When does a polynomial generate a radical ideal?

Let $k$ be a field and consider the polynomial ring $A = k[x_1,...,x_n]$.

Claim: Given $f \in A \setminus \{0\}$, $(f)$ is radical if and only if $f$ factors into a product of irreducible polynomials of multiplicity $1$.

Proof:

$\Leftarrow$: We know $A$ is a UFD. So, let $f = f_1\cdots f_m$ be a product of $f$ into irreducible factors such that for all $i \neq j$, $(f_i) \neq (f_j)$. Then $$(f) = (f_1\cdots f_m) = (f_1) \cap \cdots \cap (f_m)$$ (I am using unique factorization for the second equality). Thus, $(f)$ is an intersection of prime ideals of $A$ and hence radical.

$\Rightarrow$: Suppose $(f)$ is radical. Again, let $f = {f_1}^{e_1}\cdots {f_m}^{e_m}$ be a product of $f$ into irreducibles where $i \neq j$ $\Rightarrow$ $(f_i) \neq (f_j)$.

Our goal is to show that each $e_i = 1$. Well, suppose not. Then there exists $e_i$ such that $e_i > 1$. Then $({f_1}^{e_1}\cdots {f_i}^1\cdots {f_m}^{e_m}) \subseteq (f) \subseteq ({f_1}^{e_1}\cdots {f_i}^1\cdots {f_m}^{e_m})$. The first inclusion is because ${f_1}^{e_1}\cdots {f_i}^1\cdots {f_m}^{e_m} \in \mathrm{Rad}((f)) = (f)$, and the second inclusion follows from the fact that ${f_1}^{e_1}\cdots {f_i}^1\cdots {f_m}^{e_m}\mid f$.

But this means that there is some $u \in A^\times$ such that ${f_1}^{e_1}\cdots {f_i}^1...{f_m}^{e_m} = u{f_1}^{e_1}\cdots {f_i}^{e_i}\cdots {f_m}^{e_m}$, which contradicts unique factorization.

Let $(p_1),\dots,(p_n)$ be distinct prime ideals of a unique factorization domain, and let $k_1,\dots,k_n$ be positive integers. Then the radical of $$ (p_1^{k_1}\cdots p_n^{k_n}) $$ is clearly $$ (p_1\cdots p_n). $$