Prove or disprove $<6x,10>=<2x,10>$ where they are ideals of $\mathbb Z[x]$.
My group teammate said it is, I say it is not
$$<6x,10>=\{0,10,6x,6x+10, \dots,q_1(x)6x+10f_2(x):f_1(x),f_2(x) \in \mathbb Z[x] \}$$ and
$$<2x,10>=\{0,10,2x,2x+10, \dots,q_1(x)2x+10f_2(x):f_1(x),f_2(x) \in \mathbb Z[x] \}$$
Assume that $<6x,10>=<2x,10>$ so, $2x \in <2x,10> \Rightarrow 2x \in <6x+10>$
so, $\exists f_1(x),f_2(x) \in \mathbb Z[x]$ where $2x=f_1(x)6x+f_2(x)10$.
$\mathbb Z[x]$ has no zero divisors. and $x=f_1(x) 6x+f_2(x)5$ which is a contradiction such $f_1,f_2 \notin \mathbb Z[x]$
I don't think I own his "proof" for it to be critiqued here.
Am I wrong? Is true that $<6x,10>=<2x,10>$?
$$6x=3\cdot\underline{2x}+0\cdot \underline{10}\in(2x,10)$$ $$2x=2\cdot\underline{6x}+(-x)\underline{10}\in(6x,10)$$