It is known that one cannot expect the extension of the contraction of an ideal under a ring homomorphism $f:A\to B$ to be the original ideal, except in very special scenarios like surjections or localizations. In this example we see this does not even happen if the map is faithfully flat; however in the example given (i.e. $k[x^2]\subseteq k[x]$ and the ideal $xk[x]$) the extension of the contraction has the original ideal as its radical. This the leads to the question
Are there sufficient conditions (e.g. faithfully flat, (faithfully) finite etale) which could ensure that $\sqrt{(\mathfrak b\cap A)B}$ equals (or is otherwise related to in a reasonable way) $\mathfrak b$ for radical ideals $\mathfrak b\subseteq B$?
