AA@factorization theorem

factorization theorem

• In the algebra we learned in middle school, we have learned some methods to decompose a polynomial into the product of factors that cannot be divided further.
• But there is no in-depth discussion of this issue. The so-called indivisibility often just means that we can't see how to further divide it, and it does not strictly demonstrate that they are indeed indivisible
• The so-called indivisible concept is actually not absolute, but relative to the number field where the coefficients are located .
• For example, over the field of rational numbers, put $x^4-4$ decomposes into$x^4-4=(x^2-2)(x^2+2)$ can no longer be divided.
• But in the field of real numbers, it can be further decomposed into $x^4-4=(x-\sqrt{2})(x+\sqrt{2})(x^2+2).$
• On the field of complex numbers, it can be further decomposed into $x^4-4=(x-\sqrt{2})(x+\sqrt{2})(x-\sqrt{2}i)(x+\sqrt{2}i).$
• It can be seen that the so-called non-dividable has a definite meaning after the coefficient domain must be clarified.
• In the following discussion, still select a number field Р as the coefficient field, consider the polynomial ring P [ x ] P[x] on the number field PFactorization of polynomials in$P$$[$$x$$]\; .$

Number fields and divisibility relations

• The divisibility relationship between two polynomials will not change due to the expansion of the number field

irreducible polynomial

• The number of times s ⩾ 1 s\geqslant{1} on the number field Р$s⩾$A polynomial of$1$ cannotbe expressed as a ratio of two degreesp ( x ) p(x)The product of polynomials witha lowdegree of$p$$($$x$$)$$p\left(x\right)$ is called an irreducible polynomialover the field P
  • Zero-degree polynomials (constants) are not irreducible polynomials, but can be used as factors of irreducible polynomials
• It can be seen that the first-degree polynomial is always an irreducible polynomial, because the polynomial lower than the first-degree polynomial is a zero-degree polynomial (constant), and the product of the constant is still a zero-degree polynomial, which cannot be equal to the first-degree polynomial
• Whether a polynomial is reducible depends on the field of coefficients
  • For example, $x^2+2$ is an irreducible polynomial in the real number field, but it can be decomposed into the product of two first-degree polynomials inthecomplexnumber field(reducible)

Factors of irreducible polynomials

• irreducible polynomial p ( x ) p(x)There are only two possible factors of$p$$($$x$$)$

  • "non-zero constant c"
  • “$The\; non-zero\; constant\; multiple\; of\; p\left(x\right)$ itself$c^{-1}p(x)(c\ne 0)$”

• If the factor only has the above two $A\; polynomial\; of\; degree\; k$ ($k⩾1$ ) must be an irreducible polynomial

• Therefore, the irreducible polynomial $p\left(x\right)$ and any polynomialf ( x ) f(x)There are only two possible relationships between$f$$($$x$$)\; :$

  • $p(x)|f(x)$且$(p(x),f(x))=cp\left(x\right),c\ne 0$
  • $p(x)\nmid f\left(x\right)$ and$\left(p\right(x),f(x))=1$
  • The two cases can be further concentrated description, let $(p(x),f(x))=d(x)$,$d\left(x\right)$ may be$1$ or$cp\left(x\right),c\ne 0$ , the latter corresponds to$p(x)|f(x)$

• Analogous to "prime numbers" in integers, it can be understood as "prime polynomials"

theorem

• If $p\left(x\right)$ is an irreducible polynomial, then any$f(x),g(x)$,若$p\left(x\right)|f(x\left)g\right(x)$ then$p\left(x\right)|f(x)$ or$p(x)|g(x)$

  • If $p\left(x\right)|f(x)$ , obviously the conclusion holds
  • 若$p(x)\nmid f\left(x\right)$ , then$(p(x),f(x))=1$ and because$p\left(x\right)|f(x\left)g\right(x)$ , then$p(x)|g(x)$

• Generalization: If the irreducible polynomial $p\left(x\right)$ divides s polynomials$f_1(x),f_2\left(x\right),\cdots ,f_s\left(x\right)$ product${\sum }_{i=1}^s{f}_i\left(x\right)$ , then there existsk ∈ { 1 , 2 , ⋯ , s } k\in\{1,2,\cdots,s\}k∈{ 1,2,⋯,s } ,let$p(x)|f_k(x)$

