关于多项式整除性与根集合关联性的技术问询

Hey there! Great question—this is a super common point of confusion when working with polynomials, so let’s unpack it step by step.

First, let’s clarify the core question I think you’re asking: If the set of roots of polynomial ( P ) is a subset of the set of roots of polynomial ( Q ), does ( Q ) divide ( P ) (meaning ( P ) is a multiple of ( Q ))? If that’s not exactly what you had in mind, feel free to follow up, but let’s run with this framing for now.

The short answer is no, this isn’t always true—there are key conditions you need to add for the statement to hold. Let’s break down the scenarios:

1. Working over an algebraically closed field (like the complex numbers ( \mathbb{C} )) #

Even here, we can’t ignore root multiplicities. For example:

• Let ( Q(x) = (x-1)^2 ) (root at ( x=1 ), multiplicity 2)
• Let ( P(x) = (x-1) ) (root at ( x=1 ), multiplicity 1)

The roots of ( P ) are a subset of the roots of ( Q ) (both are just {1}), but ( Q ) does not divide ( P )—dividing ( P ) by ( Q ) gives ( 1/(x-1) ), which isn’t a polynomial.

The correct rule here is: If every root of ( P ) is a root of ( Q ), and the multiplicity of each root in ( P ) is less than or equal to its multiplicity in ( Q ), then ( Q ) divides ( P ).

2. Working over a non-algebraically closed field (like the real numbers ( \mathbb{R} )) #

Here, we have to account for irreducible polynomials that don’t have roots in the field. For example:

• Let ( P(x) = x^2 + 1 ) (no real roots—its root set is empty)
• Let ( Q(x) = x - 2 ) (root at ( x=2 ))

The empty set is a subset of {2}, but ( Q ) does not divide ( P )—the result of the division isn’t a polynomial.

Conversely, if ( Q(x) = (x^2 + 1)(x-2) ) and ( P(x) = x^2 +1 ), ( P )’s roots (empty set) are a subset of ( Q )’s roots ({2}), and ( P ) does divide ( Q )—but that’s the reverse direction of your original question.

3. Don’t forget constant polynomials! #

Non-zero constant polynomials have no roots (their root set is empty). If ( Q ) is a non-zero constant, it divides every polynomial ( P ), no matter what roots ( P ) has. If ( P ) is a non-zero constant, its root set is empty (a subset of any root set), but only other constant polynomials can divide it.

Quick summary #

To recap:

• The reverse statement always holds: If ( Q ) divides ( P ), then all roots of ( Q ) (counting multiplicities) are roots of ( P ) (counting multiplicities).
• Your original statement (subset of roots implies divisibility) only holds if:
  • We’re working over an algebraically closed field,
  • The multiplicity of each root in ( P ) is ≤ its multiplicity in ( Q ),
  • Or ( P ) is a constant polynomial.

备注：内容来源于stack exchange，提问作者yyzr