AA@rational coefficient polynomial@integral coefficient polynomial@primitive polynomial@rational polynomial reduction problem

polynomial with rational coefficients

• Factorization of unary polynomials over the field of rational numbers.
• As a special case of the factorization theorem , we have the conclusion:
  • Every polynomial with rational coefficients of degree greater than 1 can be uniquely decomposed into a product of irreducible polynomials with rational coefficients .
  • In the rational number field version, from the general number field to the " rational coefficient "
  • When we discuss polynomials, we assume that polynomials are in a certain number field P. For example, the conclusion in the general number field P is also true in the special number field (rational number field Q).
• related problems
  • For any given polynomial, it is a very complicated problem to make its decomposition formula concretely
  • Even to determine whether a polynomial with rational coefficients is reducible is not an easy problem
    • This is the difference between the field of rational numbers and the field of real numbers and complex numbers.
    • Only polynomials of first degree are irreducible over the field of complex numbers
    • However, irreducible polynomials in the field of real numbers have only one degree and some two degrees .
• Here we mainly point out two important facts about polynomials with rational coefficients.
  • The problem of factorization of polynomials with rational coefficients can be reduced to the problem of factorization of polynomials with integer (number) coefficients, and then solve the problem of finding rational roots of polynomials with rational coefficients .
  • There are irreducible polynomials of any degree in the ring of polynomials with rational coefficients .

primitive polynomial

General Polynomial to Integer Coefficient Polynomial

• 设$f(x)={\sum }_{i=1}^n{a}_i{x}^i$ is a rational number polynomial, take the appropriate integer$c$ times$f\left(x\right)$ , it is always possible to make$G(x)=cf\left(x\right)$ is apolynomial with integer coefficients(for example, c takes$a_i,i=1,2,\cdots ,least\; common\; multiple\; of\; n$ )
• If G ( x ) G(x)Each coefficient of$G$$($$x$$)$$d$ , it can be extracted:$G(x)=dg\left(x\right)$ , so$f(x)=\frac{d}{c}g(x)$
  • Remember $r=\frac{d}{c}$,$f\left(x\right)=rg(x)$
  • where $g\left(x\right)$ isa polynomial with integer coefficients, and the coefficients are not different from$The\; common\; factor\; of\; \pm 1$ (the common factor of each coefficient is only$\pm 1$)
  • For example: $f(x)=\frac{2}{3}{x}^4-2x{\_}^2-\frac{2}{5}x$=$\frac{2}{15}(5x^4-15x^2-3x)$
    • where c=15,d=2

primitive polynomial definition

• If a non-zero integer coefficient polynomial $g(x)={\sum }_{i=0}^n{b}_i{x}^i$ -like coefficient$b_i,i=1,2,\cdots ,$The common factor of$n$$\pm 1$ , that is to say$b_i,b_j,i\ne j$ isrelativelyprime

• From the discussion in the previous section, we know that any non-zero rational coefficient polynomial $f\left(x\right)$ can be expressed as arational number$r$ and aprimitive polynomialg ( x ) g(x)The productof$g$$($$x$$)$ , ie$f(x)=rg(x)$

  • This notation is unique except for a difference of a plus or minus sign

  • 若$f(x)=rg(x)=r_1{g}_1\left(x\right)$ , of which$g(x),g_1\left(x\right)$ are primitive polynomials, then$r=\pm r_1$,$g(x)=\pm g_1(x)$

  • Since $f\left(x\right)$ and$g\left(x\right)$ is only aconstant multiple, so$The\; factorization\; problem\; of\; f\left(x\right)$ can be reduced to the primitive polynomialg ( x ) g(x)Factorization problem of$g$$($$x$$)$

• The problem of whether a primitive polynomial can be decomposed into the product of two polynomials with rational coefficients of lower degree is the same as whether it can be decomposed into the product of two polynomials with integer coefficients of lower degree

