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Hilbert’s Irreducibility Theorem

June 8, 2018


A pretty neat paper about a pretty neat theorem

[ GLL edited ]

Mark Villarino, William Gasarch, and Kenneth Regan are terrific writers. Bill is a co-author of a famous blog located here. Ken is a co-author of another blog that we will not mention.

Today I would like to talk about one of their recent articles that just appeared in the Math Monthly.

The Math Monthly is one of the premier journals for exposition of math results. It is a great honor for them to have published a paper there and I thought it would be nice to write about it. Congrats to Mark, Bill, and Ken.

Their paper is of course about the famous Hilbert irreducibility theorem (HIT). The paper states the theorem and proves it from first principles. But it does much more. They also explain two important ideas that are seen through mathematics:

  1. Theorems are often not an end in themselves. They are proved to solve some problem.

  2. Theorems are often an end in themselves. They are proved for they own stake.

This sounds like a paradox, but it is not. Hilbert had a particular goal in mind in proving HIT. He wanted to show the following:

Theorem 1 For every integer {n\geq 1} there exist infinitely many polynomials {p(x)} in {\mathbb{Z}[x]} of degree {n} such that {p(x)} has the symmetric group {S_n} as its Galois group.

He used HIT to prove exactly this. This theorem is a special case of the Inverse Galois Problem:

Does every finite group appear as the Galois group of some Galois extension of the rational numbers?

This problem first stated two centuries ago—in the 19th century—is still open. For example many finite simple groups can be constructed as extensions of the rationals. All solvable groups can also be a Galois group, but the general case is still open.

The Theorem

Quoting them:

In number theory, Hilbert’s irreducibility theorem, conceived by David Hilbert, states that every finite number of irreducible polynomials in a finite number of variables and having rational number coefficients admit a common specialization of a proper subset of the variables to rational numbers such that all the polynomials remain irreducible. This theorem is a prominent theorem in number theory.

The statement, quoting them again, who are quoting Hilbert’s own words is:

Theorem 2 If {f(x,t)} is an irreducible polynomial in the two variables {x} and {t} with integral coefficients

\displaystyle  f(x,t) = T x^n + T_1 x^{n-1} +\cdots+ T_n, \ \ \ \ \ (1)

where {T,T_1,\dots,T_n} are integral polynomials in {t}, it is always possible, indeed in infinitely many ways, to substitute an integer for {t} in {f(x,t)} such that the polynomial {f(x,t)} becomes an irreducible polynomial of the single variable~{x}.

This statement has several issues as they point out. One other contribution of their paper is explain some of the differences in clarity and perhaps from Hilbert’s time to today. The above statement is a bit imprecise and they explain how to fix it. Read their paper for the details.

Applications

As we explained already Hilbert proved the HIT in order to make progress on the inverse Galois problem. Roughly he constructed a “generic” polynomial that had the required Galois group. Then he invoked his HIT to show that the generic polynomial could be changed into a rational polynomial that had the same Galois group. This is a powerful method that is still used today. However, it does not seem to be enough to solve the entire inverse problem.

Another neat application of HIT is a short proof of the following theorem.

Theorem 3 Suppose that {g(x)} is a polynomial over the integers. Suppose also that {g(n)} is a square for all numbers {n} large enough. Then there is a polynomial {h(x)} so that {g(x) = h(x)^{2}}.

Look at the polynomial {Y^{2}-f(x)} over the field {\mathbb{Z}[x,Y]}. This theorem shows that the “obvious” way to for a polynomial to always be a square is the only way. Actually proving this theorem without using HIT is possible, but not as simple as applying HIT.

The Proof

There are many known proofs of HIT. What they do in their article is to show how Hilbert originally proved the theorem. See their paper for the details which are quite interesting. Quoting them, a last time:

Hilbert remains one of the greatest mathematicians of all time. His original proof still contains insights and arguments that are well worth study even today. We offer the reader a detailed exposition of this proof in hope of saving it from the oblivion of history.

An interesting point about Hilbert’s proof is that it relied on a type of Ramsey theorem. It needed a key lemma that is now called Hilbert’s cube lemma. Define an {m}-cube as the set

\displaystyle  \{ a + \sum_{k=1}^{m} \epsilon_{k} a_{k} \mid \epsilon_{k}=0,1 \}

Lemma 4 For any positive integers {m} and {r} there exists a least integer {H} such that if the set {\{1,2,\cdots,H\}} is colored with {r} colors, then some color class must contain an {m}-cube.

This is clearly a type of Ramsey theorem: It shows that any coloring of a large enough object must have a color class that has some given regularity. They point out, as others have too, that Hilbert’s lemma is probably one of the first “Ramsey-like” theorems. It is interesting, I think, to note that even the great Hilbert missed an opportunity here. He could have invented Ramsey theory many years earlier.

Open Problems

Is there hope to solve the inverse Galois problem? Is every finite group a Galois group over the rationals? Hilbert was beyond brilliant, yet he missed creating the whole field of Ramsey Theory. Are we today missing some large new class of math that is really right in front of us?

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3 Comments leave one →
  1. test permalink
    June 9, 2018 11:31 am

    test

  2. Jhon Hard permalink
    June 11, 2018 1:00 pm

    Inverse Galois problema.

    Every polynomial with rational coefficients is associated with a group of Galois.

    The only finite group that is Galois group and solves the polinomial that contains it is (the Elliptic curve).

    if we have:

    ax^{n} + bx^{n-1} + cx^{n-2} + ….. + x^{3} + rx^{2} + sx + d = 0.

    As (x^{3} + rx^{2} + sx + d) is an elliptic curve and we have an equation to determine the value (x). We will know the value of (x) as the only posible solution of the polynomial.

    Title: Equation to solve elliptical curve.

    x = [b^{2} – (a/2)^{2}] / 4[c – (b – (a/2)^{2})(a/2)]

    Published in: Universal Journal of Computational Mathematic Vol 1 (1) 2013

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