$c(kp) = kc(p)$ where $c$ is the content of a polynomial, $p$ a polynomial in $R[x]$ and $k \in R-\{0\}$ (without assuming Gauss' Lemma)

Question

I have been trying to prove (without assuming Gauss' Lemma) that for $R$ a UFD and $p(x) \in R[x]$ then for $k \in R- \{0\}$ we have $c(kp) = kc(p)$.

My approach is as follows: $$ k\cdot c(p) \cdot p''(x) = kp(x) = c(kp)\cdot p'(x) \qquad c(p') = c(p'') = 1 $$ Since $k$ divides the coefficients of $kp(x)$, it divides $c(kp)$ thus $c(kp) = kq$ for some $q \in R$. So canceling $k$'s in the above equation gives $$ c(p)\cdot p''(x) =p(x) = q \cdot p'(x) $$ At this point I'm a bit stuck. I feel as though this little lemma should be a lot simpler than I'm making it. Any advice on how to move forward would be much appreciated.

Answer 1

We can use the $\gcd$ law: $\gcd(ma,mb) = m(a,b)$: Let \begin{align*} p'(x) &= a_0 + a_1x + \cdots + x_nx^n \\ p''(x) &= b_0 + b_1x + \cdots + b_nx^n \end{align*} (note that they must have the same degree). We have \begin{align*} c(qp') &= \gcd(qa_0, qa_1,\dots, qa_n) = q\gcd(a_0,\dots, a_n) = q \\ c\Big(c(p)p''\Big) &= \gcd(c(p)b_0,\dots, c(p)b_n) = c(p)\gcd(b_0,\dots, b_n) = c(p) \end{align*} Thus $c(p) = q$ giving us our result.

EDIT: (For completeness, for a proof that $\gcd(ma,mb) = m\gcd(a,b)$ in a UFD see Bill Dubuque's argument here )