How to show if $(f)=(c)(g)$ where $c \in R$ (UFD) and $g$ is primitive , then $(c)=(\mathrm{cont}_f)$ where $\mathrm{cont}_f$ is the content of $f$.

Question

From Algebra: Chapter $0$ by Aluffi

I know how to show the first bullet. I am not sure how to prove the second.

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If $(f)=(c)(g)=(cg)$, then $uf=cg$ where $u$ is a unit in $R$. We can also write, by the first bullet, $u \cdot \mathrm{cont}_f \underline{f}=cg$.

Where do we go from here?

Answer 1

$(f) = (\text{cont}_f)(\underline f)$, where $\underline f$ is primitive.

The statement above means, there exists a primitive polynomial $h\in R[X]$ such that $(f) = (\text{cont}_f)(h)$.

As $h$ is dependent on $f$, the author of the textbook choose the symbol $\underline f$ to denote $h$ instead.

We could have used any one of $\bar f$, $f'$, ${}_cf$, etc. instead of $\underline f$, if they have not been reserved for other purpose.

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Since $g$ is primitive, $g=g_0 +g_1X+\cdots+ g_nX^n$ where $g_i\in R$, $(\gcd(g_1, g_2, \cdot, g_n))=(1)=R$.

Since $(f)=(c)(g)=(cg)$, $f=vcg=vcg_0 + vcg_1x +\cdots +vcg_nx^n$ for some unit $v$.

$(\text{cont}_f)=(\gcd(vcg_0, vcg_1, vcg_n))=(vc\cdot\gcd(g_0, g_1, \cdots, g_n))=(vc) = (c)$