• The proof of this conclusion requires some preparatory knowledge

Gauss Lemma

• The product of two primitive polynomials is still a primitive polynomial
• Proof: prove by contradiction
  • 设$f(x)={\sum }_{i=0}^n{a}_i{x}^i$,$g(x)={\sum }_{i=0}^m{b}_i{x}^i$ are two primitive polynomials
    • 则$h(x)=f(x)g(x)={\sum }_{i=0}^{n+m}{d}_i{x}^i$
  • 令 D a = { 0 , 1 , 2 , ⋯   , n } D_a=\{0,1,2,\cdots,n\} Da​={ 0,1,2,⋯,n},$D_b=\{0,1,2,\cdots ,m\}$, D s = { 0 , 1 , 2 , ⋯   , n + m } D_s=\{0,1,2,\cdots,n+m\} Ds​={ 0,1,2,⋯,n+m}
  • if $h\left(x\right)$ is not primitive, i.e.$h(x)$的系数$d_i,\forall i\in D_s$have non $\pm 1$ common factor, there exists a prime number$p$,满足$p|d_i,\forall i\in D_s$
  • Since $f\left(x\right)$ is primitive, so$p$不满足$p|a_i,\forall i\in D_a$, $a_i$is the first one that cannot be $The\; coefficient\; of\; p$ divisibility ($p\nmid a_i$),而$(p|a_k,k=0,1,2,\cdots ,i-1)$
  • Similarly, $g\left(x\right)$ is also primitive, let$b_j$is the first one that cannot be ppcoefficient of divisibility by$p$$p\nmid b_j$,$(p|b_k,k=0,1,2,\cdots ,j)$
  • while $h(x)$的$s=i+The\; coefficient\; ds\; of\; the\; jth$ term$d_s={\sum }_{r_1+r_2=s}{a}_{r_1}{b}_{r_2}$,$r_1\in D_n,r_2\in D_m$,
    • To be more intuitive, expand the right side of the equation $(a_i{b}_j+a_{i+1}{b}_{j-1}+\cdots )$+$(a_{i-1}{b}_{j+1}+a_{i-2}{b}_{j+2}+\cdots )$,
    • Separate note: $T(x)=a_i{b}_j$,$U(x)=(a_{i+1}{b}_{j-1}+a_{i+2}{b}_{j-2}+\cdots )$;$V(x)=(a_{i-1}{b}_{j+1}+a_{i-2}{b}_{j+2}+\cdots )$
    • For $T(x)$,$p\nmid a_i,p\nmid b_j$, and $p$ is a prime number, so$p\nmid a_i{b}_j$
    • against $U(x)$,因为$b_{j-1},b_{j-2},\cdots$ are all divisible by p, so$U\left(x\right)$ can also beppdivisible by$p$
    • For $V(x)$,因为$a_{i-1},a_{i-2},\cdots$ Can be$p$ divides evenly, so$V\left(x\right)$ is also divisible by p,
    • The right side of the equation is written as $RHS=T(x)+U(x)+$Whether$V$$($$x$$)$$p$ divisibility depends on$T\left(x\right)$ , so$p\nmid RHS$
    • And by assumptions: $p|d_s$Therefore, the left side of the equation is divisible by p, which contradicts that the right side of the equation is not divisible by p.
  • So $h\left(x\right)$ must be a primitive polynomial

Decomposition Theorems of Polynomials with Integer Coefficients

• If the non- zero integer coefficient polynomial $f\left(x\right)$ can be decomposed into the product oftwolowerdegreerational coefficients$f\left(x\right)$ must be decomposed intothe multiform producttwointeger coefficientswith lower degree
• prove:
  • Let $f\left(x\right)$ has the decomposition formula$f(x)=g\left(x\right)h\left(x\right)$ , of which$g(x),f\left(x\right)$ are polynomials with rational coefficients
  • $\partial (g(x)),\partial (h(x))<\partial (f(x))$
  • 令$f(x)=a{f}_1(x)$,$g(x)=r{g}_1(x)$,$h(x)=s{h}_1\left(x\right)$ , here:
    • $f_1(x),g_1\left(x\right),h_1\left(x\right)$ are primitive polynomials
    • $a\in \mathbb{Z}$,$r,s\in \mathbb{Q}$
  • 则$a{f}_1(x)=rs\cdot g_1(x)h_1\left(x\right)$
• By Gaussian Lemma, $G(x)=g_1\left(x\right)h_1\left(x\right)$ is a primitive polynomial, so$rs=\pm a$ , so$rs\in \mathbb{Z}$
• So $f(x)=(rs\cdot g_1(x))h_1\left(x\right)$ , where$rs\cdot g_1(x),h_1\left(x\right)$ are all integer multi-coefficient forms, and the degree is lower thanf ( x ) f(x)number of times$f$$($$x$$)$

inference

• Let $f(x),g\left(x\right)$ is a polynomial with integer coefficients, where$g\left(x\right)$ is stillprimitive, if$f(x)=g\left(x\right)h\left(x\right)$ and$h\left(x\right)$ isa polynomial with rational coefficients, then$h\left(x\right)$ must still bea rational polynomialwith integer coefficients
• Proof: In the following proof, the first two methods are similar and valid

Idea 1:

