Can a squarefree polynomial in K[x,y] not be squarefree in K[[x]][y]? In a UFD, as usual one says that $f$ is square-free if it is not divisible by the square of any irreducible element, i.e., if it has no multiple factor.
An polynomial $f\in k[x,y]$ can have more factors in the factorization over $k[[x]][y]$ than over $k[x,y]$. 

Is it possible for $f\in k[x,y]$ to be square-free in $k[x,y]$ but not in $k[[x]][y]$?

It looks like an easy question but I cant find a simple argument/example.
Thank you!
 A: Here is a sketch of an argument in case that $k$ is a perfect field. Since a complete answer has already been given in the comments, a complete answer here is superfluous.
First you prove the polynomial $f$ is square free if and only if the ideal $I = (f_x, f_y)$ has finite colength in $k[x, y]$. You can define the colength simply as the dimension of $k[x, y]/I$ as a $k$-vector space. Also $f_x$ is the partial derivative of $f$ with respect to $x$ and similarly for $f_y$. To see the statement is true, observe that the curve $C = V(f)$ in affine $2$-space is reduced if and only if it has finitely many singular points. Here we use that that ground field is perfect, so nonsingular points of $C$ are smooth points of $C$.
OK, but now if $f = g^2h$ in $A = k[[x]][y]$ where $g$ is not a unit, then we see that the ideal $IA$ does not have finite colength in $A$ because $IA$ is contained in $(g)$ since $f_x = 2gg_xh + g^2 h_x$ and $f_y = 2gg_yh + g^2h_y$ and $A/(g)$ has infinite dimension as a $k$-vector space. We leave this to the reader as an exercise.
To conclude we need one final fact: if $I$ has finite colength in $k[x, y]$, then $IA$ has finite colength in $A$.
Two suggestions for proving the final fact come to mind: (1) you can prove this by classifying ideals having finite colength (do primary decomposition) and prove it for powers of maximal ideals by direct computation, or (2) you can use that the the image of $k[x, y] \to A/IA$ is dense in the $(x)$-adic topology, and hence if $IA$ has infinite colength, then $I$ must also have infinite colength. Cheers!
