Show existence of a polynomial

Question

Let $R$ be a ring and consider $Z[X]$ en let $M$ be the $R$-module $Z[X]/(X,2)$. Let $f,g \in Z[X]$ with $2f = Xg$. Show that there is a $h \in Z[X]$ with $f = Xh$ and $g = 2h$. How to show this? I don't know how to start, any hints?

Answer 1

If $2f = Xg$ then $X$ divides $Xg$ so $X$ divides $2f$. $X$ does not divide $2$ so $X$ must divide $f$. This is equivalent to saying $f = Xh$ for some $h \in \Bbb Z[X]$. Now can you do the other one?

Answer 2

Put $\,x=0\,$ in $\, 2f = xg\,\Rightarrow\, f(0)=0\,\Rightarrow\, x\mid f $

Therefore: $\ \ \ f/x = g/2\, =:\, h\in\Bbb Z[x]$