Primitive polynomial (algebra)

In algebra, a primitive polynomial is every polynomial $f(x)\in R[x]$ ${\ displaystyle f (x) \ in R [x]}$ $f (x) \ in R [x]$ where $R$ ${\ displaystyle R}$ $R$ - an associative-commutative ring , with a unique factorization, the coefficients of which do not have non-trivial common factors.

Any polynomial $g(x)\in R[x]$ ${\ displaystyle g (x) \ in R [x]}$ $g (x) \ in R [x]$ can be written as $g(x)=c_{g}\cdot f(x)$ ${\ displaystyle g (x) = c_ {g} \ cdot f (x)}$ $g (x) = c_ {g} \ cdot f (x)$ where $f(x)$ ${\ displaystyle f (x)}$ $f (x)$ Is a primitive polynomial, a $c_{g}$ ${\ displaystyle c_ {g}}$ $c_ {g}$ Is the largest common factor divisor of the polynomial $g(x)$ ${\ displaystyle g (x)}$ $g (x)$ . Element $c_{g}\in R$ ${\ displaystyle c_ {g} \ in R}$ $c_ {g} \ in R$ defined up to multiplication by invertible elements from R, it is called the content of the polynomial $g(x)$ ${\ displaystyle g (x)}$ $g (x)$ .

Gauss Lemma

If $g_{1}(x),g_{2}(x)\in R[x]$ ${\ displaystyle g_ {1} (x), g_ {2} (x) \ in R [x]}$ then $c_{g_{1}g_{2}}=c_{g_{1}}c_{g_{2}}$ ${\ displaystyle c_ {g_ {1} g_ {2}} = c_ {g_ {1}} c_ {g_ {2}}}$ . In particular, the product of primitive polynomials is again primitive.