• From the theorem and deduction conditions in this section, we know that $f\left(x\right)$ can be decomposed into the product of two integer coefficient polynomials, set$f\left(x\right)=g\left(x\right)b{h}_1\left(x\right)$ , of which$h(x)=b{h}_1(x)$,$h_1\left(x\right)$ is a primitive polynomial
• By Gaussian Lemma $G(x)=g(x)h_1\left(x\right)$ is a primitive polynomial, and$f\left(x\right)$ is a polynomial with integer coefficients, it can be seen$b\in \mathbb{Z}$
• Thus $h\left(x\right)=b{h}_1\left(x\right)$ is a polynomial with integer coefficients

Idea 2:

• Use the property that general rational polynomials can be expressed as a rational number multiplied by primitive polynomials to reason

• 任$f(x)=a{f}_1(x)$,$h(x)=b{h}_1\left(x\right)$ , of which$f_1(x),h_1\left(x\right)$ are primitive polynomials

  • while $f\left(x\right)$ is a polynomial with integer coefficients, so$a\in \mathbb{Z}$
  • And $b\in \mathbb{Q}$

• 此时$f\left(x\right)=g\left(x\right)h\left(x\right)$ is rewritten as$a{f}_1(x)=g\left(x\right)(b{h}_1(x))$=$bg\left(x\right)h_1\left(x\right)$

  • By Gaussian Lemma, $G(x)=g(x)h_1\left(x\right)$ is a primitive polynomial and$|G(x)/f_1(x)|=1$
  • It can be seen that $a=\pm b$ (or$|a|=|b|$ ), so$b\in \mathbb{Z}$

• So $h(x)=b{h}_1\left(x\right)$ is a polynomial with integer coefficients

Idea 3: (Abandoned)

• 设$g(x)={\sum }_{i=0}^m{a}_i{x}^i$,$b_i\in \mathbb{Z}$,$h(x)={\sum }_{i=0}^n{b}_i{x}^i$

• According to polynomial multiplication, let $f(x)=g(x)h(x)={\sum }_{i=1}^{m+n}{c}_i{x}^i$,$m,n$ respectively$g(x),h\left(x\right)$ times

• Since $g\left(x\right)$ is primitive, so$a_i\in \mathbb{Z}$,且$a_i,i=0,1,2,\cdots ,$The common factor of$m$$\pm 1$

• If $The\; coefficients\; of\; h\left(x\right)$ are not all integers, then$c_i={\sum }_{r_1+r_2=i}{a}_{r_1}{b}_{r_2}\in \mathbb{Z}$

• if $b_k$not an integer

  • 若$b_r,r\ne k$ is an integer, then$c_i$not an integer
  • 若$b_r,r\ne K$ is not all integers, the situation is more complicated, because the result of the sum of multiple rational numbers may still be an integer, and it cannot be directly determined that$c_i$not an integer

Rational Root Theorem and Integral Root Theorem of Polynomials with Integer Coefficients

• 设$f(x)={\sum }_{i=0}^n{a}_i{x}^i$ is aninteger coefficient polynomial, if$\frac{r}{s}$is f ( x ) f(x)Arational root(r , sr,s$of\; f$$($$x$$)$$r,s$ is a relatively prime integer), then there is an integer divisibility relationship:$r|a_0$,$s|a_n$

  • if $r,s$ is not only relatively prime but also$|s|=1$ , then the rational root$\frac{r}{s}\in \mathbb{Z}$

• The theorem shows that if f ( x ) f(x)The prime minister coefficient of$f$$($$x$$)$$a_n=1$ , then by$s|1$ , that is,$|s|=1$ , therefore,f ( x ) f(x)The rational rootsof$f$$($$x$$)$ are allwhole roots, and are$a_0$factor _

  • Extensive, $g(x)=x^n+{\sum }_{i=0}^{n-1}{a}_i{x}^i$ has a rational root, so it is impossible to directly determine whether it has a whole root
  • But we can try to judge g ( x ) g(x)The constant terma 0 a_0$of\; g$$($$x$$)$$a_0$factor of , because in this case, if the whole root exists, it must be $a_0$factor of ; in particular $a_0$When is a prime number, it is easy to quickly check whether the equation has a whole root

• prove:

  • Because $\frac{r}{s}$is $A\; rational\; root\; of\; f\left(x\right)$ , then$(x-\frac{r}{s})|f(x)$ (remainder theorem), so$(sx-r)|f(x)$
  • because $r,s\; is$ relatively prime,$t(x)=sx-r$ is a primitive polynomial of degree one
  • According to the inference of integer coefficient decomposition in the previous section, $f(x)=(sx-r)({\sum }_{i=0}^{n-1}{b}_i{x}^i)$,其中$b_i\in \mathbb{Z},i=0,1,\cdots ,n-1$
  • Compare f ( x ) f(x)Two expansion forms of$f$$($$x$$)$${\sum }_{i=0}^n{a}_i{x}^i$=$(sx-r){\sum }_{i=0}^{n-1}{b}_i{x}^i$ ,$a_n=s{b}_{n-1}$,$a_0=-r{b}_0$
  • Therefore, $s|a_n$,$r|a_0$