Factorization and uniqueness theorem

• Existence of decomposed formula: each degree $⩾$Polynomialf ( x ) f(x)$of\; 1$$f\left(x\right)$ can be uniquely decomposed into theproduct ofirreducible polynomials$f(x)={\sum }_{i=1}^s{p}_i\left(x\right)$
• Uniqueness: If two decomposition formulas $f\left(x\right)={\sum }_{i=1}^s{p}_i(x)$,$f\left(x\right)={\sum }_{i=1}^s{q}_i\left(x\right)$ , then there must be$s=t$ , and after properly arranging the order of factors,$p_i(x)=c_i{q}_i(x)$,$i=1,2,\cdots ,s$,$c_i$is a non-zero constant
  • if f ( x ) f(x)The coefficient of the highest term of all factors of$f$$($$x$$)\; is\; set\; to\; 1,\; and\; the\; constant\; coefficient\; is\; proposed\; as\; a\; term\; of\; degree\; 0$
  • ${\sum }_{i=1}^s{p}_i^{\prime }\left(x\right)={\sum }_{i=1}^t{q}_i^{\prime }\left(x\right)$
    • Then through the appropriate permutation factor, then $p_i^{\prime }\left(x\right)=q_i^{\prime }\left(x\right)$

prove

• For f ( x ) f(x)Degree of$f$$($$x$$)$$\partial (f(x))=n$ and decompositionf ( x ) f(x)The number of items after$f$$($$x$$)$$s$ is proved by mathematical induction
• The proof is divided into two parts: existence and uniqueness of decomposition

Decomposed existence

• Clear definition of irreducible polynomials shows that irreducible polynomials obviously satisfy the conclusion

• $n=1$

  • Because first-degree polynomials are irreducible, so $n=1$ when the conclusion is established

• 设$n<The\; conclusion\; is\; true\; when\; k$ , that is,$\partial \left(f\right(x))<The\; conclusion\; is\; valid\; when\; k$ , more specifically,$n=1,\cdots ,k-1$ when the conclusions are established

  • If $f\left(x\right)$ is an irreducible polynomial, the conclusion obviously holds
  • If $f\left(x\right)$ is reducible (not irreducible), ie$f(x)=f_1(x)f_2\left(x\right)$ , and let$n=k$
    • Reducible polynomials are expanded into several irreducible polynomials. For the convenience of the article, these irreducible factor expressions are divided into two parts by any kind, which are recorded as $f_1(x),f_2(x)$
    • 若$f_i\left(x\right)$ can be decomposed into an irreducible polynomial, then it can be expressed as:$f_i(x)={\sum }_{j=1}^s{p}_j^{[i]}\left(x\right)$
    • 其中$\partial (f_1(x)),\partial (f_2(x))<k$
    • By the inductive hypothesis $f_1(x),f_2\left(x\right)$ can be decomposed into some on the number field P (set to be$s_1,s_2$) product of irreducible polynomials
      • $f_1(x)={\prod }_{i=1}^{s_1}{p}_i^{[1]}\left(x\right)$
      • $f_2\left(x\right)={\prod }_{i=1}^{s_2}{p}_i^{[2]}\left(x\right)$
    • Put $f_1(x),f_2\left(x\right)$ decomposed to getf ( x ) f(x)A decomposition formula of$f$$($$x$$)$$f(x)={\prod }_{i=1}^{s_1}{p}_i^{\left[1\right]}\left(x\right){\prod }_{i=1}^{s_2}{p}_i^{\left[2\right]}\left(x\right)$
    • That is, for $n=k$ , the conclusion still holds
  • And $n=1,2,3,\cdots$
  • k=2时,$n<The\; possible\; values\; of\; k$ are$n=1$ , and the conclusion is valid when this value is taken, so the conclusion is valid when n=2
  • k=3时,$n<The\; possible\; values\; of\; k$ are$n=1,2$ , the conclusions under these values ​​are all valid, so$n=3$ when the conclusion is established
  • k=4时,$n<The\; possible\; values\; of\; k$ are$n=1,2,3$ , the conclusions under these values ​​are all valid, so$n=4$ when the conclusion is established
  • …
  • First $n-The\; establishment\; of\; one$ result can lead to the establishment of the nth result. According to the principle of induction, the conclusion is true when n takes any value.

uniqueness

• In order to facilitate the judgment of the polynomial f ( x ) f(x)The irreducible polynomial decomposition formula of f ( x ) is unique, and the position of the irreducible factor can be arranged according to the degree of the term of the decomposed multiplication chain, and then compare whether the factors of the$same$$order$$are$$equal$

