Showing that if an ideal $I \subset R$ is square-free, $I$ is radical. For some reason we've rarely used the idea of "square-free" in my Algebra courses (unless simple roots of polynomials counts?) so I'm trying to figure out this auxiliary fact about radical ideals, not sure if this is correct.

Suppose $I \subseteq R$ is an ideal of $R$, a UFD. We want to show that if $I$ is square-free, then $I$ is radical.

That is, we want to show that if $r^k\in I$ for some $k>1$, then $r \in I$. This might be unnecessary but I split this into two cases:
If $k=2$ then $r^2 \in I$. But $r^2$ divides itself, but $I$ is square-free so $\exists r^{-1}\in R$, hence $r^{-1}r^2 = r \in I$.
In general for $k > 2$, we have that $r^2\cdot r^{k-2} \in I$. But $r^2 \ | \ r^k \in I$, so $I$ being square free implies $r$ is a unit, hence multiplying by its inverse; $(r^{-1})^{k-1}\cdot r^{k}  = r \in I$ as required.
 A: For any commutative ring $R$ with ideal $I$, the following are equivalent:
a) $I$ is radical;
b) for all $r\in R$, $r^2\in I\implies r\in I$.
This is the closest thing I can think of to what you are talking about. In particular that shows that a nontrivial principle ideal $(x)$ in a UFD is radical iff $x$ is square free.
The proof isn't hard. Obviously if $I$ is radical then $r^2\in I\implies r\in I$.
In the other direction, suppose $r^2\in I\implies r\in I$.  We now work to show $I$ is radical by contradiction. To that end, suppose $x^n\in I$, $n$ minimal with that property, such that $x\notin I$.  Obviously $n=1$, and $n=2$ would imply a contradiction immediately. So WLOG $n > 2$.
Now there is a $k$, $1<k<n$ such that $x^{2k}\in I$, namely $k=\lceil n/2\rceil$ (that's the ceiling operator, or 'greatest integer' operation.). By our assumption that would imply $x^k\in I$, but this contradicts the minimality of $n$.  Therefore the assumption that $I$ could be a non-radical ideal must be false, and $I$ is indeed radical.