Proof

First, we prove that the product of primitive polynomials is a primitive polynomial. To do this, it is enough to verify that if a simple element $p$ ${\ displaystyle p}$ rings $R$ ${\ displaystyle R}$ divides all coefficients of a polynomial $f(x)g(x)$ ${\ displaystyle f (x) g (x)}$ , then it is a common divisor of all coefficients of the polynomial $f(x)$ ${\ displaystyle f (x)}$ or a common divisor of all coefficients of the polynomial $g(x)$ ${\ displaystyle g (x)}$ . Let be $f(x)=a_{0}+a_{1}x+\ldots +a_{n}x^{n}$ ${\ displaystyle f (x) = a_ {0} + a_ {1} x + \ ldots + a_ {n} x ^ {n}}$ , $g(x)=b_{0}+b_{1}x+\ldots +b_{m}x^{m}$ ${\ displaystyle g (x) = b_ {0} + b_ {1} x + \ ldots + b_ {m} x ^ {m}}$ , $n=\operatorname {deg} f,m=\operatorname {deg} g$ ${\ displaystyle n = \ operatorname {deg} f, m = \ operatorname {deg} g}$ - degrees of these polynomials. We carry out induction on $m+n$ ${\ displaystyle m + n}$ . If $m+n=0$ ${\ displaystyle m + n = 0}$ then $m=0$ ${\ displaystyle m = 0}$ and $n=0$ ${\ displaystyle n = 0}$ , $f(x)=a_{0},g(x)=b_{0}$ ${\ displaystyle f (x) = a_ {0}, g (x) = b_ {0}}$ . If $p$ ${\ displaystyle p}$ divides $a_{0}b_{0}$ ${\ displaystyle a_ {0} b_ {0}}$ then since the ring $R$ ${\ displaystyle R}$ factorially $p$ ${\ displaystyle p}$ divides $a_{0}$ ${\ displaystyle a_ {0}}$ or $p$ ${\ displaystyle p}$ divides $b_{0}$ ${\ displaystyle b_ {0}}$ , that is, in this case the statement is true. In general $f(x)g(x)=a_{0}b_{0}+(a_{0}b_{1}+a_{1}b_{0})x+\ldots +a_{n}b_{m}x^{n+m}$ {\ displaystyle f (x) g (x) = a_ {0} b_ {0} + (a_ {0} b_ {1} + a_ {1} b_ {0}) x + \ ldots + a_ {n} b_ { m} x ^ {n + m}} . Suppose some simple element $p$ ${\ displaystyle p}$ rings $R$ ${\ displaystyle R}$ divides all coefficients of a polynomial $f(x)g(x)$ ${\ displaystyle f (x) g (x)}$ . As $p\mid a_{n}b_{m}$ ${\ displaystyle p \ mid a_ {n} b_ {m}}$ and ring $R$ ${\ displaystyle R}$ factorially then $p\mid a_{n}$ ${\ displaystyle p \ mid a_ {n}}$ or $p\mid b_{m}$ ${\ displaystyle p \ mid b_ {m}}$ . Let for definiteness $p\mid a_{n}$ ${\ displaystyle p \ mid a_ {n}}$ . If $n=0$ ${\ displaystyle n = 0}$ then $p$ ${\ displaystyle p}$ divides all coefficients of a polynomial $f(x)$ ${\ displaystyle f (x)}$ . If $n>0$ ${\ displaystyle n> 0}$ , then note that $p$ ${\ displaystyle p}$ will be a common divisor of all coefficients of the polynomial $f_{1}(x)g(x)$ ${\ displaystyle f_ {1} (x) g (x)}$ where $f_{1}(x)=f(x)-a_{n}x^{n}=a_{0}+a_{1}x+\ldots +a_{n-1}x^{n-1}$ ${\ displaystyle f_ {1} (x) = f (x) -a_ {n} x ^ {n} = a_ {0} + a_ {1} x + \ ldots + a_ {n-1} x ^ {n- one}}$ . Indeed, all coefficients of the polynomial $f(x)g(x)-f_{1}(x)g(x)=a_{n}x^{n}g(x)$ ${\ displaystyle f (x) g (x) -f_ {1} (x) g (x) = a_ {n} x ^ {n} g (x)}$ are divided into $a_{n}$ ${\ displaystyle a_ {n}}$ , and therefore on $p$ ${\ displaystyle p}$ . By the hypothesis of induction $p$ ${\ displaystyle p}$ divides all coefficients of a polynomial $f_{1}(x)$ ${\ displaystyle f_ {1} (x)}$ or all coefficients of the polynomial $g(x)$ ${\ displaystyle g (x)}$ . In the first case $p$ ${\ displaystyle p}$ also divides all the coefficients of the polynomial $f(x)=f_{1}(x)+a_{n}x^{n}$ ${\ displaystyle f (x) = f_ {1} (x) + a_ {n} x ^ {n}}$ . By the principle of mathematical induction, the statement is proved for all values $m$ ${\ displaystyle m}$ and $n$ ${\ displaystyle n}$

Let us prove that $c_{f}c_{f}=c_{fg}$ ${\ displaystyle c_ {f} c_ {f} = c_ {fg}}$ . Let be $f=c_{f}f_{0}$ ${\ displaystyle f = c_ {f} f_ {0}}$ , $g=c_{g}g_{0}$ ${\ displaystyle g = c_ {g} g_ {0}}$ where $f_{0}$ ${\ displaystyle f_ {0}}$ , $g_{0}$ ${\ displaystyle g_ {0}}$ - primitive polynomials. Then $fg=c_{f}c_{g}f_{0}g_{0}$ ${\ displaystyle fg = c_ {f} c_ {g} f_ {0} g_ {0}}$ . Since the polynomial $f_{0}g_{0}$ ${\ displaystyle f_ {0} g_ {0}}$ by what is proved primitive, then $c_{fg}=c_{f}c_{g}$ ${\ displaystyle c_ {fg} = c_ {f} c_ {g}}$ . The lemma is proved.

Literature

Zarissky O., Samuel P., Commutative algebra, trans. from English, vol. 1, M.

Primitive polynomial (algebra)

Gauss Lemma

Proof

Literature

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