POLYNOMIALS OVER THE FIELD OF RATIONAL NUMBERS

​ABSTRACT

The article considers polynomials over the field of rational numbers and their solution.

Keywords: polynomials, rational numbers.

Objectives of the article: to study the concept of a polynomial with rational coefficients and the signs of polynomials over the field of these numbers, and to show on problems how to define polynomials over the field of rational numbers.

Objectives of the article:

-analyze the educational literature on this topic;

- systematize the received material;

- show the solution of problems about polynomials over a field of rational numbers;

- summarize all the material and make a conclusion.

Before considering polynomials over the field of rational numbers, you need to know the definition of a polynomial. And so, a polynomial is an expression that consists of variables (they can be any) and coefficients, and includes all the operations of addition, subtraction, multiplication, as well as exponentiation (most importantly, to a non–negative degree). An example is the following polynomial: . And now let's move on to the main topic of the article.

Any polynomial f(x) with rational coefficients can be represented as:

In order to talk about the field of rational numbers, we must also talk about the field with integer coefficients. [2]

A polynomial f(x) with integer coefficients is called primitive if all its coefficients together are mutually simple, i.e., their NODE is equal to 1.

For integer primitive polynomials, the Gauss lemma holds: "The product of two integer primitive polynomials is itself a primitive polynomial."[1]

Proof.

Let be given primitive integer polynomials

and let

If this product is not primitive, then there exists such a prime number p that serves as a common divisor for all coefficients . Since all coefficients of the primitive polynomial  cannot be divisible by , then let the coefficient  be the first one not divisible by p; similarly, by  we denote the first coefficient of the polynomial , which is not divisible by . Multiplying the points   and  and collecting the terms containing , we get:

The left side of this equality is divided by . All the terms of the right part are also obviously divided into it, except for the first one; indeed, due to the conditions imposed on the choice of ⅈ and j, all coefficients , , …, as well as ,  , …,are divided by . It follows that the product  is also divisible by , therefore, due to the simplicity of the number , at least one of the coefficients  must be divisible by , which, however, is not the case. [3]

Eisenstein's criterion on the irreducibility of polynomials over the field of rational numbers

We know that over the field of complex numbers we give any polynomial whose degree is greater than one, and over the field of real numbers we give a polynomial whose degree is greater than two. The case with the field of rational numbers is quite different: for any , you can specify a polynomial of the nth degree with rational (even integer) coefficients, irreducible over the field of rational numbers. The proof of this statement follows from Eisenstein's criterion.

The Eisenstein criterion. Let be given a polynomial

with integer coefficients. If in at least one way it is possible to find a prime number ; satisfying the following requirements:

1)the highest coefficient  is not divisible by ,

2)all other coefficients are divisible by ,

3)the free term, dividing by , is not divisible by , then the polynomial  is irreducible over the field of rational numbers.

In fact, if the polynomial  is given over the field of real numbers, then it decomposes into two factors of a lesser degree with integer coefficients:

,

where

 . Hence, comparing the coefficients in both parts of this equality, we get:

                                                          (1)

It follows from the first of the equalities (1), since  is divisible by , and the number  is prime, that one of the factors  must be divisible by . Both of them cannot simultaneously divide by this number, since  is conditionally not divisible by .

Let, for example,  be divisible by p and therefore  is mutually simple with . We now turn to the second of the equalities (1). Its left part, as well as the first term of the right part, is divided by , so the product  is divided by ; since, however, l is not divisible by , then  will be divided by . Similarly, from the third equality (1) we get that  is divisible by . Finally, from the th equality it will be obtained that  is divisible by ; but then from the last of the equalities (1) it follows that  is divisible by , which contradicts the assumption.[2]

The consequence follows from Eisenstein's criterion: over the field of rational numbers there are polynomials of any degree irreducible over this field.

The Eisenstein criterion shows that from the point of view of the structure of irreducible polynomials, the field of rational numbers is strikingly different from the fields of real and complex numbers. In fact, as we saw above, every irreducible polynomial over the field of complex is linear, and every irreducible polynomial over the field of real has degree ≤2. At the same time, by virtue of the Eisenstein criterion, the degree of a polynomial irreducible over the field of rational numbers can be any. For example, irreducible over the rational field is the polynomial , where n is an arbitrary natural number (it satisfies the premise of the Eisenstein criterion for p = 2).[3]

Problem 1. Using the Eisenstein criterion, prove that the polynomial is irreducible over the field of rational numbers.

Decision.

Having carefully considered the coefficients of the polynomial from the problem condition, we note that in this case such a prime number exists. This number is 3. Therefore, the polynomial is irreducible over the field of rational numbers.

Problem 2. Find all the rational roots of the equation 

Decision.

The polynomial from the left side of this equation has no fractional rational roots, and integer roots can only be divisors of a free term, namely, they must be searched among the numbers 1,-1, 2, -2, 5, -5, 10, -10.

By direct substitution, we obtain that for x = 1, the left side of the equation is equal to -9, and for x = -1: -11.

Using the Gorner scheme, we find the value of the left part at x=2

And for x=-2

we find that the number -2 is the root of the equation and the left part can be represented as a product:

.

The integer roots of the second factor must be sought among the divisors of the number -2. The numbers 1 and -1 are no longer exactly the roots of this polynomial, and with the help of the Gorner scheme, you can check that 5 and -5 are not roots.

Therefore, the only rational root of this equation is the number -2.

Answer: -2

Conclusion. Summing up, we can conclude that any polynomial f(x) with rational coefficients can be represented as: . In the article, the Eisenstein Criterion was considered, which shows that from the point of view of the structure of irreducible polynomials, the field of rational numbers is strikingly different from the fields of real and complex numbers.  Several problems were also analyzed in various ways to show which polynomials are in the field of rational numbers.

References:

1. Zinchenko N. A. Algebra of polynomials: a textbook for students. higher. studies. institutions / N. A. Zinchenko, N. N. Motkina, M. V. Shevtsova. – Belgorod: Publishing house "Belgorod", 2014. – 100 p.
2. Kurosh, Alexander Gennadievich. Course of higher algebra: textbook for universities / A. G. Kurosh. – 22nd ed., erased. – St. Petersburg: Lan, 2021. – 432 p.
3. Prasolov V. V. Polynomials. - 3rd ed., corrected. — Moscow: ICNMO, 2003. -336 p.