In my lecture notes we have the following:
Definition: $f, g \in \mathbb{C}[x, y]$
$f \sim g \Leftrightarrow \exists c \in \mathbb{C}, c \neq 0$ such that $g=cf$
Example: If $f \sim g \Rightarrow V(f)=V(g)$. The converse doesn't stand.
But from Nullstellensatz we have that $$V(f)=V(g) \Leftrightarrow Rad(\langle f\rangle )=Rad(\langle g\rangle )$$
If $$f=f_1^{a_1} \cdot f_2^{a_2} \cdot \dots \cdot f_s^{a_s}$$ where $f_i$ are irreducible polynomials of $\mathbb{C}[x, y]$ then, $$Rad(\langle f\rangle )=\langle f_1, f_2, \dots , f_s \rangle$$
So $V(f)=V(g)$ $\Leftrightarrow $ the irreducible factors of $f$ and $g$ are the same.
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Can you explain to me why it stands that $$Rad(\langle f\rangle )=\langle f_1, f_2, \dots , f_s \rangle$$
This is false. What is true is that $$Rad(\langle f\rangle )=\langle f_1f_2 \dots f_s \rangle$$ when you assume in addition that the $f_i$ are not associated elements ('associated' is the $\sim$ realation you recalled).
To see this note that if $h^n \in \langle f \rangle$, then $f \mid h^n$ and so $f_i \mid h^n$ for each $i$. Since $f_i$ is irreducible and thus a prime element, it follows that $f_i \mid h$ for each $i$ and further, again using 'prime', that $f_1 \dots f_s \mid h$ so the radical ideal is contained in $\langle f_1f_2 \dots f_s \rangle$.
Conversely let $a$ be the maximum of the $a_i$, then $f \mid (f_1f_2 \dots f_s)^a$ showing that $f_1f_2 \dots f_s$ is in the radical ideal.