Decomposing a Polynomial into a Sum of Two Irreducible Polynomials

将多项式分解为两个不可约多项式之和 苏剑林 · 2014-12-22 · Mathematics algebra polynomial decomposition

In the chapter on polynomials in advanced algebra textbooks, there is often an exercise of the following type:

Prove that any polynomial over the field of rational numbers can be expressed as a sum of two irreducible polynomials over the rational numbers.

This is a simple exercise with several possible proofs. Let's start with a clever one.

A Clever Proof

Working with polynomials over the field of rational numbers is equivalent to working with polynomials over the integers, so it suffices to prove the statement over the integers (this reduction lets us use Eisenstein's criterion). Let $f(x)$ be a degree-$n$ polynomial over the integers:

$$f(x)=a_n x^n+a_{n-1} x^{n-1}+\dots+a_1 x+a_0$$

We only need to note that

$$p f(x)=\left[p f(x)+x^n+p\right]-(x^{n}+p)$$more

where $p$ is a prime, so that $x^n+p$ is an irreducible polynomial over the integers. It then suffices to consider $p f(x)+x^n+p$, whose leading term is $(pa_n +1)x^n$, while all the remaining terms might be divisible by $p$. By Eisenstein's criterion, we only need to find $p$ such that $p^2$ does not divide the constant term. The constant term is $p a_0+p$; if $p^2|p a_0+p$, then $p|a_0+1$, i.e., $a_0+1$ is a multiple of $p$. But $a_0+1$ cannot possibly be divisible by every prime, so we can always find a prime $p$ such that $p\nmid a_0+1$, giving $p^2\nmid (pa_0+p)=\text{constant term}$. Hence, by Eisenstein's criterion, $p f(x)+x^n+p$ is an irreducible polynomial over the integers, and therefore $f(x)$ admits the decomposition

$$f(x)=\left[f(x)+\frac{1}{p} x^n+1\right]+\left[-\frac{1}{p}(x^n +p)\right]$$

The polynomial inside the brackets is precisely an irreducible polynomial over the field of rational numbers.

An Erroneous Attempt

The content below is essentially incorrect!!

In fact, the statement above might seem to generalize as follows:

1. Any polynomial over the field of real numbers can be expressed as a sum of two irreducible polynomials over the reals. (This does not actually hold!!)
2. Any polynomial over the field of rational numbers can be expressed as a sum of two irreducible polynomials over the rationals, where the degrees of both irreducible polynomials do not exceed the degree of the original polynomial.

Below, using an approach the author found reasonably simple, I'll attempt to prove both statements.

A General, Concise-Looking Proof

Generalization 1

Any polynomial over the field of real numbers can be expressed as a sum of two irreducible polynomials over the reals.
(This does not actually hold!!)

Let $f(x)$ be a polynomial over the field of real numbers:

$$f(x)=a_n x^n+a_{n-1} x^{n-1}+\dots+a_1 x+a_0$$

Then

$$f(x)=\left(x^{2n}+f(x)+2q^{2n}\right)-\left(x^{2n}+2q^{2n}\right)$$

where, when $q > 0$, $x^{2n}+2q^{2n}=0$ clearly has no roots in the reals, and is hence irreducible over the reals. Meanwhile, $x^{2n}+f(x)+2q^{2n}$ is a polynomial with leading term $x^{2n}$ (of even degree), so $x^{2n}+f(x)+2q^{2n}$ is bounded below. Thus, for $q$ large enough, $x^{2n}+f(x)+2q^{2n}$ is always positive, so $x^{2n}+f(x)+2q^{2n}=0$ can also have no real roots, making $x^{2n}+f(x)+2q^{2n}$ irreducible over the reals. This would seem to prove the claim.

Generalization 2

Any polynomial over the field of rational numbers can be expressed as a sum of two irreducible polynomials over the rationals, where the degrees of both irreducible polynomials do not exceed the degree of the original polynomial.

This statement was already proved by the clever argument at the beginning of the article. Now, based on the proof idea used for "Generalization 1," let's give another proof of Generalization 2. The proof of "Generalization 1" cannot be applied directly, because it relies on adding a term of higher degree, which would make the degree of the decomposed pieces exceed that of the original polynomial. In fact, "Generalization 2" fails over the reals.

Let $f(x)$ be a monic polynomial over the integers:

$$f(x)=x^n+a_{n-1} x^{n-1}+\dots+a_1 x+a_0$$

We consider cases:

1. When $n$ is even, $f(x)$ is bounded below over the reals, so for a sufficiently large integer $q$, $f(x)+q$ is always positive. Hence $f(x)+q=0$ has no real roots, and is therefore irreducible over both the reals and the integers. Thus $f(x)$ admits the decomposition:

$$f(x)=\left[f(x)+q\right]-q$$

and $-q$ is also irreducible over the integers.

1. When $n$ is odd, first suppose $a_{n-1}\neq 0$, and consider

$$f(x)=\left[f(x)-x^n+ 2q^n\right]+(x^n-2 q^n)$$

When $q$ is an integer, $x^n-2 q^n$ is irreducible over the integers (it has no rational roots), so we only need to consider $f(x)-x^n+ 2q^n$, which is a degree-$n-1$ (even) polynomial. If $a_{n-1} > 0$, then it is bounded below over the reals, so for sufficiently large $q$, it is always positive, making it irreducible over the reals; if $a_{n-1} < 0$, then it is bounded above over the reals, so for sufficiently small (very negative) $q$, it is always negative, again making it irreducible over the reals. This case is thus settled.

1. When $n$ is odd and $a_{n-1}=0$, consider

$$f(x)=\left[f(x)-x^n+2x^{n-1}+2(2q+1)\right]+\left[x^n-2x^{n-1}-2(2q+1)\right]$$

By Eisenstein's criterion, for any integer $q$, $x^n-2x^{n-1}-2(2q+1)$ is irreducible over the integers, so we only need to consider $f(x)-x^n+2x^{n-1}+2(2q+1)$, which is a degree-$n-1$ (even) polynomial whose leading term is $2x^{n-1}$. Hence it is bounded below over the reals, so for sufficiently large $q$, it is always positive, making it irreducible over the reals. This case is thus settled.

Summary of the Approach

Most of the reasoning in this piece boils down to trying to construct an even-degree polynomial containing a free constant term, such that for a sufficiently large (or sufficiently small/negative) choice of that constant, the polynomial is always positive (or always negative), and hence irreducible over the reals — and, in particular, over the rationals. The remaining piece is simply an obviously irreducible polynomial, completing the proof.

(Editor's note, so to speak: as flagged above, this "proof" of Generalization 1 is in fact flawed — a boundedness argument alone does not guarantee irreducibility, since a polynomial with no real roots can still factor over the reals into lower-degree real factors, e.g., a quartic with no real roots can still split into two real quadratics. The reader should treat the "General Proof" section as an instructive counterexample-in-waiting rather than a valid theorem.)