In the chapter on polynomials in Higher Algebra, there is usually an exercise like this:
Prove that any polynomial over the field of rational numbers can be expressed as the sum of two irreducible polynomials over the field of rational numbers.
This is a simple exercise, and there are several ways to prove it. First, let us introduce an ingenious proof.
An Ingenious Proof
A problem regarding polynomials over the field of rational numbers is equivalent to a problem regarding polynomials over the ring of integers. Therefore, we only need to prove it for polynomials over the ring of integers (this transformation allows us to use Eisenstein’s Criterion). Let f(x) be an n-th degree polynomial over the ring of integers: f(x)=a_n x^n+a_{n-1} x^{n-1}+\dots+a_1 x+a_0 We only need to notice that p f(x)=\left[p f(x)+x^n+p\right]-(x^{n}+p) where p is a prime number. Then x^n+p is an irreducible polynomial over the ring of integers. Thus, we only need to consider p f(x)+x^n+p. The highest degree term of this polynomial is (pa_n +1)x^n, and all other terms are potentially divisible by p. According to Eisenstein’s Criterion, we only need to find a p such that p^2 does not divide the constant term. The constant term is p a_0+p. If p^2 \mid (p a_0+p), then p \mid (a_0+1), meaning a_0+1 is a multiple of p. However, a_0+1 cannot be divisible by all prime numbers. Therefore, we can always find a prime p such that p \nmid (a_0+1), which implies p^2 \nmid (pa_0+p) = \text{constant term}. Thus, by Eisenstein’s Criterion, p f(x)+x^n+p is an irreducible polynomial over the ring of integers. Consequently, f(x) has the decomposition: f(x)=\left[f(x)+\frac{1}{p} x^n+1\right]+\left[-\frac{1}{p}(x^n +p)\right] The terms inside the square brackets are irreducible polynomials over the field of rational numbers.
Incorrect Process
The following content is basically incorrect!!
In fact, the above proposition can be extended:
1. Any polynomial over the field of real numbers can be expressed as the sum of two irreducible polynomials over the field of real numbers. (This does not hold!!)
2. Any polynomial over the field of rational numbers can be expressed as the sum of two irreducible polynomials over the field of rational numbers, where the degrees of the two irreducible polynomials do not exceed the degree of the original polynomial.
Below, using a relatively simple line of thought conceived by the author, we will attempt to prove the two conclusions above.
A Universal Concise Proof
Extension One
Any polynomial over the field of real numbers can be expressed as the sum of two irreducible polynomials over the field of real numbers. (This does not 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) When q > 0, x^{2n}+2q^{2n}=0 clearly has no roots in the field of real numbers, making it an irreducible polynomial in the field of real numbers. Since x^{2n}+f(x)+2q^{2n} is a polynomial with a leading term x^{2n} (even degree), x^{2n}+f(x)+2q^{2n} has a lower bound. Therefore, when q is sufficiently large, x^{2n}+f(x)+2q^{2n} is always greater than 0, so x^{2n}+f(x)+2q^{2n}=0 can also have no roots in the field of real numbers. In this case, x^{2n}+f(x)+2q^{2n} is an irreducible polynomial in the field of real numbers. Thus, the proposition is proven.
Extension Two
Any polynomial over the field of rational numbers can be expressed as the sum of two irreducible polynomials over the field of rational numbers, where the degrees of the two irreducible polynomials do not exceed the degree of the original polynomial.
This proposition was already proven in the ingenious proof at the beginning of the article. Now, based on the reasoning of "Extension One," another proof for Extension Two is provided. The proof of "Extension One" cannot be used directly because it relies on adding a higher-degree term, which prevents the degree of the decomposed expressions from being no greater than the original degree. In fact, "Extension Two" does not hold in the field of real numbers.
Let f(x) be a monic polynomial over the ring of integers: f(x)=x^n+a_{n-1} x^{n-1}+\dots+a_1 x+a_0
Discussion by cases:
1. When n is even, f(x) has a lower bound in the field of real numbers. Thus, for a sufficiently large integer q, f(x)+q is always greater than 0. Therefore, f(x)+q=0 has no roots in the field of real numbers, making it an irreducible polynomial over the field of real numbers and also over the ring of integers. Thus, f(x) has the decomposition: f(x)=\left[f(x)+q\right]-q -q is also an irreducible polynomial over the ring of integers.
2. When n is odd, first assume a_{n-1} \neq 0. In this case, 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 an irreducible polynomial over the ring of integers (it has no rational roots). So we only need to consider f(x)-x^n+ 2q^n. It is an (n-1)-th degree (even degree) polynomial. If a_{n-1} > 0, it has a lower bound in the field of real numbers; thus, for a sufficiently large q, it is always greater than 0 and is an irreducible polynomial over the field of real numbers. If a_{n-1} < 0, it has an upper bound in the field of real numbers; thus, for a sufficiently small q, it is always less than 0 and is an irreducible polynomial over the field of real numbers. This case is thus proven.
3. 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 an irreducible polynomial over the ring of integers. Thus, we only need to consider f(x)-x^n+2x^{n-1}+2(2q+1). It is an (n-1)-th degree (even degree) polynomial with the leading term 2x^{n-1}, so it has a lower bound in the field of real numbers. For a sufficiently large q, it is always greater than 0 and is an irreducible polynomial over the field of real numbers. This case is thus proven.
Summary of Ideas
Most of the ideas in this article are simple: find a way to construct an even-degree polynomial with a free constant term, so that for a sufficiently large (or small) constant, it is always greater than 0, making it irreducible over the field of real numbers and consequently over the rational numbers. The other part consists of polynomials that are obviously irreducible, thereby completing the proof.
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