• Let $f\left(x\right)$ can be decomposed into the product of irreducible polynomials$f(x)={\sum }_{i=1}^s{p}_i(x)$,

• and suppose $f\left(x\right)$ has another decomposition formula written as$f\left(x\right)={\sum }_{i=1}^t{q}_i\left(x\right)$,

• 其中$q_i\left(x\right),i=1,2,\cdots ,t$ is an irreducible polynomial, regardless off ( x ) f(x)Whether the two decomposition forms of$f$$($$x$$)\; are\; unique\; and\; their\; results\; are\; equal,\; that\; is$

  • $$\sum _{i=1}^s{p}_i(x)=\sum _{i=1}^t{q}_i\left(x\right)\text{\hspace{1em}((1))}$$

• to $s$ for induction

  • when $s=1$ ,$f\left(x\right)$ is an irreducible polynomial, defined:$s=t=1$ and$f(x)=p_1\left(x\right)=q_1\left(x\right)$
  • Suppose f ( x ) f(x)The number of irreducible factors in the decomposition formula of$f$$($$x$$)$$r=s-1$ When the uniqueness of the decomposed form is established
  • Obviously $p_1\left(x\right)|f(x)$ , by$(1)$,$p_1(x)|{\sum }_{i=1}^t{q}_i(x)(2)$
    • p 1 ( x $p_1(x),q_i\left(x\right)$ are irreducible polynomials, there must ber ∈ { 1 , 2 , ⋯ , t } r\in\{1,2,\cdots,t\}r∈{ 1,2,⋯,t } , as:$p_1(x)|q_r(x)$ , otherwise with$\left(2\right)$ Contradiction
    • Since the multiplication operation satisfies the commutative law, it is always possible to convert $q_1(x),q_r\left(x\right)$ exchange positions such that$q_r\left(x\right)$ is the first irreducible factor of G, which is convenient for discussion
    • So $p_1\left(x\right)|q_r(x)$ expressed as$p_1(x)|q_1(x)$
    • 设$p_1\left(x\right)=c_1{q}_1\left(x\right)$ , substitute into$(1)$,$p_1\left(x\right){\sum }_{i=2}^s{p}_i\left(x\right)=q_1\left(x\right){\sum }_{i=2}^t{q}_i\left(x\right)$
      • $c_1{q}_1\left(x\right){\sum }_{i=2}^s{p}_i\left(x\right)=q_1\left(x\right){\sum }_{i=2}^t{q}_i\left(x\right)$
      • ${\sum }_{i=2}^s{p}_i(x)=c_1^{-1}{\sum }_{i=2}^t{q}_i\left(x\right)$
      • 记$U(x)={\sum }_{i=2}^s{p}_i(x)$,$V(x)={\sum }_{i=2}^t{q}_i(x)$,
      • $U\left(x\right)$ has$s-1$ item, from the inductive hypothesis,U ( x ) U(x)The decomposition formula of$U$$($$x$$)$$U(x),V\left(x\right)$ contains the same number of irreducible factors, that is,$s-1=t-1$ , that is,$s=t$
      • Similarly
        • After proper arrangement, $p_2(x)=c_2^{\prime }{c}_1^{-1}{q}_2\left(x\right)$ , record$c_2=c_2^{\prime }{c}_1^{-1}$, namely $p_2\left(x\right)=c_2{q}_2\left(x\right)$
        • $p_i\left(x\right)=c_i{q}_i(x)$,$i=3,\cdots ,s(4)$
    • By $(2),(3),(4)$可知$p_i(x)=c_i{q}_i(x)$,$i=1,2,\cdots ,s$ and$s=t$ , the uniqueness is proved

• Note: Example

  • $(2x\_+3)(3x-3)=\left[1\right(2x+3)][3(x-1)]$ , where$p_1(x)=2x\_+3$,$p_2\left(x\right)=3x\_-3=3(x-1)$,$p_1(x),p_2\left(x\right)$ are all irreducible (extracting constant coefficients cannot be counted as reducible, they still meet the definition of irreducible)

  • $(x^2+3)(x^2+4)$ The two factors of ) are irreducible within the range of real numbers, and the degree is 2, so it is impossible to determine the unique decomposition formula only by arranging according to the degree, and other conditions need to be emphasized, such as the (highest) of two unequal factors When the number of times is equal, compare the next highest item until the lowest item

  • The factorization theorem has its fundamental importance in theory

  • But for the general case, there is no universally feasible way to factorize polynomials