some rational $t_0$ that the specialization $g(x) = f(x,t_0)$ is separable in ${\mathbf Q}[x] = (A/(t-t_0))[x]$ implies the prime $(t-t_0)$ is unramified in $A'$. Let $E$ be the Galois closure of $F'/F$ and $B$ be the integral closure of $A$ in $E$. A prime ideal that is unramified in a finite extension is unramified in the Galois closure, so $(t-t_0)$ is unramified in $B$. For any prime $\mathfrak P$ in $B$ that lies over $(t-t_0)$, the residue field $B/\mathfrak P$ is a finite extension of $A/(t-t_0) = \mathbf Q$ and since $E/F$ is Galois the field $B/\mathfrak P$ is normal over $A/(t-t_0)$. These residue fields have characteristic 0, so they're separable: $B/\mathfrak P$ is a finite Galois extension of $\mathbf Q$. I leave it to you to check that $B/\mathfrak P$ is the Galois closure of $g(x) = f(t_0,x)$ over $\mathbf Q$. Then the isomorphism of $D(\mathfrak P|(t-t_0))$ with $\text{Aut}((B/\mathfrak P)/\mathbf Q) = \text{Gal}((B/\mathfrak P)/\mathbf Q)$ provides (by looking at the inverse map) an embedding of the Galois group of $g$ over $\mathbf Q$ into the Galois group of $f(x,t)$ over $F = {\mathbf Q}(t)$.

I agree with Damiano that there are problems when the specialization is not separable. In that case what happens is that the Galois group of the residue field extension is identified not with the decomposition group (a subgroup of the Galois group of $E/F$) but with the quotient group $D/I$ where $I = I(\mathfrak P|\mathfrak p)$ is the inertia group, and you don't generally expect a proper quotient group of a subgroup to naturally embed into the original group.

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Very nice answer. Blurbworthy, perhaps? –  Pete L. Clark Aug 4 '10 at 21:31
Pete: I did leave it to Adam to check $B/\mathfrak P$ is the Galois closure of $g(x)$, so I will not be posting this elsewhere with all details. –  KConrad Aug 4 '10 at 22:04
The SGA1 version: For infinite field $L$ of any char. (e.g., $\mathbf{Q}$) & $f \in L[t,x]$ with $x$-deg $n > 0$ & $f$ sep'ble over $L(t)$, let $c, d \in L[t]$ resp. be lead $x$-coeff. and $x$-discr, $A := L[t][1/cd]$. Let $K/L(t)$ be finite Galois ext'n, Gal. gp $G$, splitting $f$, so splits $L(t)[x]/(f)$. Since $A[x]/(f)$ is finite etale $A$-alg, int. closure $B$ in $K$ is finite etale $G$-torsor over $A$ splitting $f$. Specialize at $L$-pt $t_0$ of $A$ (!) to get finite etale $G$-torsor $B_0$ over $L$ splitting $f(t_0,x)$. A factor field of $B_0$ is Galois over $L$ with Gal. gp in $G$. QED –  BCnrd Aug 5 '10 at 2:19
Dear Keith, Are you sure that $A'$ is integrally closed in general? What if $f(x,t)$ were $x^3 - t^2$? The rest is okay though, since, as Brian notes in this comment, $A'$ will be integrally closed in a n.h. of a separable specialization $t_0$, since it will be etale over $A$ there. –  Emerton Aug 5 '10 at 5:16
Matt: That was an accidental omission. I meant to assume that $f(x,t)$ is smooth so that $A'$ is int. closed. But since the question being asked is local on the choice of $t_0$, such a global hypothesis is not really necessary and the soln. I wrote doesn't even require knowing $A'$ concretely: let $F'$ be $F$ adjoined with one $x$-root of $f(x,t)$ and then let $A'$ be int. closure of $A$ in $F'$. The soln. barely uses $A'$ or $F'$ (just to know unram. prime stays unram. in Galois closure) and it'd be better to have defined $A$ after picking $t_0$ as the localn. of $\mathbf Q[t]$ at $t-t_0$. –  KConrad Aug 5 '10 at 7:30

You can find the first statement, for instance, in van der Waerden, Modern Algebra I, Section 61. In particular, under the inclusion of the Galois group of the reduction in the Galois group of the original polynomial, the cycle structure matches up.

For the statement when you drop the separability assertion, the answer is no. Here is an example in which the cycle structure of the two groups are incompatible. While this does not completely settle your second question, it might convince you that the answer is "no".

The polynomial $f=x^4+x^2+9$ is irreducible over $\mathbb{Q}$ and has discriminant equal to $420^2$. Therefore the Galois group of $f$ is contained in the alternating group of order 4 and contains no transposition. The reduction of the polynomial modulo 3 is $x^2(x^2+1)$ and therefore has Galois group that is a transposition. Even though the Galois group of $f$ contains elements of order two, the cycle structures do not match up.

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PARI says 3 factors in the quartic field as two primes with residue field degree 2, and that does correspond to a cycle structure in the Galois group (two disjoint 2-cycles). There is an example of a degree 6 number field generated by the root of some polynomial $f(x)$ such that neither the factorization type of $f(x) \bmod 3$ nor the way (3) factors in the number field correspond to the cycle structure of a permutation in the Galois group. See my answer to another question of Adam's at mathoverflow.net/questions/21247/… –  KConrad Aug 5 '10 at 1:43
Thanks for the link: your answers both here and there are very interesting! –  damiano Aug 5 '10 at 9:05
Thankyou both for your help! I hadn't even realized that this was basically the same result as the one I was asking about in my previous question, but using the ideal $(t-t_0)$ as opposed to $(p)$. Just to be sure, am I right in thinking that if in the last part of my question $k=1$ the result definitely holds? –  Adam Aug 5 '10 at 16:14
Yes, but make sure that you check that $h(x)$ is not $x$! (This is a silly comment, but it is not implied by what you wrote that $x$ and $h(x)$ are coprime.) –  damiano Aug 5 '10 at 16:25

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