• Example: $f(x)$=$2x{\_}^4-x^3+2x\_-3=$Rational root solution of$0$

  • According to the theorem of this section,

    • Solution $a_0=Factors\; of\; -3$ in the integer range:$-3,-1,1,3$ , abbreviated as$\pm 1,\pm 3$

    • Solution $a_n=2$在整数范围内的因子:$-2,-1,1,2$,简写为$\pm 1,\pm 2$

    • 所有可能的有理根候选:$\pm 1$,$\pm \frac{1}{2}$,$\pm 3$,$\pm \frac{3}{2}$

    • 有两种方式检验这些根,通常推荐先检验整根

      • 逐个代入这8个根检验
      • 利用带余除法来检验
      • 混合两种方法,先用代入法检验整根,然后用带余除法检验非整根

    • 经检验,$x=-1,\pm 3$不是根,$x=1$ root,$f(x)=(x-1)(2x{\_}^3+x^2+x+3)$

    • 对$f_1(x)=2x{\_}^3+x^2+x+3$ Continue to test the remaining candidate roots

    • only $x=1$ is the root, i.e.$f\left(x\right)=$The only rational root of$0$$x=1$

• Example: $f\left(x\right)=x^3-5x-1$ is irreducible over the field of rational numbers

  • Prove that, according to the rational root theorem of polynomials with integer coefficients, $a_0=1$ , the factor has$\pm 1$,$a_n=1$ , the factor is$\pm 1$
  • Therefore, the rational root of the equation can only be $\pm 1$
  • After testing, $f(1)=-5\ne 0$,$f(-1)=3\ne 0$ , so the equation has no rational roots, so$f\left(x\right)$ is irreducible in the field of rational numbers

Eisenstein discriminant

• 设$f(x)={\sum }_{i=0}^n{a}_i{x}^i$ is apolynomial with integer coefficients, if there isa prime number$p$ makes the following three conditions true, then$f\left(x\right)$ is irreduciblein the field of rational numbers
  • (1)$p\nmid a_n$
  • (2)$p|a_i,i=0,1,\cdots ,n-1$
  • (3)$p^2\nmid a_0$
• prove:
  • Prove by contradiction here, assuming $f\left(x\right)$ is reducible on the rational number field, and then it is shown that reducibility leads to a contradiction, thus proving the irreducibility
  • If $f\left(x\right)$ is reducible on the rational number field, then$f\left(x\right)$ can be decomposed into the product of two polynomials withinteger coefficients:
  • $f(x)={\sum }_{i=0}^l{b}_i{x}^i{\sum }_{j=0}^m{c}_j{x}^j$ ,其中$l,m<n$;$l+m=n$
    • 显然$a_n=b_l{c}_m$,$a_0=b_0{c}_0$
  • By condition (2), $p|a_0$, namely $p|b_0{c}_0$, so $p|b_0$or $p|c_0$
  • By condition (3), $p^2\nmid a_0$, namely $p^2\nmid b_0{c}_0$, so $p$ cannot divide$b_0,c_0$
    • Let $p|b_0$And $p\nmid c_0$(4),
    • By condition (1), $p\nmid a_n$There are $p\nmid b_l,c_m$,即$b_i,i=1,2,\cdots ,At\; least\; one\; of\; l$ is not divisible by p
    • 假设$b_i,i=1,2,\cdots ,$The first one in$l$$p$ is divisible by$b_k$(5)
    • Compare f ( x ) f(x)xkx^kof$f$$($$x$$)$$x^{}$Coefficient of${}^k$$a_k$,$a_k={\sum }_{i=0}^k{b}_i{c}_{k-i}$
    • 由于$a_k,b_i(i=0,1,\cdots ,k-1)$ can be$p$ integer division,${\sum }_{i=0}^{k-1}{b}_i{c}_{k-i}$can be $p$ divides evenly, so$b_k{c}_0$can also be ppdivisible by$p$
    • while $p$ is a prime number, so$b_k,c_0$At least one of them can be $p$ divisible by (6)
    • (4,5) and (6) produce a contradiction, so

Construct irreducible polynomials with rational coefficients of any degree

• Judgment $f(x)=x^n+2$ Is it reducible in the field of rational numbers
• $a_n=1$,$a_0=2$
• According to Eisenstein's discriminant method, we might as well take the prime number p=2, then $p\nmid a_n$,$p^2\nmid a_0$且$p|a_i,i=0$ , so that$f\left(x\right)$ is irreducible in the field of rational numbers
• This shows that, over the field of rational numbers, there exist irreducible polynomials of any degree