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I've got a couple of theorems that I must prove, but I'm completely stumped. Could someone help me out?

  1. Prove: If P(x) can be divided by Q(x), then it can be divided by cQ(x), where c - any non zero number.
  2. Prove: If the degrees of P and Q are the same ( equal), then P(x) can be divided by Q(x) then and only then, if Q(x)=cP(x), where c - any non zero number.
  3. Prove: If c - any non zero number. then polynomials P(x) and cP(x) have the same divisors

I know from simple logic, that they are all true, but I'm bad at writing logic in math proofs.


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up vote 3 down vote accepted

I'm assuming that by "number" you mean either "real number" or "complex number"; in fact, most of these statements are true in much greater generality (other classes of numbers).

  1. The statement "$P(x)$ can be divided by $Q(x)$" means that there exists a polynomial $R(x)$ such that $Q(x)R(x) = P(x)$. In order to show that $P(x)$ can also be divided by $cQ(x)$, you need to find a polynomial $S(x)$ such that $(cQ(x))S(x) = P(x)$. Try using the $R(x)$ you know exists, and the fact that $c\neq 0$.

  2. Again, you know that there is a polynomial $R(x)$ such that $Q(x)R(x)=P(x)$. What do you know about the degree of a polynomial product? That is, if the degree of $f(x)$ is $n$ and the degree of $g(x)$ is $m$, what is the degree of $f(x)g(x)$?

  3. Show that if $Q(x)$ divides $P(x)$, then it also divides $cP(x)$. Then use this to show that anything that divides $cP(x)$ also divides $\frac{1}{c}\left(cP(x)\right)$ to get that $P(x)$ and $cP(x)$ have the same divisors: that is, that $Q(x)$ divides $P(x)$ if and only if it divides $cP(x)$. For the second part (that a divisor of $cP(x)$ is a divisor of $P(x)$) you'll need to use that $c\neq 0$.

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The sought divisibility results are special cases of this simple

Divides Product Rule $\ $ Let $\rm\ a,b,c,d\in R $ any ring.

Then $\rm\quad\ \ a|b,\ \ c|d\ $ in $\rm\:R\ \ \Rightarrow\ \ ac|bd\ $ in $\rm\:R\:. $

Proof $\rm\ \ \ b/a,\ d/c \ \in\ R\ \Rightarrow\ (bd)/(ac)\in R\:.\quad$ QED

Now let's apply this Rule to derive the sought results in $\rm\ R = K[x],\ K\ $ a field.
Below I give the main steps, omitting trivial cases such as when $\rm\:p\:$ or $\rm\:q = 0\:.$

  1. $\rm c|1,\ q|p\ \Rightarrow\ cq|p\ $ by the Rule (in a field $\rm\ K,\ c\ne 0\ \Rightarrow\ 1/c\in K\ \Rightarrow\ c|1\ $ in $\rm\:K\:).$

  2. If $\rm\ deg\ p = \deg\ q\ $ then $\rm\ q|p \iff\ q = c\: p\ $ for some $\rm\ 0 \ne c\in K\:.$
    $\rm(\Leftarrow)\ \ c|1,\ p|p\ \Rightarrow\ cp|p\ $ by the Rule (or $\rm\ q \to p\ $ in 1).
    $\rm(\Rightarrow)\ \ p = q\:r\ \Rightarrow\ \deg\ r = \deg\ p - \deg\ q = 0\ \Rightarrow\ r\in K\:.$

  3. $\rm\ q|p\ \iff\ q|cp\ $
    $\rm(\Leftarrow)\ \ q|cp|p\ $ by $\rm q \to p$ in 1 and "divides" transitivity ($\rm b = c\ \ne 0\:$ in Rule)
    $\rm(\Rightarrow)\ \ 1|c,\ q|p\ \Rightarrow\ q|cp\ $ by the Rule.

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Assuming you are working over complex numbers,

Hint: Try using the fundamental theorem of algebra.

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And that is...? – Janis Peisenieks Jan 16 '11 at 23:45
Algebra , 1st semester, 1st year. Someone pointed me to this theorem, but we've not looked at it before. – Janis Peisenieks Jan 16 '11 at 23:54
@Janis: It implies (along with Remainder theorem) that every polynomial can be written as $C(x−a_1)(x−a_2)\dots(x−a_n)$ where the $a_i$ are the roots of the polynomial. What course is this, btw? – Aryabhata Jan 16 '11 at 23:55
The question title did begin with "Proofs". To the downvoter: If you don't understand/have an issue, please comment and things can be clarified. – Aryabhata Jan 17 '11 at 5:57

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